Some questions of PIDs I'm stuck on a couple of practice problems relating to PIDs, they are paraphrased below:
Given a PID $R$ with $a$ and $b$ in $R$ and gcd$(a,b)=1$ I need to show that:
1) There are elements $s$ and $t$ of $R$ such that $sa+tb=1$
2) The $R$-module $R/(a) \bigoplus R/(b)$ is isomorphic to the $R$-module $R/(ab)$ (where $(a)$ denotes the ideal generated by $a$)
I know that for 1 if I can assume that $(a)+(b)=R$ because gcd$(a,b)=1$ then I am done, but I'm not sure I can assume that (i.e. I can't remember if the poof of that fact uses 1).
For 2 I'm not really sure where to start and would appreciate a full explained answer as the test I am practicing for is tomorrow.
 A: For 1), use the definition of $\gcd(a,b)$.
For 2), consider the map $\phi:R\to R/(a)\oplus R/(b)$ defined by $\phi(x)=(x\bmod a, x\bmod b)$. Using question 1, you can show that it is surjective and that its kernel is $(ab)$. Then use the first isomorphism theorem to conclude.
A: For 1) you need to show that $(a) + (b) = (gcd(a,b))$, or use this fact if you already know it.
2) is a special case of the Chinese remainder theorem. First, try to show that the kernel of the map $R \rightarrow R/(a) \oplus R/(b)$ given by $r \mapsto (r+(a), r+(b))$ is $(a) \cap (b)$. Now try to show that $(a) \cap (b) = (ab)$. Then by the first isomorphism theorem it suffices to show surjectivity. To show surjectivity, you need to show that given $(x_1 + (a), x_2 + (b)) \in R/(a) \oplus R/(b)$, there exists $r \in R$ such that $r + (a) = x_1 + (a), r + (b) = x_2 + (b)$. For this you need to use the fact that there exist $s \in (a)$ and $ t \in (b)$ such that $s + t = 1$. For $r$ try an element of the form $ce + df$ where $c, d, e, f$ are from among the elements $s, t, x_1, x_2$ such that $r + (a) = x_1 + (a)$ and $r + (b) = x_2 + (b)$. The last part is probably the hardest. Good luck!
