# How to show $S$ is a free left-module over itself of rank $n$ for any $n \in \mathbb{N}$?

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 ?

• The rank 2 case is treated here or here. Dec 16, 2021 at 16:28

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

• Thanks. ahh, it is so easy. But I think it is not easy to explicitly calculate the basis elements
– MAS
Dec 17, 2021 at 2:11
• Do you mean to give an explicit basis for $S^n$? Dec 17, 2021 at 7:04
• 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, 2021 at 12:23
• 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. Dec 17, 2021 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, 2021 at 13:37