How to show that $\operatorname{Hom}_A(A/(b), K/A)$ is isomorphic to $A/(b)$?

Question

Let $A$ be a integral domain and let $K$ be its field of fractions.

Show that for any $b \in A$,

$\operatorname{Hom}_A(A/(b), K/A)$ is isomorphic to $A/(b)$.

So I have a few questions.

1. I don't quite understand the notation $K/A$. Can someone explain it to me?

2. How do I begin the first step, that is to construct a homomorphism ?

Answer 1

The following general fact is well known: $\operatorname{Hom}_R(R/I,M) = \{m \in M ~|~ Im=0\}$.

Let $b \neq 0$.

In our case, this yields $\operatorname{Hom}_A(A/(b),K/A) = \{x \in K/A ~|~ bx=0\}$.

$bx=0 \in K/A$ means $bx \in A$, i.e. $x = \frac{a}{b} \in \langle \frac{1}{b} \rangle$ for some $a \in A$. Thus we have shown the desired

$$\operatorname{Hom}_A(A/(b),K/A) = \{x \in K/A | bx=0\} = \langle \frac{1}{b} \rangle \cong A/(b).$$

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For the sake of completion: For $b=0$, we have

$$\operatorname{Hom}_A(A/(b),K/A) = \operatorname{Hom}_A(A/,K/A) = K/A \not\cong A = A/(b),$$

i.e. the statement is true for $b \neq 0$, but false for $b=0$.

Answer 2

You have to assume $b\ne0$, because in this case the result is false.

Given $a\in A$, you can define $\tilde{a}\colon A\to K/A$ by $$ \tilde{a}(x)=\frac{ax}{b}+A $$ (when $b\ne0$). It is clear that $(b)\subseteq\ker\tilde{a}$, so $\tilde{a}$ defines $\hat{a}\colon A/(b)\to K/A$ by $$ \hat{a}(x+(b))=\frac{ax}{b}+A $$ It's easy to verify that $a\mapsto\hat{a}$ is a homomorphism $$ \varphi\colon A\to \operatorname{Hom}_A(A/(b),K/A) $$ If $a\in (b)$, then $\tilde{a}$ is the zero map, so also $\hat{a}=0$. Conversely, if $\hat{a}=0$, we have $$ \hat{a}(1+(b))=\frac{a}{b}+A=A $$ that is, $a\in(b)$. Thus $\varphi$ induces a monomorphism $$ \tilde{\varphi}\colon A/(b)\to\operatorname{Hom}_A(A/(b),K/A) $$ and it remains to show that $\varphi$ is surjective. This will follow when we show that any homomorphism $f\colon A\to K/A$ that vanishes on $(b)$ is of the form $\tilde{a}$, for some $a$. Let $$ f(1)=\frac{t}{u}+A $$ so $f(x)=\frac{tx}{u}+A$. By assumption $f$ vanishes on $b$, so $$ \frac{tb}{u}\in A $$ and we are done by choosing $a=tb/u$.