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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.

P.G
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2 Answers2

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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 $$\mathrm{Hom}_R\left(\bigoplus_i M_i, N\right) \cong \prod_{i} \mathrm{Hom}_R(M_i, N), \quad \mathrm{Hom}_R\left(N, \prod_i M_i\right) \cong \prod_{i} \mathrm{Hom}_R(N, M_i).$$

57Jimmy
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  • Ok, I think I did it. But I didn't use the hypothesis that every homomorphis from $M_i $ to $M_j$ is zero for $i \neq j$. Where should I have use it? Thank you – P.G Apr 03 '18 at 21:42
  • I'll edit the answer to include this – 57Jimmy Apr 04 '18 at 13:40
  • Oh thanks, but I still don't see why the assumption is necessary. Let $f \in Ker(\psi)$ so we have $p_i \circ f \circ q_i = 0, \forall i$, where $p_i$ is the projection and $q_i$ is the inclusion. Let $m=m_1 + ... + m_n \in M$, we have that $f(m) = f((m_1+0+...+0) + (0+m_2+0+...+0)+...+(0+0+...+m_n)) = 0$, so we have $f=0$ and then $Kerf(\psi) = 0$ and $\psi$ is injective. What am I missing? – P.G Apr 04 '18 at 22:23
  • You cannot conclude that $f(m)=0$: you only know that $p_i(f(q_i(m_i))=0$, i.e., that the image under $f$ of $q_i(m_i)$ inside $M_i$ is $0$, but the image of $q_i(m_i)$ under $f$ in other $M_j$'s might be different from $0$, i.e., $p_j(f(q_i(m_i)) \neq 0$. – 57Jimmy Apr 05 '18 at 14:20
  • Oh, I think I got it. So we would have a nonzero homomorphism $g: M_i \rightarrow M_j$: $g(m_i) = p_j(f(q_i((m_i)))$. That's right? – P.G Apr 05 '18 at 19:16
  • Yes, or if you prefer, the other way round: if your assumption does not hold, then there is such a $g$. With it we can construct $f$ that is not zero but maps to zero under $Psi$. For instance, $f(m):=q_j(g(p_i(m)))$. – 57Jimmy Apr 05 '18 at 21:18
  • Thank you so much!! – P.G Apr 05 '18 at 23:00
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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 $$q_ip_if| M_i=f|M_i$$

Rene Schipperus
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