let $M_1,..., M_n$ be left modules over R, and $M = M_1 \bigoplus ... \bigoplus M_n$.

Question

Suppose that if $i \neq j$, then every homomorphism from $M_i$ to $M_j$ are zero. Show that

$$End{}_{R}(M) \cong End{}_{R}(M_1) \times ... \times End{}_{R}(M_n)$$.

Any help would be great.

Answer 1

Define

$$\Psi : \mathrm{End}_R(M) \to \prod_i \mathrm{End}_R(M_i), \ \phi \mapsto \prod_i \phi_i$$ where $\phi_i$ does the following: given $m_i \in M_i$, $\phi_i(m_i)$ is the $i-$th component of $\phi(0 + \dots + 0 + m_i + 0 + \dots+0)$. Show that $\Psi$ is a morphism, injective and surjective.

Edit: Actually, "the $i-$th component of $\phi(0 + \dots + 0 + m_i + 0 + \dots+0)$" is just the whole image, because of the assumption. There is a much more general form of the statement, namely

$$\mathrm{End}_R(M) \cong \prod_{i,j} \mathrm{Hom}_R(M_i, M_j).$$ This directly implies what you want to prove. In particular, in your case $\Psi$ would not be injective without your assumption. I guess at this point the best thing to do is to give an idea of why the general fact holds. Define

$$\Psi : \mathrm{End}_R(M) \to \prod_{i,j} \mathrm{Hom}_R(M_i, M_j), \ \phi \mapsto \prod_{i,j} \phi_{i,j},$$

where $\phi_{i,j}$ does the following: given $m_i \in M_i$, $\phi_{i,j}(m_i)$ is the $j-$th component of $\phi(0 + \dots + 0 + m_i + 0 + \dots+0)$. An inverse of $\Psi$ is given by

$$\Psi^{-1} : \prod_{i,j} \mathrm{Hom}_R(M_i, M_j) \to \mathrm{End}_R(M), \ \prod_{i,j} \phi_{i,j} \mapsto \phi,$$

$$\text{where} \quad \phi\left(\sum_i m_i \right) := \sum_{i,j} \phi_{i,j}(m_i).$$

If we only consider $\prod_i \mathrm{End}_R(M_i)$, instead of the larger $\prod_{i,j} \mathrm{Hom}_R(M_i, M_j)$, then we can still define $\Psi$ and $\Psi^{-1}$ in a similar way, but $\Psi$ would not be injective and $\Psi^{-1}$ would not be surjective (unless your assumption holds): $\Psi \circ \Psi^{-1}$ would still be the identity, but not $\Psi^{-1}\circ \Psi$, because it would forget all the mixed terms.

Note: If the direct sum where infinite, then neither the general fact nor your specific case would hold: $\Psi$ would not be surjective and $\Psi^{-1}$ would not be well-defined. Indeed, if I choose all the $\phi_{i,j}$ to be non-zero, then $\prod_{i,j}\phi_{i,j}$ cannot be in the image of $\Psi$. Similarly, $$\Psi^{-1} \left( \prod_{i,j}\phi_{i,j} \right) \notin \mathrm{End}_R(M).$$ But we still have an even more general statement, that also holds in the infinite case: $$\mathrm{Hom}_R\left(\bigoplus_i M_i, \prod_i M_i\right) \cong \prod_{i,j} \mathrm{Hom}_R(M_i, M_j),$$ which is a direct consequence of the following two facts: for any $R-$module $N$, we have $$\mathrm{Hom}_R\left(\bigoplus_i M_i, N\right) \cong \prod_{i} \mathrm{Hom}_R(M_i, N), \quad \mathrm{Hom}_R\left(N, \prod_i M_i\right) \cong \prod_{i} \mathrm{Hom}_R(N, M_i).$$

Answer 2

Let $p_i:M\to M_i$ be the projection and $q_i:M_i\to M$ the obvious inclusion.

Then $$1=\sum q_ip_i.$$. Now if $f:M\to M$ is an endomorphism then $$f=\sum q_ip_if$$ however $$q_ip_if| M_j=0$$ if $i\neq j$ thus we have $$q_ip_if| M_i=f|M_i$$