1. Basic Algebra: Lecture Notes MA 419 IIT Bombay editedNovember 14, 2019by Rekha Santhanam

Please use these notes with caution. These notes are a repre-sentative of the content covered in class I have tried to ensure there areno typos but some may have crept in as I keep making changes. Use a referencetext for details.

1.1. Groups and their properties. One of the first binary operations that weencounter is addition while counting. What is a binary operation?

Definition 1.1. Let S be a set. Any function ◦ : S × S → S is called a binaryoperation on S.

But soon we realize that the set N with + is not convenient for addition andthat we need both a 0 (to create the set of whole numbers) and all negativeintegers (to be able to subtract more quantity than what was available). Thinkof practical situations where you need negative integers .

The set of integers with the binary operation + is an example of a group.

Definition 1.2. A set G with a binary operation · : G×G→ G is said to be agroup if it satisfies the following.

(1) ”·” is associative, that is, a · (b · c) = (a · b) · c for all a, b, c ∈ G.(2) G has an identity with respect to ”·”, that is, there exists e ∈ G such that

a · e = a = e · a for all a ∈ G.(3) All elements of G have an inverse with respect to ”·”, that is, ∀a ∈ G,∃b ∈ G such that a · b = e = b · a.

Examples:

(1) (Z,+).(2) (Q,+).(3) (R,+)(4) (C,+).(5) In fact any vector space with its addition is a group.(6) How about (Z,×), (Q,×), etc.?(7) Recall set of functions from X → R denoted by F(X,R). Then this is a

vector space over R. Why? and hence a group.(8) How about all functions form a set X to Y ? Is there any binary operation

on this set F(X, Y ).(9) Write down your favourite set and can you create two binary operations

on it?(10) Can the same set have more than one group structure?(11) How many group structures exist on a set of 2 elements? on a set of 3

elements? on a set of n elements?

Remark 1.3. Let us create a group with 2 elements. We will call this {a, b}. Weknow one of them should be identity. But why can both of them not behave likeidentity? That would mean a · b = a = b · a and b · a = b. So only one should be.Choose a to be identity. Here is a table:

1

2

· a ba a bb b ?

Before we start answering these questions we should make some consequences ofthe group axioms precise.

Proposition 1.4. Let (G, ·) be a group. Then

(1) The identity is unique and denoted by e.(2) Inverse of any element a in G is unique and denoted by a−1.(3) For any a, b, c ∈ G if a · b = a · c then, b = c. Similarly, b ·a = c ·a implies

that b = c.(4) (a−1)−1 = a.(5) (a · b)−1 = b−1 · a−1.

Proof. (1) Let a, b ∈ G be identity. Then a · b = a = b · a but b · a = b . Hencea = b.

(2) Let a · b = e = b · a and a · c = e = c · a. Then b = b · e = b · (a · c) =(b · a) · c = e · c = c.

(3) Exercise(4) Exercise(5) Exercise.

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Remark 1.5. To show that identity is unique we only used that G has a binaryoperation with a monoid.

For instance Mn(R) has a binary operation with matrix multiplication. It isassociative and has an identity. However it is not a group since every squarematrix does not have an inverse.

However, if we restrict ourselves to GLn(R) = Set of n × n real matriceswith non-zero determinant then we still get a binary operation on GLn(R) sincedet(AB) = detA detB for all A,B ∈Mn(R).

Interestingly, Mn(F ) is a group with respect to addition for F = Z,R,Q,C.However, GLn(F ) is not a group with respect to multiplication when F = Z.More generally these are both groups when F is a field.

Recall a field is defined as a set F with two associative binary operations +and · such that

(1) (F,+) is an abelian group with 0 as the additive identity.(2) (F − {0}, ·) is an abelian group and(3) (a+ b) · c = a · c+ b · c for all a, b, c ∈ F .

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Are these the only examples? Let us try to find a group with four elements.

· e a b ce e a b ca a ? ? ?b b ? ? ?c c ? ? ?

One group is of the form {e, a, a2, a3 : a4 = e}, whereas the other group is ofthe form {e, a, b, ab : ab = ba, a2 = b2 = e}. The first one is called a cyclic groupof order 4 C4 and the second is called the Klein-4 group K4.

Does this describe the group completely? and how are these two groups related?Hint: Generators and relations.

Definition 1.6. Let (G1, ·) and (G2, ?) be groups and then a function f : G1 →G2 is said to be a group homomorphism if f(a · b) = f(a) ? f(b) for all a, b ∈ G.

Further f is said to be a group isomorphism if f is a bijective homomorphismand f−1 is also a group homomorphism.

Remark 1.7. At this point it is good to compare groups we created using gen-erators and relations with something that is familiar. Can you relate C4 withsomething familiar?

Before we start comparing groups, let us rein in a bit and look at a few moreexamples.

Solve exercises from Tutorial 1More generally, we can define the Dihedral group Dn as the symmetries of a

regular n-gon. The idea is that in any regular n-gon we will have rotation by360/n degrees. If n is odd then there are n flips along the axis joining a vertexand the midpoint of the opposite edge.

If n is even then there are n/2 flips along axis joining midpoint edges and n/2flips along axis joining opposite vertices. Again giving n flips. In general thisforms a group.

Exercise: Let D4 denote the dihedral group of symmetries of a square. Letr denote the rotation by 90degrees and f denote any of the flip which fixes theright diagonal. Then

(1) The elements e, r, r2, r3, f are all distinct.(2) Further fri 6= frj if i 6= j < 4.(3) Moreover, fri = r−if for all i < 4.(4) Verify then that D4 = {e, r, r2, r3, f, fr, fr2, fr3 | rf = fr−1}.

A succinct representation of a group is given by presentation as above. Apresentation of a group G is a description in terms of generators and relations.Generators of a group are elements of G such that every element of G can bewritten as a finite product of these generators. Relations are any relations thatexist in the group with respect to the element.

Questions:

(1) Does every group have a set of generators? have a finite set of generators?(2) What happens if there are no relations?

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(3) Is Z, Q or R finitely generated?(4) If the generating set is a singleton set. What can you say about the group?

1.2. Cyclic groups.

Definition 1.8. A group G which is generated by a single element is called acyclic group, that is if there exists a ∈ G such that G = {xn | n ∈ Z} wherex0 := e is the identity in the group and G is denoted by < a >.

Examples: • Z is a cyclic group with two generators 1 and −1. • Considerthe example of a clock. It reads only 12 hours. After every 12 hours we get aremainder for the number of hours after dividing by 12. This is an example ofmodular arithmetic. The number of hours on a clock is represented by 0, 1, 2,. . ., 11, where 0 can mean from a given starting hour a multiple of 12 hours haspassed.

In general, define Z/nZ = Set of all integers modulo n. Then addition andmultiplication of integers defines binary operations on this set.

Claim:

x+ y mod n = (x mod n+ y mod n) mod n ∀x, y ∈ Z and

xy mod n = (x mod n y mod n) mod n ∀x, y ∈ Z.Proof. Let x+y mod n = r where x+y = nt+r for 0 ≤ r < n. Let x mod n = uand y mod n = v. This implies x = ns + u and y = nl + v for 0 ≤ u, v < n.Then x+ y = n(s+ l) + (u+ v).

Let (u+ v) = nz + r′ for some 0 ≤ r′ < n.We want to say r = r′.We need to show that division algorithm is unique. That is, if x = an+ b and

x = a′n + b′ for some 0 ≤ b, b′ < n. WLOG choose b ≥ b′. Then 0 ≤ b − b′ < n.But (a−a′)n = (b−b′) implies that (b−b′) is a multiple of n. Therefore, b−b′ = 0.Why? �

Proposition 1.9. The set (Z/nZ,+) is a group.

Proof. We have verified that + is a binary operation on Z/nZ. Associativityfollows since

(x+ y) + z = x+ (y + z) mod n.

Further, we have that x+ 0 mod n = x mod n = 0 +x mod n. Finally for anyx ∈ Z x+ (n− x) mod n = n mod n = 0 mod n. �

Question : Is (Z/nZ)∗ = Z/nZ− {0} a group under multiplication?

Definition 1.10. The order of a group is the cardinality of the group.

Properties

Proposition 1.11. (1) Every cyclic group is abelian.(2) If G =< x > is cyclic group of order n then x0, x, x2, . . . , xn−1 are all

distinct and order of x is n.(3) If G =< x > is of infinite order then xi 6= xj whenever i 6= j.

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(4) If G =< x > and xm = e be the least m > 0, m ∈ Z with this property,then xn = e implies m/n.

Proof.

• (4) Note n > m (why?). Then n = qm+r for some 0 ≤ r < m by divisionalgorithm. We get that

e = xn = xqm+r = (xm)qxr = xr.

Since r < m, r = 0 and m/n.

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Consider Z10 with addition. Is 1 the only generator? How can we find all thegenerators?

Definition 1.12. Let G be a group. Then if for some n ∈ Z>0, xn = e define the

order of x to be the smallest positive integer m such that xm = e. Else order ofx is infinite.

In a cyclic group we can find the order of all the elements of the group.

Proposition 1.13. If G =< x > is a cyclic group. Then

(1) if G is infinite every non-identity element has infinite order.(2) if G is finite of order n then order of xk is n/(n, k).

Proof. (1). If G has a non-identity element a of finite order m . Since G is cyclica = xr for some r. Then am = xrm = e. �

This allows us to characterize all cyclic groups up to isomorphism.

Definition 1.14. Let (G1, ∗) and (G2, ·) be two groups. They are said to beisomorphic as groups if there exists a group homomorphism G1 → G2 which is abijection and the inverse is also a group homomorphism.

Proposition 1.15. Any bijective group homomorphism between groups is a groupisomorphism.

Theorem 1.16. (1) Any infinite cyclic group is isomorphic to Z.(2) Any cyclic group of order n is isomorphic to Z/nZ.

Proof. (2). Let G be a cyclic group of order n. Then we have proved earlier inclass that if G =< x > then order of x must be n. Define φ : G → Z/nZ to bethe function φ(xi) = [i] ∈ Z/nZ. Then we claim that φ is injective. If [i] = [j]then n/i− j. Therefore i− j = nt and xi−j = xnt = e.

For any [i] ∈ Z/nZ clearly xi is a preimage.Finally to show this is an isomorphism we note.

φ(xi · xj) = φ(xi+j) = [i+ j] = [i] + [j] = φ(xi) + φ(xj)

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2. Creating more groups

RecallMm×n(R), the set of matrices is a vector space over R and in particulara group with respect to matrix addition.

Consider the following subset {A ∈Mm×n(R) | A =

(a 0b 0

)}. Is it a group?

Similarly, is { x2n| n ∈ N, x ∈ Z}, a group under addition?

Definition 2.1. A non-empty subset H of a group (G, ·) is said to be a subgroupif (H, ·) is a group.

Examples:

(1) (Z,+), (Q,+) are both subgroups of (R,+).(2) SLn(R) = {A ∈Mn(R) | detA = 1} and On(R) = {A ∈Mn(R) | detA =

1} are both subgroups of GLn(R).(3) If (G, ·) is a group, what is the smallest subgroup that an element a belongs

to? Can it be G?(4) Can some groups have no non-trivial subgroups?(5) Find all subgroups of a cyclic group.(6) Is (Z− {0}, ·) is a subgroup of (R− {0}, ·)?

Properties:

Proposition 2.2. Let (G, ·) be a subgroup.

(1) Then {e} is a subgroup.(2) Intersection of two subgroups is again a subgroup.(3) Union of subgroups need not be a subgroup.(4) H ⊆ G is a subgroup if and only if for every a, b ∈ G, ab−1 ∈ G.

Examples:

(1) Let R∗ = R− {0} and C∗ = C− {0}. Then (R∗, ·) and (C∗, ·) are groupsunder multiplication. Consider S1 = {z ∈ C | |z| = 1}. Then this is asubgroup of (C∗, ·). How to verify this?

(2) Let ω denote the primitive nth root of unity in C. Then consider a evensmaller subgroup of S1 defined as {1, ω, ω2, . . . , ωn−1}. Why is this agroup?

(3) Recall the set of all bijections from {1, 2, · · · , n} to itself form a group.We denote this group by Sn. The group of symmetries of a square can bethought of a subgroup of S4. Why?

(4) What are all subgroups of S4?(5) Let φ : G → G′ be a group homomorphism. Then Ker(G) = {x ∈

G | φ(x) = eG′} is a subgroup of G. Also Im(φ) is a subgroup of G′.(6) Let G be a group then the set of all elements of G which commute with

every other element is a subgroup of G. Its called the centralizer of G.

2.1. More about Symmetric groups.

We denote an element σ ∈ Sn as follows: σ =

(1 2 3 . . . n

σ(1) σ(2) σ(3) . . . σ(n)

).

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For example σ =

(1 2 3 42 4 3 1

)is an element of S4.

Exercises:• Write down all symmetries of an equilateral triangle as elements of a S3.• Write down the dihedral group D4 as a subgroup of S4. • Some examples ofcomposition of permutations.

Definition 2.3. A permutation σ ∈ Sn is said to be a cycle if the elements of{1, 2 . . . , n} can be ordered so that σ(ai) = ai+1 for all i = 1, 2 . . . k−1, σ(ak) = a1and σ(j) = j for all j /∈ {a1, . . . , ak}. In this case it is denoted by (a1a2 . . . ak).

Remark 2.4. (1) Every permutation can be written as a composition of cy-cles.

(2) Can you write a permutation as a composition of disjoint cycles?(3) Do disjoint cycles commute?(4) What is the order of a cycle?(5) Can you use this to find the order of any permutation?

Definition 2.5. Any two cycle in Sn is called a transposition. Any cycle can bewritten as a product of transpositions.

(a1 . . . an) = (a1an) . . . (a1a3)(a1a2)

Thus any cycle can be written as a product of transpositions.

Proposition 2.6. Every permutation is either a product of even or odd transpo-sitions and not both.

We first show that

Lemma 2.7. The identity permutation can only be written as a composition ofeven number of transpositions.

We then have a group homomorphism from s : Sn → Z/2Z defined as

s(σ) = [no. of transpositions which compose to σ].

Why is this a group homomorphism? Is it onto? Is it one one?The Kernel of this group homomorphism is a subgroup An of Sn and is called

the alternating group.

Proposition 2.8. The order of An is n!/2.

Can we classify all subgroups of Sn ?Recall D4 to be the group of symmetries of a square defined as < r, f/r4 = e =

f 2, rf = frn−i >. This acts on the corners of the square which we mark as 1, 2,3 and 4 in the following manner.

r.1 = 2, r.2 = 3, r.3 = 4, r.4 = 1, f.1 = 3, f.2 = 2, f.3 = 1, f.4 = 4

Then note that r and f induce a bijection on {1, 2, 3, 4} and hence an elementof S4. More interestingly, this action can be extended to all of elements of D4

with the feature that product of elements in D4 corresponds to composition in S4

and e ∈ D4 gets mapped to identity in S4 and inducing a group homomorphismfrom D4 → S4.

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Definition 2.9. More generally, a group G is said to act on a set S if there existsa function • : G × S → S with property that e • s = s and g • (g′ • s) = gg′ • sfor all s ∈ S and for all g, g′ ∈ G.

Examples:

(1) The group Z/nZ acts on elements of S1 ⊆ C by r • z = zr.(2) The group Z/2Z acts on R3 as a • (x, y, z) = (−x, y, z).(3) The group (Q,+) acts on R4 as q • (x1, x2, x3, x4) = (x1 + a, x2 + a, x3 +

a, x4 + a). More generally translation is a group action.(4) Come up with two different examples of your own.(5) All groups can act on themselves via translation. Let G be a group then

g • a = g· for all a, g ∈ G.(6) G acts on itself by conjugation. That is, g • a = gag−1 for all a, g,∈ G.(7) Both these actions make sense if take a subgroup of G and make it act on

G.

Proposition 2.10. A group G action on a set S is equivalent to a group homo-morphism from G→ Iso(S).

Remark 2.11. Any group action on a set S determines an equivalence relationon S.

For instance, let nZ act on Z via translation.That is nr • x = x+ nr.Then define a relation on elements of Z as x ∼ y if x = g • y for some g ∈ nZ.Why is this an equivalence relation?What is the corresponding group homomorphism from nZ to Iso(Z).

Definition 2.12. Let G be a group acting on a set S. Then define for anys the stabilizer of s to be Gs = {g ∈ G | g • s = s} and orbit of s to beOs = {s′ ∈ S | s′ = g • s for some g ∈ G}.

Note that Os is the set of all elements with the same equivalence class as s forthe relation defined earlier.

Exercise: Show thatOs partitions the set S. That is, Os∩Ot =

{∅ or

Os if Os ∩Ot 6= ∅and S = tOs.

Let us consider this in some examples. What is the stabilizer of any elementwhenG acts on itself by left translation? Gh = {g ∈ G | g•h = h} but that impliesGh = {e}. The orbit of any element h ∈ G is Oh = {g ∈ G | g = g′ • h ∈ G}.But recall g = gh−1h implies that Oh = G.

What about conjugation? What is Gh and Oh when the action is by conjuga-tion?

We want to understand the relation between Gs and Os? In general we have afunction from G→ Os which takes g to g •s. What kind of a function is it? Notethat all elements of Gs map to s. More importantly, any element of the form gg′

where g′ ∈ Gs will get mapped to g • s.What can we conclude?

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Consider the action of a group on itself by translation when G is finite. Thenwe get the following theorem.

Theorem 2.13 (Cayley). Every finite group is isomorphic to a subgroup of apermutation group.

3. Factor groups and Isomorphism theorems

Let G be a group and H ≤ G then the set gH = {gh | h ∈ H} where g ∈ G issaid to be a left coset of H in G. We can similarly define a right coset to be theset Hg = {hg | h ∈ H}.Observe that

(1) If G is abelian then gH = Hg. More generally however it is possiblethat G is not abelian but gH = Hg for some g ∈ G. Can you think ofsome example? Let H =< (123) >= {(123), (132), id}. Then (12)H ={(13), (23), (12)} and H(12) = {(13), (23), (12)}.

(2) If H has the property that gH = Hg for all g ∈ G then H is said to bea normal subgroup of G. All subgroups of an abelian group are normal.Clearly, H =< 123 > is a normal subgroup of S3.

(3) Note even if gH 6= Hg, the number of elements in the two sets are equal.Show that there is a bijection from gH to Hg.

(4) Lets enumerate number of left cosets. Clearly g1H = g2H will mean thatg−11 g2 ∈ H. Is this an equivalence relation on G?

(5) Note that there is a bijection from gH to H for all g ∈ G.(6) We define index of a subgroup H in G, [G : H] to be the number of left

cosets of H in G.

Theorem 3.1 (Lagrange). Let G be a finite group and H be a subgroup of Gthen O(H) divides O(G).

Proof. Follows from previous two observations. �

Corollary 3.2. Let a ∈ G and G be a finite group then order of a divides theorder of G.Further aO(G) = e.

Corollary 3.3. Let G be a finite group and H ≤ G then [G : H] = O(G)/O(H).

Corollary 3.4. Any group of prime order is cyclic.

Let us now consider G/H the set of all left cosets of H in G. Then this is aset and we want to know if the binary operation on G defines one on G/Hg1H · g2H 6= g1g2H in general. However if we know that

Proposition 3.5. The operation on the set of left cosets G/H defined as

g1H · g2H = g1g2H

is a binary operation on G/H if and only if g2H = Hg2 for all g2 ∈ G.

Proposition 3.6. Let G be a group and N be a subgroup of G. Then the followingstatements are equivalent.

(1) N is normal in G.

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(2) For all g ∈ G, g−1Ng = N .(3) For all g ∈ G, g−1Ng ⊆ N .

This is denoted by N E G.

Proof. �

This explains our notation Z/nZ for modulo n integers.Why? Examples:

(1) Let φ : G → G′ be a group homomorphism between groups G and G′.Then Ker(φ) is a normal subgroup of G.

(2) If a subgroup H ≤ G has index 2 then it is normal in G.(3) Is every subgroup of S3 normal?(4) Does every group have a normal subgroup? A non-trivial proper sub-

group? If they do no they are called simple groups.(5) Later we will show that An for n ≥ 5 is a simple group.(6) Show that if G is a finite group of order n and p is the smallest prime

dividing n then any subgroup of G of index p is normal in G,

Theorem 3.7. Let φ : G→ G′ be a group homomorphism between groups G andG′. Then G/Ker(φ) isomorphic to Imφ via the map φ.

Definition 3.8. For any subgroups H and K of G, the set HK = {hk | h ∈H and k ∈ K}.

Exercise: Show that |HK| = O(H)O(K)

O(H ∩K).

When is this a group? What do we need? If G is abelian then HK would bea subgroup and in fact HK would be isomorphic to H ×K. In general we onlyrequire that HK = KH.

Proposition 3.9. Let G be a group and H,K ≤ G. Then HK is a subgroup ifand only if HK = KH.

Proof. Exercise �

Note in this case the product in HK will be of the form h1k1 ·h2k2 = h1h′2k′1k2,

which is definitely far from being a direct product of subgroups. A strongercondition which would still give us a group structure on HK is if for any h ∈ H,k ∈ K there exists k′ ∈ K such that kh = h′k. That is, require that khk−1 ∈ Hfor all k ∈ K.

Definition 3.10. Let G be a group and H ≤ G. Then the normalizer of H in Gdenoted by NG(H) = {g ∈ G | g−1Hg = H}.

Proposition 3.11. Let G be a group and H,K ≤ G. Then HK ≤ G if K ⊆NG(H).

Remark 3.12. Then HK is isomorphic to a twisted product for H with K.What do we mean by this?

Let H,K ≤ G and H ≤ NG(K). Then conjugation defines an action of H onthe set K. That is ? : K ×H → H defined as k ? H = khk−1.

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Recall this is equivalent to a group homomorphism from φ? : K → Iso(H).In fact, the bijection h → khk−1 is an automorphism. Thus we have a grouphomomorphism from φ : K → Aut(H), where (φ(k))(h) = khk−1. In which casewe can write h1k1h2k2 = h1k1h2k

−11 k1k2 as h1φk1(h2)k1k2.

Exercise: Show that the map φ : K → Aut(H), defined as (φ(k))(h) =khk−1 is a group homomorphism. (Aut(H) is a group with composition.)Caution: It matters that we take φ(k) be as defined and not as φ(k)(h) =

k−1hk if we want it to be a group homomorphism. We had noted this in classand hence we need K to act on H and not the other way around!!

More generally this allows us to define what are called semi direct products ofgroups as opposed to direct products of groups as done in Tutorial 2, problem 4.

Exercise Let G1 and G2 be groups and φ : G2 → Aut(G1) be a group ho-momorphism. Then consider G1 ×G2 with binary peration · defined as (h1, k1) ·(h2, k2) = (h1φ(k1)(h2), k1k2). Show that this gives a group structure on G1×G2

which can be different from the structure of direct product.

Definition 3.13. Let G1 and G2 be groups and φ : G2 → Aut(G1) be a grouphomomorphism. Then define the semidirect product of G1 oφ G2 with respectto φ as the set G1 × G2 with binary operation · defined as (h1, k1) · (h2, k2) =(h1φ(k1)(h2), k1k2).

Exercise: There is only one non-trivial homomorphism from φ : Z/4Z →Aut(Z/6Z). Computing Z/6Z oφ Z/4Z is a group of order 24. Is it isomorphicor non-isomorphic to S4?

Theorem 3.14. Let H and K be subgroups of a group G and K ≤ NG(H). ThenHK/K is isomorphic to H/H ∩K.

Theorem 3.15. Let G be a group with normal subgroups H and K and H ≤ K.Then K/H E G/H.

4. Class equation

Let us use the idea of cosets to answer our question on orbits of a group action.Let G be a group acting on a set S. Recall Gs denotes the set of all elements of

G which fix s ∈ S and Os the orbit of S. Then note that two elements g?s = g′?sif g′−1g ∈ Gs, that is, g ∈ g′Gs.

Thus we can say that two elements in the same left coset of Gs give the sameelement in the orbit.

Thus we get a bijection from G/Gs → Os.If G is finite, this says that O(Os) = O(G)/O(Gs). Note that elements which

are of s which are fixed by every element of G will have only one element in itsorbit. We denote the fixed points of the action by SG = {s ∈ S | g?s = s ∀g ∈ G}.More generally we get the class equation.

Theorem 4.1. Let G be a finite group acting on a finite set S. Then

|S| = |SG|+∑|Os|6=1

O(G)/O(Gs)

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where the sum is taken over disjoint Os.

As an application we prove the following theorem.

Theorem 4.2. (Cauchy) Let G be a finite group and p be a prime dividing theorder of G then G has an element of order p.

Proof. Consider the set S = {(x1, x2, . . . , xp) ∈ Gp | x1 . . . xp = e}. Clearly Sis non-empty since (e, . . . , e) ∈ S. In fact for every x1, x2 . . . , xp−1 ∈ G xp =(x1 . . . xp−1)

−1 defines an element of G. Therefore |S| = |G|p−1.Now < (12 . . . p) >, cyclic group of order p acts on S. Then since the group

acting is of order p, the stabilizers Gx1,...,xp either have order 1 or p. That is,O(Os) = 1 or p. Since p divides |S| and p divides

∑|Os|6=1O(G)/O(Gs) by class

equation we have SG = or p divides it. Since (e, . . . , e) ∈ SG, SG is non-emptyand has a non-trivial element which is of the form (x, . . . , x). Hence proved. �

More particularly, the equation we called the class equation is used when Gacts on itself by conjugation. In this case we have the following equation;

O(G) = O(Z(G)) +r∑i=1

[G : CG(gi)]

where Z(G) denotes the center of G, that is, Z(G) = {g ∈ G | gg′ = g′g ∀g′ ∈ G}and g1, . . . , gr represent the distinct conjugacy classes which are not in Z(G).

Theorem 4.3. If p is a prime and G is a group of order pr then P has a non-trivial center.

Corollary 4.4. If O(G) = p2 for a prime p then G is abelian and G is isomorphicto either Zp2 or Z/pZ× Z/pZ.

Theorem 4.5. Let H and K be subrgoups of G with H ∩K = {e} and H andK be normal in G. Then HK is isomorphic to H ×K.

Proof. We have shown earlier that HK is a subgroup if either of them is normalin G. Note h1k1 = h2k2 in HK if and only if h−12 h1 = k2k

−11 ∈ H ∩ K. This

implies h−12 h1 = k2k−11 = e, showing that h1 = h2 and k1 = k2.

Thus there is a bijection from φ : HK → H ×K defined as φ(hk) = (h, k).We need to show that h1k1h2k2 = h1h2k1k2 or equivalently that h1k1h

−11 = k1

or equivalently that h1k1h−11 k−11 = e. But note K CG implies h1k1h

−11 ∈ K and

hence h1k1h−11 k−11 ∈ K. Similarly, H C G implies that k1h

−11 k−11 ∈ H and hence

h1k1h−11 k−11 ∈ H giving us the desired result.

Ex: Show that φ is a group homomorphism.�

Corollary 4.6. If O(G) = pq for distinct primes p < q and p - q − 1, then G isabelian and isomorphic to Z/pZ× Z/qZ.

Recall that a group is simple if it has no non-trivial normal subgroups. Exer-cise: What are all the simple groups?

An application of the idea of conjugacy classes and class equations can be usedto show that A5 does not have any normal subgroups. The idea is to first write

13

down all conjugacy classes and their representative in A5. Then we note that anynormal subgroup of A5 must contain either all the conjugate elements or noneof them. This together withe fact that the order of such a subgroup of A5 mustdivide 60 will give us the required claim.

We will give another proof using class equation but with icosahedral group.The Icosahedral group is the group of symmetries of a dodecahedron (regularpolyhedron with 12 sides). Symmetries can arise by the following ways :

• rotating about a axis through vertex v at an angle θ. There are 20 distinctrotations and using matrix theory one can show that they are all conjugateto each other.• rotation around the center of each face (which is a pentagon). Again

using matrices, one can show that rotation by 2π/5 are all conjugatealong different faces and rotation by 4π/5 are all conjugate. There are 12each of those (no of faces).• Finally, one can rotation along any of the edges about the center of the

edge by angle π. Here however two opposite edges induce the same spinsince π = −π and there will be 15 different spins which turn out to beconjugate

This says that Order of the Icosahedral group is

|G| = 1 + 20 + 12 + 12 + 15

since it has only the identity element in the center.Now we show that G is simple. Note that if H is a normal subgroup of G and

x ∈ H then all its conjugates are H. Thus we can write the order of H as a sumof order of conjugacy classes.

For Icosahedral group, it is easy to verify that none of the sum of orders ofconjugacy classes add up to a divisor of 60 except for 1.

Therefore, G is simple.

Theorem 4.7. The icosahedral group G is isomorphic to A5.

Proof. We first show that G acts on a set of five elements. Inscribe cubes intothe dodecahedron and then note that there are 5 of them. This gives a grouphomomorphism from φ : G → S5. Since G is simple, the kernel of this map iseither trivial or the whole group. Ex: Show that φ is non-trivial to see that thekernel is trivial. Ex: Compose with the sign map. Again use simplicity of G toshow this map is trivial.

Then this will mean image of φ is A5 to get the require isomorphism. �

Definition 4.8. Let G be a group of order prm with p - m, where p is a prime.Then a p-subgroup of G is a subgroup of order ps. If the subgroup is of order pr

then, it is said to be a p-Sylow subgroup of G.

Theorem 4.9. Let G be a group of order pm, where p is a prime not dividing m.

(1) Sylow p-subgroups of G exist,.(2) If P is a Sylow p-subgroup of G and Q is any p-subgroup of G, then there

exists g ∈ G such that Q ≤ gPg−1, i.e., Q is contained in some conjugateof P . In particular, any two Sylow p-subgroups of G are conjugate in G.

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(3) The number of Sylow p-subgroups of G, np, is of the form 1 + kp,i.e.,np = 1( mod p). Further, np is the index in G of the normalizer NG(P )for any Sylow p-subgroup P , hence np divides m.

Proof. The proof that a Sylow subgroup exists is proved by induction. The induc-tive argument splits into two cases, one when p/O(Z(G)) and when it doesn’t.For second argument we first show that if P is a Sylow p-subgroup then anyp-subgroup of G acts on all its conjugates. We use this to first show that numberof conjugates are 1( mod p). Then we show that every p-group is a subgroupof one of these conjugates and hence all Sylow p-groups are conjugates of eachother. �

But before we get into details, let us use Sylow theory to classify a few groups.Let us use this to characterize all groups of order pq, where p and q are different

primes.Let G be a group of order pq, p, q prime and let p < q. Then we know that G

has a subgroup K of order p and a subgroup H of order q.By Sylow’s theorem nq = kq + 1 for some integer k and nq/p. Since p < q this

implies nq = 1. Therefore H is a normal subgroup. Then HK is a subgroup withpq elements since H ∩K = {e}. Therefore G = HK.

To understand various types of groups we can get we first try to understandwhat structure this group will have. Recall every group homomorphism φ : K →Aut(H) will give HK the structure of a semi direct group H oφ K.

Since H is a cyclic group of order q, Aut(H) is a cyclic group of order q − 1.If p - q − 1 then there is only a trivial φ and HK ∼= H ×K is therefore cyclic

of order pq.Else, there are p−1 different morphisms φi . Since Aut(H) is cyclic of order q−1

it has a unique subgroup of order p/q−1. We get a different group homomorphismfor every choice of generator of this p-subgroup. Define φi(k) = khik where h isa fixed generator for p-subgroup of H.

Exercise: Show that H oφi K is isomorphic to H oφj K for all i, j =1, 2 . . . , p− 1.

Theorem 4.10. Let p,q be distinct primes then up to isomorphism there onlytwo different groups of order pq.

Examples

• Find all groups of order 15 up to isomorphism• Find all groups of order 21 up to isomorphism.• Find all groups of order 30 up to isomorphism.

Consider any group of order 30 = 2 · 3 · 5. By Sylow theorem there is a Sylowgroup of order 2, 3 and 5 each. Now n5 = 1 mod 5 and n5/6 implies n5 = 1or6.Now n3 = 1 mod 3 and n3/10 implies n3 = 1or10 and n2 = 1 mod 2, n2/15.Therefore n2 need not be normal.

Now if n5 = 1 or n3 = 1 then HK is a cyclic subgroup of order 15 and G hasa normal subgroup of order 15.

If n5 = 6 and n3 = 10 then G has two many elements.

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Now since HK is normal the Sylow group of order 2 which we denote by Rgives us a group HKR. We need to know what all possible actions of Z2 exist onAut(Z15).

Proof. Sylow Theorems We first show that every group G of order pkm,(m, p) = 1 has a subgroup of order pk. We will apply induction. The state-ment is vacuous for O(G) = 1. Assume the statement is true for all k < n. LetO(G) = n = prm where (m, p) = 1. If p/O(Z(G) then we use Cauchy’s theoremto get a subgroup H of order p in Z(G). Since H < Z(G), H CG. Then G/H isa group of order pr−1m and has a Sylow p-subgroup K/H where K < G.

Exercise: Show that if H C G then every subgroup of G/H is of the formK/H for some subgroup K ≤ G which contains H.

Again H CG implies HK < G and since H ∩K = {e}, O(HK) = pr.Let p - O(Z(G)). Recall the class equation

O(G) = O(Z(G)) +∑

ai /∈Z(G)

[G : CG(ai)]

where ai vary over distinct conjugacy classes. Then p/O(G) but p/ - implies thatthere is some ai for which p - [G : CG(ai)]. More interestingly, p/O(CG(ai)) andin fact, pr/O(CG(ai)). Now since ai /∈ Z(G), O(CG(ai)) < O(G) and we canapply induction to find a Sylow p-subgroup.

Now to count the number of Sylow p-subgroups we first choose P to be one andtake S be the set of all distinct conjugates of P . Note the number of elements inorbit of a conjugate Pi in S will be [Q : NQ(Pi).

First we show that Q ∩ NQ(Pi) = Q ∩ Pi. We only need to show that H =Q ∩NG(Pi) ⊆ Q ∩ Pi. Since both the sets are in Q and hence p-groups, we onlyneed to show that H ⊆ Pi. Since H ⊆ NG(Pi) we get that PiH is a subgroup.We can show that

O(PiH) =O(Pi)O(H)

O(Pi ∩H).

Then O(PiH) is a power of p which is greater than order of Pi. HoweverO(PiH) ≥ O(Pi) implies it must be a Sylow p-subgroup. This implies PiH = Piand H ≤ Pi.

Let Q be any p-group in G. Then Q acts on S via conjugation and the orbitis |Oi| = [Q : NQ(Pi) = [Q : Q ∩ Pi].

We know that |S| =∑

i |Oi| =∑

i[Q : Q ∩ Pi].Let us choose Q = P1 then |O1| = 1. However since all the conjugates are

distinct, we have Pi 6= P1 for all i 6= 1. Then, P1∩Pi � P1 implies |Oi| > 1 for alli 6= 1. Since p/O(P1), p/|Oi for all i 6= 1. Thus |S| = 1 + kp for some k ∈ N∪{0}or |S| = 1 mod p.

Let Q be any p-group acting on S. Then if Q * Pi for all i then, [Q : Q∩Pi] > 1for all i. This implies p/|S|. But the number of conjugates is fixed and we knowhas to be 1 mod p. This implies there is an i such that Q ⊆ Pi, that is, there isa g ∈ G such that Q ⊆ gPg−1.

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In particular, if we choose Q to be any Sylow p-subgroup then Q ≤ gPg−1

for some g ∈ G implies it is equal to the conjugate because there will have samenumber of element.

Thus all the Sylow p-subgroups of G are conjugates of one another. If weconsider G to act on S then there is only one orbit of this action and

|S| = |OP | = [G : GP ] = [G : NG(P )].

�

Corollary 4.11. Every finite abelian group is isomorphic to the product of itsSylow subgroups.

Thus in order to understand finite abelian groups its sufficient to understandp groups.

What we want to claim is that every finite abelian p-group of order pr has asmany isomorphism classes as partitions of r.

It is then sufficient to show the following two lemmas and then apply induction.

Lemma 4.12. Every finite abelian p group G can be written as a direct productof its maximal cyclic subgroup C and another subgroup H.

Proof. Sketch We will apply induction. Its clear if G has order p. Let C bea maximal cyclic subgroup of G where O(G) = pr is an abelian group. AssumeC 6= G else we are done.

Lemma 4.13. If an abelian p group has a unique subgroup of order p then it iscyclic.

Assuming this, we can then show that there is a subgroup K of order p which isnot in C and hence C∩K = {e}. Then CK is a subgroup of G and C+K/K ∼= Chas maximal order in G/K and by induction G/K = C + K/K × H/K whereH ≤ G. This implies G = C + K + H = C + H since K ⊆ H and H ∩ C = {e}since K ∩ C = {e} and H/K ∩ C +K/K = {e}.

Hence proved.�

To prove the other lemma, note that if we take the subgroup K of all elements oforder p in G. (why is this a subgroup?). Then H = K. (Why?) Further, G/K hasan element of order p by Cauchy’s theorem and hence a subgroup of order p. Butconsidering φ : G → G with φ(x) = px, we get that G/K ∼= φ(G) ≤ G. Hence,G/K has a unique subgroup of order p and is hence cyclic. Let G/K =< aK >.But < a >≤ G must have a unique subgroup of order p which is K, therefore Kis a subgroup of < a > which means G =< a > and is hence cyclic.

More precisely, we have the following theorem for finite abelian groups.

Theorem 4.14. Let A be a finitely generated abelian group of order n and giventhe unique prime factorsization of n = p1

α1p2α2 . . . pk

αk , then

• A ∼= G1 ×G2 × . . . Gk where Gi if order piαi.

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• Each G where G ∈ {G1, G2 . . . , Gk}, |G| = pα, we have

G ∼= Zpβ1 × . . .× Zpβrwhere β1 ≥ β2 ≥ . . . ≥ βr and β1 + β2 + . . . βr = α and r and βj dependon i.• This expression is unique up to isomorphism of invariant factors.

CAUTION: In this argument we are talking about direct products. Ideallywe should be using the notion of direct sums. When we are talking about a finitecollection, direct product and direct sum are the same. We did not discuss directsums in this course, look up Algebra by Hungerford for more details.

In summary, we completely understand all finite abelian groups up to isomor-phism. We also often can write a group as a product of its Sylow subgroups. Insuch cases, we want to understand p-groups. Again, its not always possible toclassify every p-group. But we have discussed some of the known methods.

A different situation arises when a non-cyclic group is simple, that is none of itsnon-trivial subgroups are normal. We do not discuss them here, but the methodsrequired to understand them would be completely different, so it is important tobe able to identify when a group is in fact simple. We have given three differentproofs to say that there is only one simple group of order 60 which is isomorphicto A5.

With some more work one can show that the only other non-cyclic simple groupwith order less than 200 will be a group of order 168.

5. Introduction to Rings

When we were dealing with groups we often came across groups which haveother binary operations compatible with the group structure.

For examples Z has both addition and multiplication and multiplication dis-tributes over addition. Similarly, Mn(R) the set of all n × n real matrices has agroup structure with addition and has another binary operation of multiplicationwhich is compatible with it.

Definition 5.1. A set R with binary operations + : R×R→ R and · : R×R→ Ris said to be a ring if

• The set (R,+) is an abelian group,• the operation · is associative,• and · distributes over +, that is,

(a+ b) · c = a · c+ b · c c · (a+ b) = c · a+ c · b ∀ a, b, c ∈ R.

We will denote the additive identity with 0. Further, R is said to be a ring withidentity if ∃1 ∈ R such that 1 · a = a = a · 1 ∀a ∈ R. A commutative R is a ringif · is commutative.

Examples:

• The ring of polynomials, Z[x] = the set of all polynomials over Z, that is{a0x+ a1x+ . . .+ anx

n | ai ∈ Z, n ∈ N}.

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• Set of all real valued functions from [0, 1] to R. Set of all real valuedcontinuous (differentiable/ integrable functions from [0, 1]→ R?• The group nZ forms a ring too.• The set of integers modulo n, Z/nZ is a ring without identity.• Similarly {f : [0, 1]→ R | f(1) = 0} is a ring with a identity but different

from the identity than that for all functions from R → R. Note that fora subgroup of a group the identity cannot be different.• How about the set F (R,R) of functions from R→ R. Recall composition

is a binary operation. Does + and ◦ give it a ring structure?• How about for an abelian group G, (Aut(G),+, ◦)?• Its easy to see Z,Q,R,C are all examples of commutative rings with iden-

tity. In fact, R, C, Q are examples of field which are commutative ringswith identity such that all elements which all non-zero elements have mul-tiplicative inverses.• More generally, Consider the group of rational Hamiltonian ring that is,

H = {a+bi+cj+dk | a, b, c, d ∈ Q, i2 = −1, j2 = −1 = k2, ij = −ji, jk = −kj, ki = −ik}where + is defined component wise and · is defined as (a+ bi+ cj + dk) ·(a′ + b′i + c′j + d′k) by distributing · over + and using multiplication inQ and relations above.

Then we claim this is a ring which has an identity but clearly is notcommutative. Instead of a, b, c, d ∈ Q we can get a, b, c, d ∈ R as well.

More generally, would it work if a, b, c, d ∈ Z?

Definition 5.2. A ring with identity is said to be a division ring if every non-zeroelement has a multiplicative inverse.

Are there division rings which are not fields? Are there rings without identity?What are some examples of non commutative rings with identity?

More examples

• Another examples is the subset Z[i] = {a + bi | a, b ∈ Z} check that thisis a ring where addition is component wise and multiplication is by

(a+ bi)(c+ di) = (ac− bd) + (bc+ ad)i.

Replace Z by Q and check if its still a ring? Is it a field? (Consider

(a+ bi)(a− bi)a2 + b2

= 1. )

• Consider the set of all functions from R → R which have the property

that f(1) = 0. Then this is a ring with identity as i(x) =

{1 ifx 6= 1

0 if x = 0.

Note this identity is different than that in the ring of all functions fromR → R. More generally, a subring is a subgroup A of the ring (R,+, ·)which is closed under multiplication.• Let D be a rational number such that there is not rational x with the

property x2 = D. Then Q[√D] = {a + b

√D | a, b ∈ Q}. Then this is in

fact a field. Why? Further the set unique only up to the square free partof D. (Note

√D exists in C. )

19

• What if we replace Q with Z?• Consider Z6. Is this a field? What is the issue? Here on we will use Zn

to denote Z/nZ. This ring has zero divisors. A non-zero element a of aring R is a zero divisor if ∃b 6= 0 ∈ R such that ab = 0.• A commutative ring is said to be an integral domain if it has no non-zero

divisors.• We already know that Zp is a field whenever p is a prime.

EXercise If R is a integral domain then the cancellation law hold.

Theorem 5.3. Any finite integral domain is a field.

Proof. Let R be a finite integral domain. Then let a 6= 0 be an element of Rthen consider {ai/i ∈ N}. Since R is finite there exists i, j such that ai = aj. Ormore particularly. ak = 1 for some k. Choose the smallest such k to obtain amultiplicative inverse of a. �

Note that rings of the form Q[√D] have a norm function Q[

√D] → R with

defined as N(a+ b√D) = a2 −Db2.

Exercise Check that x = a+ b√D is zero if and only if N(x) = 0. Further,

if x ∈ Z(√D), then units in Z[

√D] can be found using N(x).

Properties of rings

Theorem 5.4. Let R be a ring with identity.

• Additive identity and Multiplicative identity is unique (when it exists).• a.0 = 0 = 0.a for all a ∈ R.• (−a)b = −(ab) = a(−b) for all a, b ∈ R.• (−a)(−b) = ab for all a, b ∈ R.• −a = (−1).a for all a ∈ R.

6. Constructing more rings

We can always consider the subset of a ring which is a subgroup under theaddition operation and closed under multiplication. This gives gives us severalexamples of a ring.

Subrings

• The subgroups nZ ⊂ Z are subrings.• Z[i] is a subring of Q[i].

• Similarly Z[√D] where D doesn’t have a rational square root is a subring

of Q[√D].

• The Hamiltonians with integral coefficients form a subring of rationalHamiltonians.

Polynomial Rings We can define polynomials over any commutative ring Rwith identity as R[x] = {a0 + a1x + . . . + anx

n | n ∈ N, ai ∈ R}, where additionand multiplication are defined as

n∑i=0

aixi +

m∑i=0

bixi =

k∑i=0

cixi,

20

where m ≤ n and ci = ai + bi for all i ≤ m and ci = ai for all m < i ≤ n and

n∑i=0

aixi ·

m∑i=0

bixi =

m+n∑i=0

cixi

where ci =∑j = 0iajbi−j.

Proposition 6.1. If R is an integral domain then R[x] is an integral domain.Further units of R[x] are units of R.

Group RingsLet G = {g1, . . . , gn} be a finite group with product structure given by · and

R be a commutative ring with identity.Then show that R(G) = {

∑ni=1 rigi | ri ∈ R, gi ∈ G} is a group with addition

defined component wise.n∑i=1

rigi +n∑i=1

r′igi =n∑i=1

(ri + r′i)gi

and multiplication is defined with distributively with (rigi)(rjgj) := (rirj)(gigj).

Matrix Rings Just as we can define square matrices over R, we can definematrices over any commutative ring R with identity. and this gives a ring.

We now have a plethora of examples of rings, all of these arise out of questionsfrom different areas of mathematics.

Ring homomorphisms and Ideals

Now that we have set up all these examples of rings. We need to know how tocompare them.

Definition 6.2. Let R and R′ be rings then a ring homomorphism is a grouphomomorphism φ : R → R′ such that φ(xy) = φ(x)φ(y). Further this is anisomorphism if φ is a bijection.

Properties

Proposition 6.3. Let φ : R→ R′ be a ring homomorphism then both Kerφ andImφ are subrings of R and R′ respectively.

Examples

• Are 2Z and 3Z isomorphic?• If f : R→ R′ is a map of rings with identity and R′ is an integral domain

then either its a trivial morphism or it takes identity to identity.• Z[x]→ Z mapping

∑ni=0 aix

i to a0 is a ring homomorphism.• The quotient map Z→ Z/nZ is a ring homomorphism.• The above map induces a map from Z[x]→ Z/nZ[x] which is also a ring

homomorphism. This is especially useful, for proving that a equationdoesn’t have solution by showing there is no solution modulo n for somen.

21

We now recall that one other way of constructing new groups was by quotientingby a subgroup when it is a normal subgroup.

We try to first understand what property our subring would require.Note we want (r+ I)(s+ I) = rs+ I for all r, s ∈ R. But consider any element

in r + I,s+ I, that is, for some x, y ∈ I, (r + x)(s+ y) = rs+ ry + xs+ xy.The issue is I being a subring does not ensure that ry, xs ∈ I for all x, y ∈ I

and r, s ∈ R.

Definition 6.4. An ideal is a subring with the property that rx, xr ∈ I for allx ∈ I and r ∈ R. If only rx ∈ I for all x ∈ I and r ∈ R, then I is said to be a leftideal. If only xr ∈ I for all x ∈ I and r ∈ R, then I is said to be a right ideal.

Examples

• nZ is an ideal of Z.• {∑n

i=1 aixi | ai ∈ Z, n ∈ N} ⊆ Z[x].

• {(a 0b 0

)| a, b ∈ R} ⊆ M2(R) is a left ideal?

• {(a, 2a) ∈ Z2} ⊆ Z2?• Kernel of a ring homomorphism is an ideal.

We can define the quotient of a ring by an ideal.

Proposition 6.5. Let I ⊆ R be an of R. Then R/I the quotient group of I < Ris a ring with respect to the multiplication defined as (r + I)(s+ I) = rs+ I.

Proof. We first check this is well defined. That is If

r + I = r′ + I, s+ I = s′ + I, then rs+ I = r′s′ + I.

That is, if r−r′, s−s′ ∈ I then rs−r′s′ = rs−r′s+r′s−r′s′ = (r−r′)s+r′(s−s′) ∈I since I is a two sided ideal.

Cleary R/I is closed under multiplication and the as an exercise check thatmultiplication distributes over addition in R/I. �

Theorem 6.6. Let f : R → S be a ring homomorphism then R/I ∼= f(R) asrings.

Exercise: Is intersection of two ideals an ideal? What about union?

Remark 6.7. Let I and J be ideals then is the sum of two ideals I +J an ideal?Note sum of two subrings is not a subring!! Consider A = {a0 + a2x

2 + a4x4 +

. . .+ a2kx2k | ai ∈ Z, k ∈ N∪{0}} and B = {a0 + a3x

3 + a6x6 + . . .+ a3kx

3k | ai ∈Z, k ∈ N ∪ {0}}. The A and B are subrings but not ideals and their sum is nota subring since its not closed under multiplication. Also note that if one of themwere an ideal then the sum would be a subring!

However if you take two ideals and take product of their elements then it isnot an ideal.

Definition 6.8. The product ideal of a ideals I, J ⊆ R is defined to be IJ ={∑n

i=0 aibi | n ∈ N, ai ∈ I, bi ∈ J}.

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Exercise Check IJ is an ideal.

Theorem 6.9. • Let R be a ring and A be a subring and I be an ideal ofR. Then A + I = {x + y | x ∈ A, y ∈ I} is a subring of R and A ∩ I isan ideal of A and (A+ I)/I = A/(A ∩ I).• Let I and J be ideals of a ring R such that I ⊆ J . Then J/I is an ideal

of R/I and (R/I)/(J/I) is ring isomorphic to R/J .

More generally, what is the smallest subring containing an element a ∈ R.Clearly na ∈ R for all a ∈ R. But also, ak ∈ R for all ainN. In a ring R,< a >= {nar | n ∈ N ∪ {0}, r ∈ N}.

More particularly if we want the smallest ideal containing a ∈ R. Then Ia ={∑k

i=1 xiax′i + na+ y′ia+ yia | x, x′, y, y′ ∈ R, k ∈ N, n ∈ N ∪ {0}}.

How will this change if R is commutative or R has identity?

Definition 6.10. An ideal generated by a single element is called a prinicpalideal.

Examples

• Every ideal in Z is a principal ideal.

• In Z[x] this is no longer true. Is the ideal I = {2a0 + a1x + a2x2 + . . . +

anxn | n ∈ N, ai ∈ Z} is an ideal which is not generated by a single

element. If it is generated by a single element then 2a0 + a1x + a2x2 +

. . . + anxn = p(x)q(x) where q(x) =

∑ki=0 bix

i is our generating element.This implies 2a0 = b0c0 where b0 is fixed. This implies c0 = (2a0/b0) ∈ Z.If a0 = 1 this implies b0 = 2. Then for instance if 2 + x = q(x)p(x)will imply that either degree of p = 0 which is not possible (else everycoefficient in I will be even). If degree of p is 1 then degree of q has tobe zero and p(x) = 2+x. Its easy to verify that I is not generated by 2+x.

• In Z[√

3] is every ideal principal? What does it mean to show that ringhas non-principal ideals? Consider a finitely generated ideal say the onegenerated by < 5,

√3 > Can it be generated by a single element?

• Consider the set of all functions in F(R,R) with the property that f(1/2) =0 then its an ideal with generator g(1/2) = 0, g(x) = 1, x 6= 1/2. But thisideal is not principal in Continuous functions.

Proposition 6.11. Let I be an ideal of a ring with identity. Then I = R if andonly 1 ∈ R. This implies that a field has no proper non trivial ideals.

Properties

• Commutative rings with identity with only two ideals is a field. Thisfollows from previous argument which says that for any I =< a >∈ R,a 6= 0. I = R and hence contains 1 ∈ R. Therefore, there exists x ∈ Rsuch that xa = 1. Hence R is a field.

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• Note the previous proposition also implies that any ring homomorphismfrom a field to a ring is an injection.

Ideals of a ring, give information about the quotient ring.Let us consider R/I where R is a commutative ring with identity. Then if R/I

is an integral domain, this will imply for any a+ I, b+ I ∈ R/I,

(a+ I)(b+ I) = I =⇒ either a+ I = Iorb+ I = I.

Equivalently,ab ∈ I =⇒ a ∈ Iorb ∈ I

For instance consider the ideal pZ ⊆ Z. If ab ∈ pZ then p/ab which impliesp/a or p/b. This implies a ∈ pZ or b ∈ pZ. Note this will also tell you that Z/pZ.

Consider (x) ⊆ Z[x]. Consider p(x)q(x) ∈ (x). Then p(x)q(x) = a1x + a2x2 +

. . . + anxn = (

∑ni=1 aix

i−1)x. Let p(x) =∑r

i=0 bixi and q(x) =

∑mi=0 cix

i. Thenb0c0 = 0 implies either a0 = 0 or b0 = 0, that is, either p(x) ∈ (x) or q(x) ∈ (x).

Definition 6.12. Let R be a commutative ring with identity. An ideal I ⊆ R issaid to be prime if ab ∈ I implies a ∈ I or b ∈ I.

Examples

• Consider the ideal (x2 + 1) in R[x]. Its a prime ideal.• Consider I = {f ∈ F(R,R) | f(1/2) = 0}. This is a principal ideal. Iffg ∈ I then f(1/2) = 0 or g(1/2) = 0 . So its a prime ideal.• Consider Z/pZ or R[x]/(x2+1). They are both fields and not just integral

domains. What does this say about the corresponding ideals?

Let I be a ideal in a commutative ring R with identity and R/I be a field thenthis means for any a /∈ I, a+ I has a multiplicative inverse. That is, there exists,b /∈ I such that (a+ I)(b+ I) = 1 + I.

More interestingly, this means that R/I has a no non trivial proper ideals.Note this implies there are no proper ideals in R containing I. Exercise: Showthat there is a one to one onto correspondence between ideals of R containing Iand ideals of R/I.

This then implies I is maximal.

Definition 6.13. Let R be a ring with identity. Then a non-empty proper idealof R is said to be maximal if there is no other proper ideal containing it.

Theorem 6.14. R is commutative ring with identity then R/M is a field if andonly if M is maximal.

Theorem 6.15. Let R be a ring with identity. The R has a maximal ideal.

Proof. This argument uses Zorn’s lemma, which says that any set in in whichevery chain has a maximal element must have a maximal element.

That is in our context if take S to be the set of ideals in R and C to be anytotally ordered subset of S under inclusion. If C has a maximal element in Cthen so does S.

Let T = {Jk | k ∈ I} be totally ordered. Then ∪Jk is a maximal element C.(Show it is an ideal)

24

Then by Zorns lemma the set of all ideals has a maximal element.�

Examples

• Note a ring can have more than one maximal element. For instance inR[x] both (x) and (x2 + 1) are maximal ideals.• Every maximal ideal is prime. However every prime ideal is not maximal.

For instance, (x) in Z is a not a maximal ideal. The ideal (2, x) containsit and is a proper ideal.

Two ideals of a commutative ring R are said to be comaximal if I + J = R.We have the following theorem about them.

Theorem 6.16 (Chinese Remainder theorem). Let A1, . . . Ak be ideals . ThenR→ R/A1× . . .×R/Ak defined by R→ (r+A1, . . . , r+Ak) is a ring homomor-phism with kernel A1 ∩ A2 . . . ∩ Ak . Further if Ai + Aj = R for all i 6= j thenA1 ∩ A2 ∩ . . . Ak = A1A2 . . . Ak and

R/(A1 . . . Ak) ∼= R/A1 × . . .×R/Ak

In particular for R = Z and I = pαii Z, this says if all the pi are distinct thenwe get the following corollary if n = pα1

1 . . . pαkk then

Z/nZ ∼= (Z/pα11 )Z× . . .× (Z/pαkk Z),

are ring isomorphic and induced a group isomorphism on the group of units(invertible elements.).

7. Fraction Rings

Another way to construct new rings would be to create ring which have theappropriate number of inverses we want. What do we mean by adding inverses?Recall how Q was defined. We inverted all the non-zero integers. In fact everyelement of Q is a defined by a pair of elements (p, q) where p ∈ Z and 0 6= q ∈ Z.Moreover p/q = s/t if pt = qs.

This notion can be generalized to any commutative ring R with a multiplicativeset. A set S ⊂ R is said to be a multiplicative set if a, b ∈ S implies ab ∈ S.

Theorem 7.1. Let R be a commutative ring and D be a multiplicative set without0. Then there exists a commutative ring Q with identity with i : R → Q asa injective ring homomorphism where every element of i(D) is a unit with thefollowing universal property; for any commutative ring T and ring homomorphismφ : R→ T where φ(D) ⊂ T ∗, there exists a ring homomorphism φ′ : Q→ T suchthat φ′ ◦ i = φ.

Proof. Show that

Q = {(r, s) ∈ R×R | s ∈ D, r ∈ R}/ ∼ where ∼ is defined as (r, s) ∼ (r′, s′) ⇐⇒ rs′ = r′s

has all the requisite properties. �

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8. Polynomial rings, division algorithm, Euclidean Domains

Let us now look at the case of polynomials rings. Define degree of a polyno-mial to be the largest exponent of the monomial with a non-zero coefficient. Byconvention we declare degree of the zero polynomial to be zero.

Proposition 8.1. Let R be a commutative ring. Then deg(p(x)g(x)) ≤ deg(p(x))+deg(q(x)). This is an equality when R is an integral domain.

Theorem 8.2. If R is a integral domain then R[x] is an integral domain.

Proposition 8.3. Let I be an ideal of a commutative ring R. Then (I) = I[x]is an ideal of R[x] and

R[x]/I[x] ∼= R/I[x].

Further if I is prime in R then so is I[x] ⊂ R[x].

Example: For instance Zn[x] = Z[x]/nZ[x].

Theorem 8.4. If F is a field then for any a(x), b(x) ∈ F [x] there exist uniqueq(x), r(x) ∈ F [x] such that a(x) = q(x)b(x) + r(x) and 0 < degr(x) < degb(x).

Remark 8.5. Why is this not true for an integral domain? For instance , is therea division algorithm in Z[x]?

Theorem 8.6. Let F be a field, then every ideal in F [x] is a principal ideal.

Turns out F [x] is not the only ring with this property. For example, Z does.Apparently Z[i] does too.

Definition 8.7. Let R be an integral domain with a a function N : R→ Z≥0 withN(0) = 0. If N(a) > 0 for all a 6= 0 then it is called positive norm. An integraldomain with a norm is said to be an Euclidean domain if for every x, y ∈ R ,there exist elements q, r ∈ R such that

x = qy + r with N(r) < N(y).

EXample

• Z, R[x].• Z[i] has norm as defined before N(a + bi) = a2 + b2. It has the given

property. To show this we note the following. Consider (a+ ib), (c+ id) ∈

Z[i]. Then a+ ib =ac+ bd

c2 + d2+

(bc− ad)i

c2 + d2(c+ id) = (r + is)(c+ id) where

both r, s ∈ Q. If r, s ∈ Z then we are done, else Now we can choosex = [r], y = [s] to be the integers such that |r − [r]| ≤ 1/2 and similarly|s− [s]| ≤ 1/2. Then a+ ib = (x+ iy)(c+ id) + t+ t′i.

It then is easy to show that N(c+ id) > N(t+ it′).• Ring of quadratic integers: These arise out of considering the ring of

integers in Q[√D]. Let D be a square free integer.

Consider Z[ω] where ω =

√D if D ≡ 2, 3 mod 4

1 +√D

2else D ≡ 1 mod 4

26

All of these rings have a norm with the property N(a+b√D) = a2−Db2

when D ≡ 2, 3 mod 4 and N(a + b1 +√D

2) = a2 + ab +

1−D4

b2 for all

D ≡ 1 mod 4.However, not all of these rings are Euclidean domains. For instance

Z[√−2 will be a Euclidean domain. But Z[

√−5] is not.

• Subrings of a Euclidean Domain will be Euclidean domain.

Remark 8.8. Note in order to show that Z[√−5] is not a Euclidean domain, its

not sufficient to show that the given norm doesn’t work. We have to show thatthere is no norm which makes Z[

√−5] into a Euclidean domain.

It would be simpler to show that Z[√−5] does not satisfy some property of a

Euclidean domain.

Theorem 8.9. Every ideal in a Euclidean domain is principal

Example The ring Z[√−5] has ideas which are not principal. For example

consider the ideal generated by (3, 2 +√−5) = I. If it is principal. Then I =

(a+ b√−5). This implies 3 = α(a+ b

√−5). That is, N(3) = N(α)N(a+ b

√−5).

Thus N((a+ b√−5))/9. This implies N((a+ b

√−5) = 1, 3or9.

If N(a+ b√−5) = ±1 then (a+ b

√−5) = ±1, which will mean I is the whole

ring. This implies 1 = 3x + (2 +√−5)y. Multiplying by (2 −

√−5) we get a

contradiction. Hence proved.This means N(a+ b

√−5) = 3 or 9. It cannot be 3. If it is 9. Then 2 +

√−5 =

N(β)N(a+ b√−5). This implies β = ±1 and (a+ b

√−5) = 2 +

√−5. However,

3 = (a+ b√−5)α will mean α = ±1, which would lead to a contradiction.

In any commutative ring. We can define divisibility as follows.

Definition 8.10. Let R be a commutative ring with identity. The a is said todivide b or b is said to be a multiple of a if there exists r ∈ R such that b = ar.

The greatest common divisor of a and b is an element d such that d/b and d/aand if d′/a, d′/b then d/d′.

Equivalently, if I = (a, b) then d is a g.c.d(a,b) if and only if I ⊆ (d) andI ⊆ (d′) implies (d) ⊆ (d′).

Exercise Show that if I = (a, b) = (d), then d is the gcd of a and b.

Proposition 8.11. Let R be a integral domain. If two elements d and d′ of Rgenerate the same principal ideal then they differ by a unit. Thus g.c.d of a,b ∈ Ris unique up to a unit.

Exercise: Let R is a Euclidean Domain and x and y are non-zero elements ofR. Apply Euclidean algorithm to x and y. Show that the last non-zero remainderis the gcd for x and y. Further that d = ax+ by for some a, b ∈ R.

9. Principal Ideal Domains, irreducibility in Polynomial Rings

Definition 9.1. An integral domain where every ideal is principal is called aPrincipal Integral Domain or a P.I.D.

27

We have already given examples of PIDs which arise as Euclidean domains.However there are PIDs which are not Euclidean domains. For example , Z[1 +√−19/2] is a PID and not a Euclidean domain. Refer to text to see the argument

why this is the case.Exercise: Is Q[x, y] a principal ideal domain?

Proposition 9.2. In a PID every non zero prime ideal is a maximal ideal.

Proof. Let R be a PID. Let (x) be a prime ideal. Let there exist J ⊆ R suchthat (x) ⊆ J . Since R is a PID, J = (y) for some y ∈ R. Then x ∈ (y) impliesthere exists r ∈ R such that x = ry. This means ry ∈ (x). Since the ideal isprime r ∈ (x) or y ∈ (x). If r ∈ (x) then r = xt for some t ∈ R. This impliesx = ry = xty. This means ty = 1 and y is a unit. Thus (y) = R. If y ∈ (x) thenclearly (y) = (x).

�

For example in F [x] where F is a field will be a PID and hence any non zeroprime ideal will be maximal. Turns out these are the only kind of polynomialrings which are PIDs. That is,

Proposition 9.3. Let R be a commutative ring with identity such that the poly-nomial ring R[x] is a PID. Then R is a field.

Proof. Note that (x) has to be maximal and hence R[x]/(x) which is isomorphicto R is a field. �

Does this mean that R[x] when R is a not a field is uninteresting?Not really. Consider Z[x]. This is not a PID but, given any polynomial we can

factorize into irreducibles.

Definition 9.4. Let R be a commutative ring with identity. An element a ∈ Ris said to be irreducible if a = xy then x or y are units.

An non zero element a ∈ R is said to be prime if (a) is a prime ideal in R, thatis if a/xy then a/x or a/y.

Two elements in a, b ∈ R are said to be associate if there exists a unit u ∈ R∗such that a = bu.

Example:

• In Z are all prime numbers are irreducibles.• In Z[x], x is a irreducible. If x = p(x)q(x), then one of p, q ∈ Z[x] has

zero constant coefficient. Further degp + degq = degx = 1. Therefore ifp has a non-zero constant coefficient then it must have degree 0 (since qhas zero constant coefficient it must have degree 1.) Thus x = a0(b1x).This implies a0b0 = 1 which means p(x) = a0 is a unit.• What about x2 + 1 in Z[x]?• In Z[i] 3 is irreducible but 2 and 5 are not.

Proposition 9.5. In any integral domain, every prime is a irreducible. Furtherif R is a PID every irreducible generates a maximal ideal and is hence prime.

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Proof. Let x ∈ R be a prime. Then let x = ab ∈ R. This implies x/ab and hencex/a or x/b. Let x/a. Then a = xy and x = xyb. This implies x(yb − 1) = 0.Since R is a integral domain. This implies yb = 1 since x 6= 0. Thus b is a unitand x is irreducible.

Let R be a PID and x ∈ R be a irreducible. Let I be any ideal containingx. Then I = (a) since R is a PID. Now, x ∈ (a) implies that x = ra for somer ∈ R. However since x is irreducible this means r or a is a unit. If r is a unitthen I = (x) and if a is a unit then I = R. This means any ideal containing x iseither (x) or R. This implies (x) is maximal (why?) and hence prime.

�

Example Note 3 ∈ Z[√−5] is irreducible. This is because if 3 = ab. Then

N(3) = N(a)N(b). This means 3/N(a) or 3/N(b). This will mean a = 3(x +y√−5) and (x+ y

√−5)b = 1. This will imply b is a unit.

But 3/(2 +√−5)(2−

√(−5) and does not divide either of them.

Definition 9.6. Let R be a integral domain. Then R is said to be a UFD if(a) every element x ∈ R can be written as a product x = up1 . . . pk where pi areirreducibles and (b) any two such factorizations differ by associates and order ofthe factors.

Examples

• Z.• We will show that every PID is a UFD and hence F [x] for a field F andZ[i], etc are all UFDs.• We will also show that if R is a UFD, so is R[x]. Thus Z[x] is an example

of a UFD which is not a PID.• Z[√−5] is not a UFD. 6 = 2.3 = (1 +

√−5)(1−

√−5).

• Is x2 + 1 irreducible in R[x]?

Proposition 9.7. In a UFD an element is prime if and only if it is irreducible.

Proof. We show that if x is irreducible then it is prime. Let x/ab then xy = ab.Using factorization, we can write ab as a product of irreudcibles and then x hasto be the associate of one of them. Since they divide either a or b so does x. �

Proposition 9.8. Every PID is a UFD.

Proof. Let R be a PID. Let x ∈ R. If its irreducible there is nothing to prove. Ifit is reducible then let r = s1r1. WLOG let r1 be reducible. Consider (r1), thisideal contains r and hence (r) ⊆ (r1). Now let r1 = s2r2 where r2 is reducible.Then (r1) ⊆ (r2). We thus get a ascending chain of prinicpal ideals in R, (r) ⊆(r1) ⊆ (r2) . . .. Consider the union of all these ideals I. Then I is a ideal. (Why?)But I is principal which means I = (a). This implies a ∈ (rk) for some k. Whichmeans I ⊆ Ik. Therefore this chain can be only finite. This implies eventuallywe should be able to write r as a product of irreducibles (why?)

For uniqueness we use the fact that in a PID primes as irreducibles and vicev-ersa. We can show that each factor can divide a factor in the different factorizationand is therefore an associate.

�

29

Remark 9.9. Note this implies that in Z every element can be unique factorizedinto primes up to sign.

Lemma 9.10. Let R be a UFD and F be its fraction field. If p(x) is reduciblein F [x] then p(x) is reducible in R[x].

Proof. Let p(x) be reducible in F [x] then p(x) = A(x)B(x) where A,B ∈ F [x].We can always take the product of denominators and write dp(x) = a(x)b(x),where d ∈ R and a(x), b(x) ∈ R[x]. Now either d is a unit or a product ofirreducibles since R is a UFD. If its a unit we are done we can write p(x) =(d−1a(x))b(x) in R[x]. Let d = p1 . . . pk where pis are irreducibles and henceprime in R (since its a UFD). Recall (p1)R[x] is a prime ideal in R[x] since (p1)is a prime ideal in R. This implies R[x]/p1R[x] is an integral domain.

Therefore a(x)b(x) = 0 modulo p1R[x] implies a(x) = 0 modulo p1R[x] orb(x) = 0 modulo p1R[x]. Thus, p1/a(x) or p1/b(x).

We can cancel them out to write p2 . . . pkp(x) = a1(x)b1(x) in R[x]. Repeatingthe same way for all pi’s i = 2, . . . n, we get that p(x) = an(x)bn(x) in R[x] andis hence reducible.

Hence proved. �

Note that this says that p(x) can be written as a product of F -multiples of A(x)and B(x) since we are dividing A(x) by some primes in R to get the requiredelements.

For instance if we write x3 = kx21

kx in Q[x], then we need to write it as

x3 =1

k(kx2)k(

1

kx) to get it as a product of elements in Z[x].

In fact elements of R all become units in F [x]. So elements which are reduciblein R[x] need not be reducible in F [x] if there are products of elements in R. Butnote the only units in F [x] are these non-zero elements of R.

Corollary 9.11. Let R be a UFD with fraction field F . Let p(x) ∈ R[x] have allcoefficients which which have gcd 1. Then p(x) is irreducible in R[x] if and onlyif it is irreducible in F [x].

Proof. If p(x) is a reducible in F [x] then it is reducible in R[x]. Since all thecoefficients of p(x) have gcd 1, there is no element of R which divides p(x) Ifp(x) reducible in R[x] and there is no element of R which divides p(x) thenp(x) = a(x)b(x), where a(x), b(x) are non-constant polynomials in R[x] and hencenon units in F [x].

�

Theorem 9.12. R is a UFD if and only R[x] is a UFD.

Proof. If R[x] is a UFD then R is a UFD. Conversely if R is a UFD then considerany p(x) ∈ R[x] and write p(x) = dp′(x) where, d is the greatest common factorof the coefficients. Now factor p′(x) in F [x] into irreducibles. By Gauss’s lemmap′(x) can be written as Fmultiples of these irreducibles in R[x]. Since the gcdof all these factors has to be 1, previous corollary says these F -multiplies ofirreducibles in F [x] are irreducible in R[x].

30

We need to show this factorization is unique. Its clear that factorization of dis unique. Also factorization of p′ in F [x] is unique. Therefore, two factorizationscan differ by an element of F −{0}. That is p1(x) = (a/b)p1

′(x) for some a/b ∈ Fwhere coefficients of p1, p1

′ have gcd 1.But this implies bp1(x) = ap1

′(x). Therefore b = ua for some u ∈ R∗ sincegcd of coefficients in bp1(x) and ap1

′(x) can be same up to a unit u ∈ R∗. Henceproved.

�

Remark 9.13. This is not true for PIDs. The ring Z is a PID but Z[x] is not.Note this means that a polynomial ring on any number of variables on a UFD

is a UFD. For example Q[x, y].

10. Irreducibility in Polynomial rings

Proposition 10.1. Let F be a field and let p(x) ∈ F [x]. Then p(x) has a factorof degree one if and only if p(x) has a root in F , that is, there is a α ∈ F suchthat p(α) = 0.

Proof. Use division algorithm. �

Corollary 10.2. A polynomial of degree 2 or 3 is reducible in F [x] for a field Fif it has a root.

Proof. It follows from the fact that such polynomials should have a degree onefactor. �

Proposition 10.3. Let p(x) =∑n

i=0 aixi be a polynomial with integer coeffi-

cients. If r/s ∈ Q with (r, s) = 1 is a root of p(x) then r/a0 and s/an. Thus, ifp(x) is monic ( that is, an = 1) then if p(d) 6= 0 for all divisors d of a0 then p(x)has no roots in Q.

Proof. Follows from plugging in r/s as a root. �

Proposition 10.4. Let I be a proper ideal in an integral domain R and let p(x)be a nonconstant monic polynomial in R[x]. If the image of p(x) in R/I[x] cannotbe factored into product of smaller degree factors then p(x) is irreducible in R[x].

Proof. Show reducible in R[x] means reducible in R/I[x]. �

Proposition 10.5. Let P be a prime ideal in an integral domain. and f be amonic polynomial with coefficients ai. If an−1, . . . , a0 ∈ P and a0 is not in P 2

then f is irreducible in R[x].

Corollary 10.6 (Eisenstein’s criterion). Let p be a prime in Z and let f(x) =xn + an−1x

n−1 + . . . + a0, n ≥ 1 and ai ∈ Z. Let p/ai for all i but p2 does notdivide a0. Then f(x) is irreducible in both Z[x] and Q[x].