Past exam of the commutative algebra course of the University of Oxford / 2017 / 1 / a

A commutative ring $R$ is a Noetherian ring if every ideal has a finite generating set of an ideal. Equivalently, it satisfies the ascending chain condition on ideals. Indeed, the union of an ascending chain is an ideal, and finitely many generators all belong to one term, after which the chain stabilizes. Conversely, an ideal without a finite generating set of an ideal would allow us to choose successively $a_{n+1}\notin (a_1,\dots ,a_n)$, producing a strictly ascending chain of ideals.

Let $\varphi :R\to S$ be a surjective ring homomorphism, where $R$ is a Noetherian ring. For any ideal $J$ of $S$, its inverse image is an ideal of $R$, say ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$. Taking images givesThus homomorphic images of Noetherian rings are Noetherian. This also applies to the image of an arbitrary ring homomorphism, by taking its image as the codomain.

The Hilbert basis theorem states that if $R$ is a Noetherian ring, then the polynomial ring $R[t]$ is a Noetherian ring. Repeating this gives the result for finitely many variables.

To prove it, fix an ideal $K\subseteq R[t]$. For $n\ge 0$, define the coefficient ideals of a polynomial idealHere the zero polynomial is allowed. Each $A_n$ is an ideal: addition and multiplication by constants preserve the required degree bound. Multiplying a representing polynomial by $t$ gives $A_n\subseteq A_{n+1}$. The ascending chain condition therefore supplies $N$ such that $A_n=A_N$ for every $n\ge N$.

For each $0\le n\le N$, choose a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$ for $A_n$, omitting zero generators, and choose $f_{nj}\in K$ of polynomial degree $n$ with leading coefficient $a_{nj}$. If $A_n=0$, choose no such polynomials. We claim that the finitely many $f_{nj}$ generate $K$.

Take a nonzero $f\in K$ of polynomial degree $d$. Put $n=\min (d,N)$. Its leading coefficient belongs to $A_d=A_n$, so write it as ${\sum }_j{b}_j{a}_{nj}$. The polynomialbelongs to $K$ and is either zero or has polynomial degree strictly below $d$. Induction on polynomial degree now expresses $f$ as a combination of the chosen $f_{nj}$; when $d=0$, the displayed difference is zero. Hence $K$ is finitely generated. Since $K$ was arbitrary, $R[t]$ is a Noetherian ring.