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.
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.