How to show that two left ideals over $R$ are not isomorphic $R$-modules?

Question

Let $R=\begin{bmatrix}F&0&0\\ F&F&0\\F&0&F\end{bmatrix}$ with a field $F$. Show that

$$\begin{align} I_1=\begin{bmatrix}0&0&0\\0&0&0\\F&0&0\end{bmatrix}, I_2=\begin{bmatrix}0&0&0\\0&F&0\\0&0&0\end{bmatrix} \end{align}$$

are not isomorphic $R$-modules.

My ideas:

$I_1$ is nilpotent and $I_2$ is not. Therefore, I thought that they can't be isomorphic as $R$-modules. But the left ideals $I_1$ and

$$\begin{align} I_3=\begin{bmatrix} 0&0&0\\0&0&0\\0&0&F\end{bmatrix} \end{align}$$

are isomorphic as $R$-modules, even though $I_1$ is nilpotent and $I_3$ isn't.

So, how do I show that $I_1$ and $I_2$ aren't isomorphic as $R$-modules? And why can't I use the nilpotent property?

Answer 1

Be careful

For a noncommutative ring like this you have to specify what side you are working on (left or right modules.) The answer will hinge on this.

Q+A

Why can't I use the nilpotent property?

The nilpotency of a submodule of $R$ is not useful because module isomorphisms aren't multiplicative, and so the property isn't relevant for modules.

how do I show that $I_1$ and $I_2$ aren't isomorphic as $R$-modules?

One thing useful for disproving isomorphisms are the annihilators, since module isomorphisms preserve annihilators.

The left annihilator of $I_1$ is $\begin{bmatrix}F&0&0\\F&F&0\\F&0&0\end{bmatrix}$ and the right annihilator is $\begin{bmatrix}0&0&0\\F&F&0\\F&0&F\end{bmatrix}$.

The left annihilator of $I_2$ is $\begin{bmatrix}F&0&0\\F&0&0\\F&0&F\end{bmatrix}$ and the right annihilator is $\begin{bmatrix}F&0&0\\0&0&0\\F&0&F\end{bmatrix}$.

The left annihilator of $I_3$ is $\begin{bmatrix}F&0&0\\F&F&0\\F&0&0\end{bmatrix}$ and the right annihilator is $\begin{bmatrix}F&0&0\\F&F&0\\0&0&0\end{bmatrix}$.

From this we can see that none of the three ideals are isomorphic as right $R$ modules, and that $I_1$ and $I_2$ are not isomorphic as left $R$ modules. Now you see the importance of side...

The disposition of the left annihilators doesn't immediately mean that $I_1\cong I_3$, but upon further investigation we do confirm that right multiplication by $\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}$ furnishes a left-$R$ linear isomorphism of $I_1$ to $I_3$.

But being a minimal ideal is preserved by module isomorphisms, right?

I think you want to say that being a simple module is certainly preserved by isomorphisms.

And if I have a two-sided minimal ideal, then it is simple as a left and right module. So, an isomorphic module must also be simple as a left and right module. Thus, also minimal two-sided ideal, isn't?

No, the statement I boldfaced is false. In a simple ring like $M_2(\Bbb R)$, the entire ring is the unique minimal (nonzero) ideal, but it has infinitely many minimal left ideals and minimal right ideals. It is not simple as a left or right module. The other direction is true though: if a two-sided ideal is simple as a left module or a right module, then it is a minimal nonzero two-sided ideal.

I'm also trying to find a third left ideal which is isomorphic to $I_1$ but which is neither $I_1$ nor $I_3$. I looked at ideals which have the same annihilator but since $F$ is field, I always come back to $I_1$ and $I_3$. Is it even possible to find another one?

$L_\lambda:=\{\left[\begin{smallmatrix}0&0&0\\0&0&0\\\lambda x&0& x\end{smallmatrix}\right]\mid x\in F\}$ for any nonzero $\lambda$ in $F$ provides many examples.

Answer 2

From your own observations and the comments it seems that it is already clear to you why nilpotency of the ideal is not preserved under module isomorphisms.

Something that is preserved under module isomorphisms, is the annihilator ideal: For a left $R$-module $M$, let $\operatorname{ann}(M) = \{\, r \in R \mid rx=0 \text{ for all }x \in M \,\}$. This is an ideal of $R$, and it is straightforward to check that $M \cong N$ implies $\operatorname{ann}{M}=\operatorname{ann}{N}$.

Now, $I_1$ is annihilated by $$ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}, $$ but $I_2$ is not. Therefore $I_1 \ncong I_2$ as left $R$-ideals.