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I am working on the following problem:

Let $R$ be a PID and let $a,b \in R$ be such that $\gcd(a,b) = 1$. Prove that there are $s,t \in R$ such that $sa+tb = 1$, that the $R$-module $R/\langle a \rangle \oplus R/\langle b \rangle$ is isomorphic to the $R$-module $R/\langle ab \rangle$, and that $R$-module $R/\langle a \rangle \otimes R/\langle b \rangle$ is isomorphic to the trivial $R$-module $0$.

I am thinking to use a well-known theorem on tensor products involving the gcd but I don't recall what the theorem is. I would greatly appreciate any help with this. Thank you.

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Oops! That's a REALLY low accept rate! –  DonAntonio Dec 3 '12 at 20:58
    
That $R/(a) \otimes R/(b) = R/(gcd(a,b))$? –  Rankeya Dec 3 '12 at 21:59
    
You have asked $6$ questions on $MSE$ so far. You should accept answers to the previous questions you are satisfied with. –  Rankeya Dec 3 '12 at 22:03
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@ DonAntonio and Rankeya: I have now accepted the answers. I am new to this site and still learning how it works. Thanks for letting me know about this. –  user49097 Dec 3 '12 at 23:17
    
@Rankeya: Yes, that is the theorem I am thinking of using, thanks. –  user49097 Dec 3 '12 at 23:34
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1 Answer 1

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So we have that $\,\exists \,s,t\in R\,\,\,s.t.\,\,\,sa+tb=1$ , and then putting $\,I_a:\langle a\rangle\,\,,\,I_b:=\langle b\rangle\,$ , we get for any $\,x,y\in R\,$:

$$(x+I_a)\otimes(y+I_b)=xy(1+I_a)\otimes(1+I_b)=(xysa+xytb)(1+I_a)\otimes(1+I_b)=$$

$$=\left[xysa(1+I_a)\otimes(1+I_b)\right]+\left[xytb(1+I_a)\otimes(1+I_b)\right]=$$

$$xys\left[(a+I_a)\otimes(1+I_b)\right]+xyt\left[(1+I_a)\otimes(b+I_b)\right]=xys(0)\otimes(1+I_b)+xt(1+I_a)\otimes(0)=$$

$$=0+0=0$$

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