# Two ideals both alike in dignity, in fair Paris where we lay our scene. (proving ideals are isomorphic)

Let $A$ be an integral domain. I have to show that two ideals $\mathfrak a$ and $\mathfrak b$ are isomorphic as $A$-modules if and only if there exist $a$ and $b$ such that $a\mathfrak b=b\mathfrak a$.

I gather that for "$\Leftarrow$" the isomorphism is $x\rightarrow a^{-1}bx$ or something, but I can't prove that $a$ and $b$ have inverses.

My question is: why are $a$ and $b$ invertible?

• They need not be. Let $K$ the field of fractions of $A$. Jan 25, 2014 at 22:55
• Great choice in title! Jan 25, 2014 at 22:58
• @Asinus: $a$ has an inverse in $K$, the field of fractions. It doesn't matter whether $a^{-1} \in A$, it only matters if it is a homomorphism. In your question, you are missing a "nonzero" before "$a$ and $b$" since $0\mathfrak{a}=0\mathfrak{b}$ for all ideals. Jan 25, 2014 at 23:35
• I can't fault the title's cleverness, but the problem is that it's totally unhelpful for people searching for the same question... Oct 16, 2014 at 10:15
• @rschwieb Too clever for me. I don't get it, what does Paris have to do with the question? Also, Paris doesn't scan as well as Verona, you might have used Lutetia.
– bof
Nov 14, 2014 at 8:21

They need not be. Let $K$ the field of fractions of $A$. – Daniel Fischer Jan 25 at 22:55

and

$a$ has an inverse in $K$, the field of fractions. It doesn't matter whether $a^{−1}∈A$, it only matters if it is a homomorphism. In your question, you are missing a "nonzero" before "$a$ and $b$" since $0\mathfrak{a}=0\mathfrak{b}$ for all ideals. – Jack Schmidt Jan 25 at 23:35

• This is not really an answer. It is just a comment how the claim has to be modified. Nov 14, 2014 at 9:07
• @MartinBrandenburg As I understood it, the question said 'why are $a$ and $b$ invertible?', and the response was 'they might not be but that doesn't matter because...' Nov 14, 2014 at 11:45

The correct statement is: Let $I,J \subseteq A$ be two ideals of an integral domain $A$. Then $I,J$ are isomorphic $A$-modules if and only if there are $a,b \in A \setminus \{0\}$ such that $aI = bJ$.

$\Leftarrow$: Let $aI=bJ$. For every $i \in I$ there is a unique $j \in J$ such that $ai=bj$. It is now easy to check that $i \mapsto j$ induces an isomorphism of $A$-modules. Alternatively, you may look at the $A$-module isomorphism $Q(A) \to Q(A), ~ x \mapsto b^{-1} a x$ and observe that it maps $I$ onto $J$.

$\Rightarrow$: Let $I \cong J$ be an isomorphism of $A$-modules. It extends to an isomorphism of $Q(A)$-modules $I \otimes_A Q(A) \cong J \otimes_A Q(A)$. Notice that $I \otimes_A Q(A)$ embeds into $A \otimes_A Q(A) = Q(A)$, so that its dimension over $Q(A)$ is at most $1$. It follows that the isomorphism of $Q(A)$-modules is given by multiplying with an element of $Q(A)^*$ when we embed both into $Q(A)$, say $\frac{a}{b}$. It now follows that $aI=bJ$.