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Let $R$ be a PID. How can one show that for all non zero $a,b \in R$ we have $R/(a) \oplus R/(b) \cong R/\gcd(a,b) \oplus R/\operatorname{lcm}(a,b) $.

I have no idea how to define such an isomorphism. What I tried is to define $f([x],[y])=([\gcd(x,y)], [\operatorname{lcm}(x,y)])$. I didn't get anywhere with that my map is probably not even well defined I got lost in the calculations. So does anyone know how to find the desired isomorphism.

Thanks in advance

user26857
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TheGeometer
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    HINT: when $\gcd(a,b)=1$, then the statement is simply the Chinese Remainder Theorem. Now, since every PID is a UFD, use prime factorizarion and apply Chinese Remainder Theorem to conclude. – Crostul Feb 01 '18 at 08:01

1 Answers1

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Consider the exact sequence $0\rightarrow S\rightarrow R^{2}\rightarrow V\rightarrow 0.$ Write the canonical basis of $R^{2}$ as $e_{1},e_{2}$, the generators of $S$ as $\{f_{1},f_{2}\}$.

Let $(f_{1},f_{2})=(e_{1},e_{2}) \left(\begin{matrix} a&0\\ 0&b \end{matrix}\right)$, then $V=R/(a)\oplus R/(b).$

The Smith Normal Form of $\left(\begin{matrix} a&0\\ 0&b \end{matrix}\right)$ is $\left(\begin{matrix} \gcd(a,b)&0\\ 0&\operatorname{lcm}(a,b) \end{matrix}\right)$ (it can be achieved by multiplying on the left and right by invertible square matrices), so we have:

$$R/(a)\oplus R/(b)\cong V\cong R/(\gcd(a,b))\oplus R/(\operatorname{lcm}(a,b)).$$

user26857
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Erica
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