TITLE: Why is $M=\{(x_i)\in R^n: \sum r_ix_i=0\}$ a projective module?
QUESTION [5 upvotes]: Suppose $R$ is a commutative unital ring with generators $r_1,\dots, r_n$. How can we see that the submodule
$$
M=\{(x_1,\dots,x_n)\in R^n:\sum_{i=1}^n r_ix_i=0\}
$$
is a projective submodule?
I noticed that the homomorphism $f: R^n\to R$ defined by $f(x_1,\dots,x_n)=\sum r_ix_i$ is a surjective map with kernel $M$, so $R^n/M\cong R$. Does this somehow suggest that $R^n\cong M\times R$, or possibly that $M\cong R^{n-1}$? I know it would for vector spaces at least. That is my hope, since then $M$ would be a direct summand of the free $R$-module $R^n$, hence projective. Thanks.

REPLY [4 votes]: Consider the surjection $R^n \to R$ described by the OP, namely
$$(x_1,\ldots,x_n) \mapsto \sum_i r_i x_i.$$
This is a surjection $R^n \to R$.  Since $R$ is a free as a module over iself,
we may split this surjection.  (Concretely, write $1 = \sum_i a_i r_i,$
and define a splitting via $r \mapsto (ra_1,\ldots,r a_n).$)
Thus $R^n \cong M \oplus R,$ and so $M$ is a direct summand of a free $R$-module,
proving that it is projective.