I am asking if my proof is correct as follows:
$\textbf{Theorem}$ Let $R$ be a ring with unity and $S$ a nonempty set. Then the polynomial ring $R[S]$ is projective.
$\textit{proof }$ First of all, $R$ is a subring of $R[S]$, so $R[S]$ is a unitary $R-$module Let $Y:= \{1_R\}\cup\{f\in R[S]\mid \text{ f is a monomial}\}$ Let us show that $Y$ is a basis of $R[S]$. For $f\in R[S]$, we know that $f = \Sigma_{j = 1}^nr_jM_j$, where the $M_j$ are distinct elements of $Y$ and $r_j\in R$ (f is a map $f:T:= \{\phi\in \mathbb{N}^S\mid \phi(s_1) = 0\text{ for all but finitely many $s_1\in S$}\}\to R$ s.t. $f(\phi) = 0_R$ for all but finitely many $\phi\in T$, so we may choose $M_1,...,M_n$ to be the elements $\phi \in T$ s.t. $f(\phi)\neq 0_R$). So $Y$ generates $R[S]$. But also, if $\exists a_1,...,a_m\in R$ and $N_1,...,N_m\in Y$ s.t. \begin{equation*} a_1N_1+...+a_mN_m = 0_{R[S]}, \end{equation*} then every $a_i = 0$ (two polynomials are equal iff they have equal coefficients). Hence, $R[S]$ has a nonempty basis ($Y$ for instance) and thus a free $R-$module. So $R[S]$ is projective. $\blacksquare$
I know that the univariate polynomial ring $R[x]$ is a free $R-$module (by a somewhat similar argument), I am just wondering if I phrased all of this rigorously.