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This is the exercise IV.2.12 of Algebra book by Thomas W. Hungerford, page 190.

Let $R$ be a ring with unity and let $M$ be a free $R$-module with countably infinite basis $e_1,e_2, e_3, \cdots$.

Then $S=\text{End}_R(M)$ is a ring over $R$.

I need to show that for any positive integer $n$, $S$ is a free left-module over itself of rank $n$ i.e., $$S \cong S \oplus S \oplus \cdots \oplus S \ (\text{n times}).$$ The fact that is astonishing to me that $S \cong \bigoplus_{i=1}^{n}S$ for any $n$. That is, \begin{align}&S \cong S \oplus S, \\ &S \cong S \oplus S \oplus S, \\ & S \cong S \oplus S \oplus S \oplus S, \\& \text{so on} \end{align} How can this happen ?

The hint says:

$\{1_S\}$ is a basis of one element,

$\{f_1,f_2\}$ is a basis of two elements, where $f_1(e_{2n})=e_n,~f_1(e_{2n-1})=0$ and $f_2(e_{2n})=0,~f_2(e_{2n-1})=e_n$,

It is obvious that $S$ is a free $S$-module of rank $1$.

What about the other cases ?

I think we need to use induction. For,

if $n=1$, then $S$ is a free $S$-module with basis $\{1_S\}$,

Asumme $S$ is free $S$-module with basis say $\{f_1,f_2, \cdots, f_n\}$, that is, assume $S \cong S^n$, then we need to show $S \cong S^{n+1}$.

How does the above hints help ?

user26857
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MAS
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1 Answers1

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If $S\simeq S\oplus S$ then $S\oplus S\simeq S\oplus S\oplus S$. Can you take it from here?

user26857
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  • Thanks. ahh, it is so easy. But I think it is not easy to explicitly calculate the basis elements – MAS Dec 17 '21 at 02:11
  • Do you mean to give an explicit basis for $S^n$? – user26857 Dec 17 '21 at 07:04
  • Yes. I mean what would be the basis elements of $S^n$ ? How do we find that ? For $n=1$, it is trivial, for $n=2$ it is given in the hint. What about cases or for general case $n$ ? – MAS Dec 17 '21 at 12:23
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    This shouldn't be that hard: just use the way to build a basis for the direct sum of two free modules whose bases are known. – user26857 Dec 17 '21 at 12:50
  • Thanks for the idea. I have just question. As you noticed I have tried to use induction for $n=2$, i.e., I assumed $S \cong S \oplus S$ and the rest follows easily. But is it a good way ? Because any ring $R$ is isomorphic to itself i.e., $R \cong R$. Now if we assume $R \cong R \oplus R$ (which both you and me assumed), then for any ring $R$, we would have $R \cong R^n$ for any $n$, positive integer. If yes, then there nothing speciality in the given ring $S=\text{End}_K(M)$. Am I right ? – MAS Dec 17 '21 at 13:37
  • Well, finding a ring such $R$ that $R\simeq R^2$ is the main point here. – user26857 Dec 17 '21 at 19:58