# Prove that any free module with finite basis is projective

Let $$P$$ be a free R-module such that $$\{p_1,\cdots,p_n\}$$ is a basis. To show that $$P$$ is projective we need to show that if exists $$a: M \rightarrow P$$ surjective and R-linear, then $$M = \ker(a) \oplus P_1$$ with $$M$$ as a R-module and $$P_1 \cong P$$.

Here's what I've got by now:

Assume that there is such R-homomorphism $$a$$. Since it is onto, it follows that $$p_i = a(m_i)$$ for $$m_i \in M$$. Now, consider the following homomorphism: \begin{align*} b:P &\rightarrow M\\ \sum r_ip_i &\mapsto \sum r_im_i \end{align*} That homomorphism exists since there are at least $$n$$ different $$m_i$$'s in $$M$$ by our previous conclusion, and it is well defined since $$P = Rp_1 \oplus \cdots \oplus Rp_n$$.

It follows that: $$ab(p_i)=a(m_i)=p_i \implies ab(p) = p \quad \forall p \in P \implies ab(p) = \mathbb{1}_P$$ Now, see that if $$m \in \ker{a}$$ then $$m = m - ba(m)$$ because $$a\big(m - ba(m)\big) = a(m) - aba(m) = a(m) - a(m) = 0$$. Therefore we can write any element of $$m$$ as: $$[m - ba(m)] + ba(m)$$ and the first term is in $$\ker(a)$$ and second term is in $$b(P)$$, hence: $$M = \ker(a) + b(P)$$ Now suppose $$m \in \ker(a) \cap b(P)$$. It follows from $$m \in \ker(a)$$ that $$m = m-ba(m)$$. It follows from $$m \in b(P)$$ that $$m = \sum r_im_i$$. Therefore: \begin{align*} m &= m - ba(m)\\ &= \sum r_im_i - ba\left(\sum r_im_i\right)\\ &= \sum r_im_i - b\left(\sum r_ip_i\right)\\ &= \sum r_im_i - \sum r_im_i = 0 \end{align*} And we conclude that $$\ker(a) \cap b(P) = \{0\} \implies M = \ker(a) \oplus b(P)$$. Now it remains to show that $$b(P) \cong P$$. For that, see that: $$b(P) \cong M/\ker(a) \cong a(M) = P \implies b(P) \cong P$$

Can someone please check my proof? The part of show that $$M = \ker(a) + b(P)$$ was a bit tricky but it eventually came.

Thanks and any constructive critics about the proof is highly appreciated.

I think you have a confused definition of projective modules. An $$R$$-module $$M$$ is called projective if whenever $$\pi : A \to B$$ is a surjective $$R$$-module morphism, and $$g : M \to B$$ is some $$R$$-module morphism, then there exists some $$R$$-module morphism $$\tilde{g} : M \to A$$ with $$\pi \circ \tilde{g} = g$$. Your definition is that $$(M / N) \oplus N \cong M$$ for all submodules $$N$$ of $$M$$, which is equivalent but this is usually stated as a theorem rather than the definition.

I think your proof is okay, but you should look up something called "the splitting lemma" which encapsulates a lot of what your proof is doing.

Just for the avoidance of doubt, the fact that the basis of $$P$$ is finite is not required. There is the much more general proposition:

$$\textbf{Claim:}$$ Let $$R$$ be a ring, then an $$R$$-module $$M$$ is projective if and only if is the direct summand of a free $$R$$-module.

$$\textit{Proof.}$$ $$(\Rightarrow):$$ Suppose first that $$M$$ is projective, and let $$F$$ be the free $$R$$-module on the elements of $$M$$. Formally $$F$$ is the direct sum of copies of $$R$$ indexed by the elements of $$M$$, and for $$m \in M$$ let $$1_m \in F$$ be the basis element of $$F$$ indexed by $$m$$. Then there is a surjective $$R$$-module morphism $$\pi : F \to R$$ mapping $$1_m \mapsto m$$. Since $$M$$ is projective there is an $$R$$-module morphism $$\tilde{\pi} : M \to F$$ lifting the identity morphism $$M \to M$$ through the surjection $$\pi$$, which is to say $$\pi \circ \tilde{\pi} = \operatorname{Id}_M$$. But then $$\tilde{\pi}$$ is injective, and

$$F = \tilde{\pi}(M) \oplus \operatorname{\ker}(\pi) \cong M \oplus \operatorname{\ker}(\pi),$$ and so $$M$$ is a direct summand of a free module.

$$(\Leftarrow):$$ Suppose now that $$M$$ is the direct summand of a free $$R$$-module, say $$F$$. Then $$F = M \oplus N$$ for some $$R$$-module $$N$$. Then suppose that $$\pi : A \to B$$ is a surjective morphism of $$R$$-modules, with $$g : M \to B$$ some morphism of $$R$$-modules. Then $$F$$ has some basis $$\left\{ f_i \mid i \in \mathcal{I} \right\},$$ say, where $$\mathcal{I}$$ is some indexing set. Then for each $$i \in \mathcal{I}$$, there is some unique $$m_i \in M, n_i \in N$$ such that $$f_i = m_i + n_i$$. Now we can extend $$g$$ to a map $$g' : F \to B$$ by letting $$g'(f_i) = g(m_i)$$. Then since $$\pi$$ is surjective there is some $$a_i \in A$$ with $$\pi(a_i) = g'(f_i)$$. Then define $$\tilde{g} : F \to A$$ such that $$f_i \mapsto a_i$$. Then the restriction of $$\tilde{g}$$ to $$M$$ satisfies $$\pi \circ \tilde{g} = g$$, and so $$M$$ is indeed projective.

The fact that $$P$$ is free with basis $$\{p_1,p_2,\dots,p_n\}$$ means that, given any module $$M$$ and any choice of $$x_1,x_2,\dots,x_n$$, there exists a unique homomorphism of $$R$$-modules $$f\colon P\to M$$ such that $$f(p_i)=x_i$$, for $$i=1,2,\dots,n$$.

Thus, if you're given $$a\colon M\to P$$ surjective, you can choose $$x_1,\dots,x_n\in M$$ such that $$a(x_i)=p_i$$, $$i=1,2,\dots,n$$.

Thus we get $$b\colon P\to M$$ with the property that $$b(p_i)=x_i$$, $$i=1,2,\dots,n$$.

Consider $$a\circ b\colon P\to P$$; then $$a\circ b(p_i)=a(x_i)=p_i$$. By the uniqueness property for free modules, you conclude that $$a\circ b=1_P$$ (the identity on $$P$$).

In particular, $$b$$ is injective, so $$P\cong P_1=b(P)$$. Moreover, if $$x\in M$$, then, setting $$f=b\circ a$$, we have $$x=(x-f(x))+f(x)$$ Note that $$a(x-f(x))=a(x)-a\circ b\circ a(x)=a(x)-a(x)=0$$, that is, $$x-a(x)\in\ker a$$, and that $$f(x)=b\circ a(x)\in P_1$$. Therefore $$M=\ker a+P_1$$.

In order to finish we need to show that $$\ker a\cap P_1=\{0\}$$. Suppose $$x\in\ker a\cap P_1$$. Then $$x\in P_1$$ implies $$x=b(y)$$ for some $$y\in P$$; then $$0=a(x)=a(b(y))=y$$ and, in particular $$x=0$$.

That's essentially your own proof, but with much less computations.