2
$\begingroup$

Let $A\subset M_2(k)$ for a field $k$, be a matrix subring consisting of matrices of the form $$A=\left(\begin{matrix}*&*\\0&* \end{matrix} \right). $$ Then, consider the maximal right ideals $$m_1=\left( \begin{matrix}1&1\\0&0 \end{matrix} \right) \text{ and } m_2=\left( \begin{matrix}0&1\\0&1 \end{matrix}\right).$$ Then, $A/m_1 \cong k$ with right action defined by $$1 \cdot \left(\begin{matrix}a&b\\0&c \end{matrix} \right)=b. $$ Additionally, $A/m_2 \cong k$ with right action defined by $$1 \cdot \left(\begin{matrix}a&b\\0&c \end{matrix} \right)=a. $$ However, these are not isomorphic modules, despite them both being isomorphic to $k$. Indeed, let $$f:k \to k $$ be non trivial. Then, if the module were isomorphic, $$f\left(1 \cdot \left(\begin{matrix}a&b\\0&c \end{matrix} \right) \right)=f(1)\left( \begin{matrix}a&b\\0&c \end{matrix} \right), $$ $$\Rightarrow f(b)=f(1)\cdot a $$ which clearly can't hold for every $a,b \in k$, and so the modules cannot be isomorphic.

$\textbf{Finally, my question:}$ When we consider the isomorphism classes of simple right modules (both of the right modules presented above are simple), what are we to do about the fact that both of the modules are isomorphic to $k$, but not to each other as modules. This is simply a consequence of how we defined the action on $A$ in their cases, and so my intuition is to say they both fall into the same isomorphism class, but I'm not actually sure whether this is correct. What's the standard convention in this case?

$\endgroup$

1 Answer 1

1
$\begingroup$

what are we to do about the fact that both of the modules are isomorphic to $𝑘$, but not to each other as modules.

Here might be a good time to pause and ask "isomorphic to $k$... as what type of object?" An isomorphism is always in the context of a particular category. This case is a good example: the two modules are isomorphic as sets, but not isomorphic as modules.

and so my intuition is to say they both fall into the same isomorphism class, but I'm not actually sure whether this is correct.

Modules can't be both nonisomorphic and fall into the same isomorphism class, no. The two contradict each other.

Perhaps you are just experiencing a little cognitive dissonance around one set $k$ having disparate module structures?

The two structures are completely different. $\begin{bmatrix}0&k\\ 0&0\end{bmatrix}$ annihilates the first but not the second, and $\begin{bmatrix}k&0\\ 0&0\end{bmatrix}$ annihilates the second and not the first. The two module structures behave completely differently. Of course, isomorphic modules have equal annihilators. The two are not in the same isomorphism class of $R$ modules.

But as sets, isomorphism just means a bijection, so the two are in the same isomorphism class of sets.

$\endgroup$
3
  • $\begingroup$ If one were to go about systematically finding all of the isomorphism classes if left modules (simple or otherwise), is there a good way to go about this? For example, I'm pretty sure that all of the modules isomorphic to $k^2$ are module isomorphic, but this is only from checking by hand. And I don't know that I've checked every possibility. Is there a systematic way to figure out what all of the isomorphism classes are? $\endgroup$
    – Ty Perkins
    Sep 7, 2023 at 16:48
  • 1
    $\begingroup$ "Isomorphic to $k^2$" has no meaning here. At least you need to be more specific. From the discussion above, we know there are at least three nonisomorphic $A$ modules with the underlying set $k^2$. $\endgroup$
    – rschwieb
    Sep 7, 2023 at 17:41
  • 1
    $\begingroup$ In general there will be infinitely many isomorphism classes of modules. I'm not sure you'd want to do that. But at least the classes of simple modules are doable. With those in hand, you can say you understand the class of semisimple $A$ modules. $\endgroup$
    – rschwieb
    Sep 7, 2023 at 17:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .