# Why is the Rees Algebra Noetherian if the underlying ring is?

Let $R$ be a commutative ring with $1$, $I \subset R$ a proper ideal. The Rees Algebra, with respect to $I$, is defined: $R[It]= \bigoplus_{n=0}^\infty I^nt^n \subseteq R[t]$. In many places I've read that if $R$ is Noetherian, then $R[It]$ is also Noetherian, but I can't find a proof of this anywhere. Can someone provide me with a link to a proof (or just a proof in general) of this? At the surface I see no reason for it to be true, $R[It]$ is just a subring of $R[t]$. $R[t]$ is Noetherian by the Hilbert basis theorem, but there is no guarantee that every subring of a Noetherian ring is Noetherian.

If $S=\{i_1,\dots,i_n\}$ generates $I$, then $R[It]$ is generated as an $R$-algebra by the elements $i_1t,\dots,i_nt$ (any element of $I^n$ is an $R$-linear combination of $n$-fold products of elements of $S$, so any element of $I^nt^n$ is an $R$-linear combination of $n$-fold products of elements of $St$). In particular, $R[It]$ is a finitely generated $R$-algebra, so since $R$ is Noetherian $R[It]$ is Noetherian by the Hilbert basis theorem.