Homomorphism and Tensor Product Over P.I.D.

Let $R$ be a P.I.D., let $(a), (b)$ be two ideals of $R$, $d=gcd(a,b)$, then

$Hom_R(R/(a),R/(b))\cong R/(d)$

Proof

Suppose $a=\prod up_i^{e_i}$, $b=\prod vp_i^{f_i}$, then by Chinese Remainder Theorem (CRT), since Bezout's Identity holds for P.I.D., the ideals of form $(p_i^{e_i})$ are pairwise co-maximal(co-prime), so we have:

$R/(a)\cong \prod R/(p_i^{e_i})$ and $R/(b)\cong \prod R/(p_i^{f_i})$. (Note that $R/(u)\cong 0$ for unit $u$).

Given $R$ modules $A_{\alpha}, B$, we have $Hom_R(\prod_{\alpha} A_{\alpha}, B) \cong \prod_{\alpha} Hom_R(A_{\alpha}, B)$ (this is also true for the second coordinate).

So, $Hom_R(R/(a),R/(b)) \cong \prod Hom_R(R/(p_i^{e_i}),R/(p_j^{f_j}))$.

(1) when $i\neq j$, $M = Hom_R(R/(p_i^{e_i}),R/(p_j^{f_j})) = 0$ because if $h\in M-{0}$, then $f(\overline{1})\neq 0$, but $f(\overline{p_i^{e_i}})=p_i^{e_i}f(\overline{1})=0$, contradiction.

(2) when $i=j$, consider $M = Hom_R(R/(p^{e}),R/(p^{f}))$

     (i) if $e\ge f$, then $M\cong R/(p^f)$ by $(\overline{1}\mapsto \overline{r})\mapsto \overline{r}$.

     (ii) if $e< f$, then $M\cong R/(p^e)$ by $(\overline{1}\mapsto kp^{f-e})\mapsto k$.

Therefore, $Hom_R(R/(a),R/(b)) \cong \prod Hom_R(R/(p_i^{e_i}),R/(p_j^{f_j})) \cong \prod R/(p_i)^{min(e_i,f_i)}$, by CRT, $Hom_R(R/(a),R/(b)) \cong R/(d)$.

$R/(a)\otimes R/(b)\cong R/(d)$

Proof

We first define an $R$-balanced map $f$ as follows:

$(x\mod (a), y \mod (b))\mapsto xy \mod (d)$

Then there is a induced homomorphism $\varphi$ that sends $(x \mod (a)\otimes y \mod (b))\mapsto xy \mod (d)$. 

Define $\psi: R/(d) \to R/(a)\otimes R/(b)$ by sending $r\mod (d) \mapsto (r\mod (a),1\mod (b))$, then $\varphi$ and $\psi$ are inverse homomorphisms hence isomorphisms.