I have come upon with the following problem: Let $M$ be an $A$-module (assume that $A$ is commutative with identity) which is both free and finitely-generated. My question is whether there is an elementary way to prove that $M$ is isomorphic to the free finitely-generated $A$-module $A^{n}$, for some $n\in\mathbb{Z}_{\geq 0}$. Here is a proof using tensor-products: First, since $M$ is free, we have an isomorphism $M\cong \bigoplus_{I}A$ for a set $I$. Then, since it is finitely-generated, say by $x_{1},\dots,x_{n}$, we get a surjective map $\phi:A^{n}\twoheadrightarrow M\cong \bigoplus_{I}A$ (sending each basis vector $e_{i}$ to the generator $x_{i}$). Let then $\mathfrak{m}\subseteq A$ be maximal ideal and consider the field $k=A/\mathfrak{m}$ which is an $A$-module. Then, since tensoring is right-exact, we have that the induced map $\phi\otimes\mathrm{id}_{k}:A^{n}\otimes_{A}k\to (\bigoplus_{I}A)\otimes_{A}k$ is surjective. Now, as tensor product commutes (distributes) naturally with direct sums we have a commutative square: $\require{AMScd}$ \begin{CD} A^{n}\otimes_{A}k @&gt;{\phi\otimes\mathrm{id}_{k}}&gt;&gt; \left(\bigoplus_{I}A\right)\otimes_{A}k\\ @VVV @VVV\\ k^{n} @&gt;{\psi}&gt;&gt; \bigoplus_{I}k \end{CD} where the vertical arrows are isomorphisms. Thus, $\psi:k^{n}\to \bigoplus_{I}k$ is a surjective map of $k$-vector spaces and so $n\geq |I|$, as desired. I am wondering though if there is a more elementary proof, e.g. by using the basis explicitly.
Say I have a module $M$ over a ring $R$ that is   $1.$ finitely-generated, i.e. $\exists$ finite $X \subset M$ such that $M = \sum_{i=1}^{n} Rx_{i}, x_{i}\in X$ (but $X$ is not necessarily free)  $2.$ free, i.e. $\exists$ a basis $Y \subset M$ a.k.a. $Y$ is free and generates M, (but $Y$ is not necessarily finite).  Now I want to show, that there exists a finite basis for $M$. For vector spaces, this would obviously be trivial, but for some  $R$-Modul, I'm not sure how to proceed.   Furthermore, is every minimal generating subset of $M$ a basis?
Travel to Ireland (Dublin) with U.K. Visitor visa I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  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