Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 101 / 1

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.

The induced map on the spectrum of a commutative ring is contraction of an ideal:To see that it is well defined, let $\mathfrak{q}$ be a prime ideal of $S$. Its inverse image is an ideal of $R$, and it is proper because $\phi (1)=1\notin \mathfrak{q}$. If $ab\in {\phi }^{-1}(\mathfrak{q})$, then $\phi (a)\phi (b)\in \mathfrak{q}$, whence $\phi (a)\in \mathfrak{q}$ or $\phi (b)\in \mathfrak{q}$. This proves that the inverse image is a prime ideal.

For any ideal $I\subseteq R$, its extension of an ideal to $S$ is $IS=\phi (I)S$. We haveThus inverse images of Zariski-closed sets are Zariski-closed sets, which proves that $f$ is a continuous map.

We first establish the nilradical as the intersection of prime ideals:Every nilpotent element belongs to every prime ideal. In the other direction, if $a$ is not nilpotent, the multiplicative subset $T=\{1,a,a^2,\dots \}$ avoids zero. By the Zorn lemma, there is an ideal $J$ maximal among those disjoint from $T$. It is a prime ideal: if $uv\in J$ and neither $u$ nor $v$ belongs to $J$, then $J+(u)$ and $J+(v)$ each meet $T$. Multiplying elements in these two intersections gives an element of $T$ lying in $J$, a contradiction. Since $a\notin J$, this proves the identity. The identity also holds for the zero ring, with the intersection over an empty family interpreted as the whole ring.

A Zariski-closed set $V_R(I)$ contains the image of $f$ exactly whenThe last equality follows because $\phi (r)$ is a nilpotent element exactly when $r^n\in \ker \phi$ for some $n\ge 1$. Consequently the smallest Zariski-closed set containing the image, namely its closure, isThis is all of $\mathrm{Spec}R$ precisely when the kernel of a ring homomorphism is contained in every prime ideal of $R$. Hence

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.