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Let $\mathbb{K}$ be a field and $A$ a finite dimensional $\mathbb{k}$-algebra with identity $1_{A}$. For two $A$-modules $M$ and $N$ we have the set of all $A$-linear maps from $M$ to $N$ denoted $\mathrm{Hom}_{A}(M,N)$. This set is not in general an $A$-module neither by the left nor by the right (unless the $\mathbb{k}$-algebra $A$ is a commutative). Recall that any $A$-module $M$ is also a $\mathbb{K}$-vector space by the action: $$\lambda\cdot m = (\lambda 1_{A}) m,\quad \forall\,m\in M\,,\lambda\in\mathbb{k}$$

In particular, the set of $A$-linear maps $\mathrm{Hom}_{A}(M,N)$ is also a $\mathbb{k}$-vector space using:

$$(\lambda\cdot f)(m)=(\lambda 1_{A})f(m).$$ It's known that $\ \dim_{\mathbb{K}}(M)<\infty\ $ and $\ \dim_{\mathbb{K}}(N)<\infty\quad$ implies $$\quad \dim_{\mathbb{K}}\left(\mathrm{Hom}_{A}(M,N)\right)<\infty.$$

Is there any formula for $$\dim_{\mathbb{K}}\left(\mathrm{Hom}_{A}(M,N)\right)$$ when $M$ and $N$ are finite dimensional $A$-modules, that is, $\dim_{\mathbb{K}}(M)<\infty$ and $\dim_{\mathbb{K}}(N)<\infty$ ?

we already know when $A=\mathbb{K}$ that: $$\dim_{\mathbb{K}}\left(\mathrm{Hom}_{\mathbb{K}}(M,N)\right)=\dim_{\mathbb{K}}(M)\dim_{\mathbb{k}}(N)$$

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    $\begingroup$ No, $\hom_A(M,N)$ will not itself be an $A$-module, unless $A$ is commutative (or at least its commutators annihilate $N$ I guess). Can you see why? In any case, yes $\hom_A(M,N)$ is a vector space, but I see no reason why its dimension would have a formula involving established invariants of modules, I feel it should instead depend very specifically on the exact nature of $A$ and its action on $M$ and $N$. $\endgroup$ – anon Jul 1 '18 at 16:23
  • $\begingroup$ Thank you, is true the set $\mathrm{Hom}_{A}(M,N) $ is not an $A$-module in general. $\endgroup$ – Hector Blandin Jul 1 '18 at 17:42
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    $\begingroup$ If $M$ and $N$ are non isomorphic simple modules, then the dimension of the Hom-space will be $0$; if they're isomorphic, the dimension can be different for each isomorphism class. $\endgroup$ – egreg Jul 1 '18 at 21:48
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I don't know if that's what you're expecting, but there can be no formula in terms of $\dim_k(M), \dim_k(N)$ in the general case.

Case 1: For instance, if $A= k[T]/(T^2)$, $M$ is a $k$-vector space with a trivial action of $T$ and $N = A$, then for $f\in\hom_A(M,N)$, $x\in M$ one must have $Tf(x) = f(Tx)$, hence $Tf(x) = 0$ and so $f(x) =\lambda_x T$ for some $\lambda_x\in k$, and clearly $x\mapsto \lambda_x$ is $k$-linear. Conversely, if $l:M\to k$ is $k$-linear, then $x\mapsto l(x)T$ is $A$-linear, so $\dim(\hom_A(M,N)) = \dim\hom_k(M,k) = \dim(M)$

Case 2: However if $N$ is a $2$-dimensional $k$-vector space with a trivial action of $T$, then an $A$-module morphism $M\to N$ is just a $k$-linear map, so $\dim\hom_A(M,N) = \dim\hom_k(M,N) = (\dim M)(\dim N) =2(\dim M)$.

This shows that in general, there is no $F_A:\mathbb{N}^2\to \mathbb{N}$ such that $F_A(\dim M,\dim N) = \dim\hom_A(M,N)$

Case 3: There is an interesting case where there is a formula : when $A$ is a field and $A/k$ is thus a (finite dimensional) field extension. Then $\dim_A$ makes sense and the following formula holds for any $A$-module $M$ ($A$-vector space): $\dim_kM = (\dim_kA)(\dim_AM)$. Thus $\dim_k\hom_A(M,N) = (\dim_kA)(\dim_A\hom_A(M,N)) = (\dim_kA)(\dim_AM)(\dim_AN)= \frac{(\dim_kM)(\dim_kN)}{\dim_kA}$

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  • $\begingroup$ Sorry, what is the trivial action of $T$ on $M$ ? $\endgroup$ – Hector Blandin Jul 12 '18 at 19:18
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    $\begingroup$ $Tx =0$ for all $x$ : it corresponds to the morphism $T\mapsto 0$, $A\to End(M)$ $\endgroup$ – Maxime Ramzi Jul 12 '18 at 19:50
  • $\begingroup$ $\ell:M\longrightarrow k$ the codomian of $\ell$ is the field $k$ non ? $\endgroup$ – Hector Blandin Jul 12 '18 at 21:25
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    $\begingroup$ Yes , $l: M\to k$, I'll edit this. $\endgroup$ – Maxime Ramzi Jul 13 '18 at 8:58
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    $\begingroup$ And no, even if $\dim_k(A)=2$, Case 3 considers only the possibility where $A$ is a field $\endgroup$ – Maxime Ramzi Jul 13 '18 at 8:58

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