If $R$ is noetherian, show that the polynomial ring of infinite variables $R[x_1,x_2,…]$ is coherent, i.e. every finitely generated ideal is finitely presented.
I don't really know how to get started. I tried to use that over a noetherian ring any finitely generated module is also finitely presented. But a finitely generated ideal over the polynomial ring may not be finitely generated over $R$. Any help is appreciated.
Best Answer
Ok, here's a sketch.
First note that $\varinjlim R[x_1,x_2,\ldots,x_j]=R[x_1,x_2,\ldots,]$. Also note that the direct limit is an exact functor (see the exercises of Atiyah-MacDonald for example).
Let $M$ be a finitely generated ideal of $R[x_1,x_2,\ldots]$.
It is generated by some $f_1,\ldots,f_n$, which all lie in some $R[x_1,\ldots,x_N]$ for some big $N$.
As an ideal in $R[x_1,\ldots,x_N]$, $\langle f_1,\ldots,f_n \rangle$ is finitely presented. Thus we have an exact sequence
$$ R[x_1,\ldots,x_N]^m \to R[x_1,\ldots,x_N]^n \to \langle f_1,\ldots,f_n \rangle \to 0. $$ of $R[x_1,\ldots,x_N]$-modules.
Now apply $\varinjlim (-)$ to get an exact sequence $$ \varinjlim R[x_1,\ldots,x_N]^m \to \varinjlim R[x_1,\ldots,x_N]^n \to \varinjlim \langle f_1,\ldots,f_n \rangle \to 0. $$ which is equal to $$ R[x_1,x_2,\ldots]^m \to R[x_1,x_2,\ldots]^n \to \langle f_1,\ldots,f_n \rangle=M \to 0 $$ as $R[x_1,x_2,\ldots,]$-modules. The above exact sequence says that $M$ is finitely presented.