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Let us consider the following sub algebra of $\mathbb{M}_4(K)$:

\begin{equation} A=\begin{pmatrix} K & 0 & 0 & 0\\ K & K & 0 & 0\\ K & 0 & K & 0\\ K & K & K & K \end{pmatrix} \end{equation}

Calculate soc($A)$ (as $A$-module).

My attempt.

By definition soc($A$) is the sum of all simple submodules of $A$. Let's consider $e_1=e_{1,1}, e_2=e_{2,2}, e_3=e_{3,3}$ and $e_4=e_{4,4}$. It is clear that it $\{e_1,e_2,e_3,e_4\}$ is a complete set of primitive orthogonal idempotents of $A$. Therefore any $A$-simple module is isomorphic to some $S(i)$, where $S(i)=$top$(e_iA)=e_iA/$rad($e_iA$). We calculate the radicals:

\begin{equation} \text{rad}(e_1A)=0, \hspace{0.1cm}\text{rad}(e_2A)=\begin{pmatrix} 0 & 0 & 0 & 0\\ K & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}, \hspace{0.1cm} \text{rad}(e_3A)=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ K & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation}

\begin{equation} \text{rad}(e_4A)=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ K & K & K & 0 \end{pmatrix} \end{equation}

So:

\begin{equation} S(1)=e_1A, \hspace{0.1cm}S(2)=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & K & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}, \hspace{0.1cm} S(3)=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & K & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation}

\begin{equation} S(4)=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & K \end{pmatrix} \end{equation}

And then

\begin{equation} \text{soc}(A)=\begin{pmatrix} K & 0 & 0 & 0\\ 0 & K & 0 & 0\\ 0 & 0 & K & 0\\ 0 & 0 & 0 & K \end{pmatrix} \end{equation}

I think that at some point in my reasoning I am wrong, since the text says that:

\begin{equation} \text{soc}(A)=\begin{pmatrix} K & 0 & 0 & 0\\ K & 0 & 0 & 0\\ K & 0 & 0 & 0\\ K & 0 & 0 & 0 \end{pmatrix} \end{equation}

Martin Sleziak
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Ghost_Math125
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2 Answers2

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What you have computed is $\operatorname{top} A$. It should be noted that the action of $A$ on $\operatorname{top} A$ is not given by multiplication of elements in $A$. Notice how the subset $$ D:= \begin{bmatrix} K & 0 & 0 & 0\\ 0 & K & 0 & 0\\ 0 & 0 & K & 0\\ 0 & 0 & 0 &K \end{bmatrix} $$ is not closed under right multiplication by elements of $A$, as for example $e_{1,1} e_{1,2} = e_{1,2} \notin D$, and thus $D$ is not even a submodule of $A$.

You can check that the textbook's suggestion is indeed semisimple. To show that it is in fact the socle you can use the observation that if $M$ is a submodule of $A$ such that $e_{i,j} \in M$ with $1<j$, then $e_{i,j} e_{j,1} = e_{i,1} \in M$; this implies that $Ke_{i,1}$ is a proper nonzero submodule of $M$, and thus $Ke_{i,1}$ are the only simple submodules of $A$.

Ost
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  • Can we always consider that some $e_{ij}$ is in $M$? – Ghost_Math125 Feb 28 '25 at 00:58
  • It seems I was a bit sloppy in the second paragraph of my answer. For instance, $K(e_{i,1}-e_{j,1})$ is also simple for $i\neq j$. Fortunately, it is not difficult to fix my error. Assume instead that $M$ does not lie in our candidate for the socle, which we denote by $S$. Then $M$ contains an element of the form $m=e_{i,j} + x$, with $x$ a linear combination of other elements from the obvious basis of $A$. Acting by $e_{j,1}$, we get that $me_{j,1}= e_{i,1}+xe_{j,1}\in M\cap S$. By construction $xe_{j,1}$ does not contain any $e_{i,1}$-term, so $m\neq 0$ and thus $0\neq M\cap S\subsetneq M$. – Ost Feb 28 '25 at 08:51
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For a finite dimensional algebra, you can just compute that $soc(R_R)=\ell.ann(J(R))$ and $soc(_RR)=\mathscr r.ann(J(R))$.

Of course, $J(R)$ in this case is just $\begin{bmatrix} 0&0&0&0 \\ a&0&0&0 \\ b&0&0&0 \\ c&d&e&0 \\ \end{bmatrix}$

By using $1$'s and $0$'s judiciously to form various elements of $J(R)$, you can easily see $soc(R_R)=\begin{bmatrix} a&0&0&0 \\ b&0&0&0 \\ c&0&0&0 \\ d&0&0&0 \\ \end{bmatrix}$

and

$soc(_RR)=\begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ a&b&c&d \\ \end{bmatrix}$

rschwieb
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