As is well known, the global dimension of $k[x_1,...,x_n]$, where $k$ is a field, is $n$ . Similarly, the projective dimension of $k$ as a module over $k[x_1,...,x_n]$ is also $n$ .
However, in the case of infinitely many indeterminates, what happens? Does the global dimension of $k[x_1,...,x_n,...]$ become infinite? And, is the projective dimension of k over $k[x_1,...,x_n,...]$ also infinite? How can this be demonstrated?
@Mariano Su'arez-'Alvarez already gave the answer, so let me explain the details and get this question out of the unanswered list. To prove that the global dimension of $R:=k[x_1,\dots,x_n,\dots]$ is infinite, it suffices to show that there are Tor-modules $\operatorname{Tor}_*^R(M,N)$ which is nonvanishing for $*\geq 0$ for arbitraruly large $*$. So it suffices for the question to show $\operatorname{Tor}_*^R(M,M)\neq 0$ for all $*>0$ where $M:=R/(x_1,x_2,\dots)$.
The Koszul complex of $M$ is a free resolution defined as follows: $$ \dots \to \wedge^r R^\infty\to \wedge^{r-1}R^\infty\to \dots\to \wedge^2 R^\infty\to R^\infty \to R\to 0 $$ where the first RHS map is given by $(x_1,x_2,\dots)$ and the differentials $\wedge^r R\to \wedge^{r-1}R$ are defined by $\sum_{i=1}^r (-1)^{k+1} x_i(e_1\wedge e_2\wedge\dots\wedge \widehat{e_i} \wedge\dots \wedge e_r)$. Now if you tensor $\otimes_R M$, all of these differentials are killed.
After tensoring the sequence by $M$ and taking homology, it is clear that $\operatorname{Tor}^R_r(M,M)\cong M\otimes_R \bigwedge^r R^\infty$. which is nonzero.
Remark: It is worth noting that this same method is how one might show $\operatorname{gldim}(k[x_1,\dots,x_n])\geq n$.