Let R be a commutative ring with a 1, and let J and K be ideals of R. Prove that $_R(R/J) \cong {_R}(R/K)$ iff $J = K$.

Question

This is a question taken from p.95 of Hartley and Hawkes: Rings, modules and Linear Algebra.

Firstly is the question correct or should it read $J \cong K$? This would seem more natural to me.

Assuming the question is correct then one of the directions is trivial so here's my attempt at the other direction.

If $R/J \cong R/K$ then there exists homomorphisms $\psi, \phi$ mapping $R \to N$ with $\ker(\psi) = J$ and $\ker(\phi) = K$ then $\text{im}(\psi) \cong \text{im}(\phi)$. Then I would have to show some how that this implies $J = K$.

I cannot even show that it implies $J \cong K$.

Many thanks

Answer 1

Notice that $(j + J)(r + J) = 0 + J$ for $j \in J$ and that $\phi$ is a module homomorphism.

Since $\phi$ is surjective there exists $r+J$ such that $\phi(r+J) = 1 + K$.

We have $0 + K = \phi(0 + J) = \phi((j+J)(r+J)) = j\cdot\phi(r+J)=j(1+K) = j+K$.

So $j \in K$ and thus $J \subset K$. By symmetry, using $\psi$, we get $J = K$.

Answer 2

The annihilator of an $R$-module $M$ is the set of $a\in R$ for which $aM=\{0\}$. Clearly isomorphic ideals have equal annihilators.

Let $R$ be a commutative ring, and $I$ an ideal. I claim that the annihilator of $R/I$ is $I$. If $a\in I$ and $b+I\in R/I$ then $a(b+I)=ab+I=0+I$ as $ab\in I$. Then the annihilator contains $I$. But if $a$ is in the annihilator then $a(1+I)=a+I=0+I$, that is $a\in I$. So the annihilator is $I$.

If $R/J$ and $R/K$ are isomorphic $R$-modules, their annihilators are equal: $J=K$.