8/10/2019 Math 230A HW 10

1/2

Math 230A HW 10

Aaron Chen

12/8/2014

1. Find d=gcd(1122211,1234321) and also integers r, s s.t. r

1122211 +s

1234321 =d.Answer: First, we have that:

1234321 = 1122211 1 + 112110

Then we have:1122211 = 112110 10 + 1111

112110 = 1111 100 + 10101111 = 1010 1 + 101

Finally:1010 = 101

10

Going backwards, we have that:1111 = 101 (10 + 1) 112110 = 101 (11 100) + 101 10 = 101 (1110) 1122211 =

101 (1110 10) + 101 11 = 101 (11111) 1234321 = 101 11111 + 101 1110.

Hence,d = 101. We solve for r, ss.t. from 101 = 11111010 101 = (1111) (112110100 1111) = 112110+101(112221111211010) 101 = 10111222111011(12343211122211) 101 = 1122211 (101 + 1011) 1011(1234321). Therefore, r = 1112, s= 1011.Remark. 1112*1122211 - 1011*1234321 = 1247898632 - 1247898531 = 101, yay!

2. LetR, S commutative rings;I , J ideals ofR, S respectively. View I Jas an ideal ofR Jproduct ring.(i) Prove I J is prime either I=R, J Sprime; or vice versa.Pf: () Let the product I Jbe prime. Then consider what the coset ring looks like when wemod - we have that (R S)/(I J)=(R/I) (S/J). Suppose instead that both ideals are nottheir entire rings. Then, their coset rings are non-trivial. Mainly, we have that R/I {0} (S/J).Therefore, we can find nonzero coset elements from both rings, (r0 +I), (s0 +J). Since the productI J is prime, we have that modding R Sby the product must be a domain; no zero divisors.But then, we have that (r0+ I, 0 + S) (0 + I, s0+ J) = (0 + I, 0 + J). And this is a contradictionto how the modular ring must be a domain. Hence, it is impossible for neither of the rings to betheir entire ring.() Suppose one of the ideals is actually its entire ring with the other one prime. Then, when wemod (R

S)/(I

J), the result is something isomorphic to R/I(WLOG let it be the J=R case)

in whichR/Iis a domain. Since (RS)/(I J) is a domain, the product ideal must be prime.

(ii) Is part (i) still true of we replace prime with maximal?The () part is still true because the modding collapses the product and will produce a field as acoset ring from one of the rings Ror S. Also, the () part is also still true because if it is not,we wll have zero divisors again and fields do not have zero divisors. So, yes, it is all still true ofpart (i) used maximal ideals instead.

8/10/2019 Math 230A HW 10

2/2

3. Show that Z[211] is not a PID.

Pf: Maybe we can find an ideal such that there are elements in the ideal where they are not involvedwith the (one) generator. Or maybe we can show it is not a UFD because PID implies UFD! So,consider the element44 Z[211]. We can write it as44 = 22 11 or as44 = (211)2.And the element 2

11Z[211]] is irreducible because the only way to write it is one times

itself. So it has two different factorizations that are not differing by just a unit, so its not even aUFD. Therefore it cannot be a PID.

4. Let R commutative ring, D a multiplicative subset ofR. In defining the ring structure ofD1R, we setr/d+s/e= (re+sd)/(de) and (r/d) (s/e) = (rs)/(de). Prove that these operationsare well defined.Pf: For well defined-ness, we need to show that the operations respect the equivalence classes ofD1R...its really the field of fractions ofR. Let two representatives of a class be (r/d), (r/d); (s/e), (s/e).We get the relations rd =rd; se =se.

(*): (r/d)(s/e) := (rs)/(de).From the relations, we have (r =rd/d); (s =se/e) - invertingis okay because d, e

D1. Plugging in: ((rd/d)

(se/e))/(de) := (rdse)/(dede) = (rs)/(de).

(+) r/d +s/e := (re + sd)/(de). Plugging in that (r = rd/d); (s = se/e) again, weget: (rd/d)/d + (se/e)/e = (rd)/(dd) + (se)/(ee) = (r/d) + (s/e) := (re+sd)/(de).