Past exam of the commutative algebra course of the University of Oxford / 2023 / 2

The image $f/1$ is a unit in the localization at a prime ideal $R_{\mathfrak{p}}$, because $f\notin \mathfrak{p}$. The universal property of localization, applied to the ring homomorphism $R\to R_{\mathfrak{p}}$ and the multiplicative subset $\{1,f,f^2,\dots \}$, gives a unique ring homomorphism fixing the image of $R$:Fixing the image of $R$ is exactly the required compatibility with the algebra over a ring structures. Uniqueness also follows directly: such a ring homomorphism must send $1/f$ to the inverse of $f/1$, and these elements together with $R$ generate $R_f$.

Choose $f=1$, so that $R_f=R$ and $f\notin \mathfrak{p}$. For an integral domain, the canonical ring homomorphism $R\to R_{\mathfrak{p}}$ is injective: if $a/1=0$, the vanishing criterion in a module localization gives $sa=0$ for some $s\notin \mathfrak{p}$. Such an $s$ is nonzero, so the absence of zero divisors gives $a=0$. Thus

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

For $n\ge 1$, let $K_n=\ker ({\varphi }^n)$, where the power denotes repeated composition of the ring endomorphism. These ideals form an ascending chain:The ascending chain condition for a Noetherian ring gives $K_N=K_{N+1}$ for some $N\ge 1$. Take $a\in \ker \varphi$. Since $\varphi$ is surjective, its iterate ${\varphi }^N$ is surjective, so choose $b\in R$ with ${\varphi }^N(b)=a$. Thenso $b\in K_{N+1}=K_N$. Therefore $a={\varphi }^N(b)=0$.

Thus the kernel of a ring homomorphism $\varphi$ is zero. This proves that a surjective endomorphism of a Noetherian ring is an automorphism, and in particular

Use the direct product of ringswith coordinatewise addition and multiplication and unit $(1,1,\dots )$. Define the shiftCoordinatewise operations show that $g$ preserves addition, multiplication and the unit, so it is a ring endomorphism. It is surjective because $(0,b_0,b_1,\dots )$ maps to any prescribed $(b_0,b_1,\dots )$. Its kernel of a ring homomorphism isso it is not injective. This is the noninjective surjective shift on a product of fields.

This ring fails the Noetherian hypothesis in c. Indeed, let $e_n$ have a single $1$ in coordinate $n$. The ideals $I_n=(e_0,\dots ,e_n)$ form the strictly ascending chainsince $e_{n+1}\notin I_n$.