Let $R$ be a ring and $T$ be a simple left $R$-module. Consider a nonzero $R$-linear map $$h:T\to\bigoplus_{i\in I}{T_i},$$ where $(T_i)_{i\in I}$ is a nonempty family of $R$-modules isomorphic to $T$. For convenience, we may identify each $T_i$ with its canonical copy in $\bigoplus_{i\in I}{T_i}$.

Conjecture. $h(T)=T_i$ for exactly one $T_i$.

Here are my attempts: Because $h$ is nonzero and $T$ is simple, by Schur's lemma, the map $h$ is injective, namely $h(T)\cong T$.

Besides, since $h$ is nonzero, we can find one $i\in I$ such that $h(T)\cap T_i\ne 0$. However, by the simplicity of $h(T)$ and $T_i$, we must have $h(T)=T_i$ in this case.

Then according to the property of direct sum, for $j\ne i$, $$h(T)\cap T_j=T_i\cap T_j=0,$$ proving the uniqueness.

Edit. My arguments above are false. Consider $T=\operatorname{Span}([1,1])\subseteq\mathbb{R}^2$, where $T_1=\operatorname{Span}([1,0])$ and $T_2=\operatorname{Span}([0,1])$ in which $$h:T\to T_1\oplus T_2=\mathbb{R}^2$$ is the inclusion map. But $$h(T)\cap T_1=h(T)\cap T_2=0$$ in this case.

Nevertheless, I still wonder if it is possible to show that

New Conjecture. $h(T)\subseteq\bigoplus_{i\in J}{T_i}$ for some finite subset $J\subseteq I$.

Any help will be appreciated.


1 Answer 1


Every simply module is generated by one element. Let's call that element $u$. Then $h(u)$ is an element of a direct sum, hence co-finitely many of its 'coordinates' are zero. Letting $J$ be the set of all $i$ for which the $i$th coordinate of $h(u)$ is nonzero provides an affirmative answer to your question.

  • $\begingroup$ Bravo! How could I ignore this trivial fact! Many thanks!!! $\endgroup$ Dec 29, 2023 at 3:20
  • 1
    $\begingroup$ I'll be honest --- I thought it was false. You could take countably many copies of Z, say, and send $u$ to $(1,1,1,...)$. But then I realized you'd said "simple module", and I had to look up the definition. And somewhere in Wikipedia it pointed out that simple modules are cyclic. And I had to look up direct sum to see whether it was the one with the "cofinite" restriction (vs direct product) -- I can never remember that(I haven't been a real mathematician for 30+ years now). And with those two lookups, the rest just fell out. So answering your question taught me a good deal as well. $\endgroup$ Dec 29, 2023 at 4:12

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