\( \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\C}{\mathbb{C}} \)

Inspired by Ravi Vakil’s summer project about algebraic geometry I wanted to make this blog post about the spectrum of polynomial rings and how to conceptualize them geometrically when your field isn’t algebraically closed.

The spectrum of a ring $R$, denoted $\Spec R$, is the set of all prime ideals of $R$ (with some additional structure). We think of a $\Spec R$ as a geometric space where the points are the maximal ideals, and the prime ideals contained in a maximal ideal represent geometric objects containing the point.

As an example, think of $\Spec \C[x, y]$: the maximal ideals are of the form $(x-a, y-b)$, and we can think of this as a point $(a,b) \in \C^2$. The other prime ideals of $\Spec \C[x, y]$ are the irreducible polynomials, and $(0)$. We can think of the irreducible polynomials as “curves” in $\C^2$, for example we think of $(y - x^2)$ as the curve where $y=x^2$. You can verify that $(x-a, y-b)$ contains $(y - x^2)$ if and only if $b=a^2$, so this matches our intuition for the geometry of $\C^2$. Lastly since every ideal contains $(0)$ it represents the entire plane.

In general when $k$ is an algebraically closed field the points of $\Spec k[x_1, \cdots, x_n]$ are exactly the points of $k^n$. Let’s warm up by proving this fact.

Hilbert’s weak nullstellensatz

Noether’s normalization lemma states that if $A$ is a finitely generated commutative $k$-algebra, then there is a finitely generated subalgebra $B \subseteq A$ such that $A$ is a finite $B$-module, and $B$ is a free commutative $k$-algebra. In other words $B$ is isomorphic to a polynomial ring in a finite amount of variables. I will assume this as known, but a proof is available at for example Wikipedia.

We can use Noether’s normalization lemma to understand the maximal ideals of $A := k[x_1, \cdots, x_n]$. Say $\mathfrak{m}$ is a maximal ideal of $A$. Then $A/\mathfrak{m}$ is a finitely generated $k$-algebra which is also a field. Applying Noether normalization we see that $A/\mathfrak{m}$ has a free subalgebra $B$. Let $B_{(0)}$ be the field of fractions of $B$ contained in $A/\mathfrak{m}$. Since finitely generated algebras over $k$ are Noetherian and $A/\mathfrak{m}$ is a finitely generated $B$-module, it follows that $B_{(0)}$ is a finitely generated $B$-module. This must imply that $B=k$, because for all other free $k$-algebras there field of fractions are infinitely generated. In conclusion $A/\mathfrak{m}$ must be a finite field extension of $k$.

If we further assume that $k$ is algebraically closed the only finite extension of $k$ is $k$ itself, so $\mathfrak{m}$ is the kernel of a $k$-algebra homomorphism $\varphi: A \to k$. Then it is clear that $\mathfrak{m} = \big(x_1 - \varphi(x_1), \cdots, x_n - \varphi(x_n)\big)$. This allows us to associate the maximal ideals of $A$ with points in $k^n$.

Galois orbits

We turn our attention to what happens when $k$ is not algebraically closed. $A/\mathfrak{m}$ is still a finite extension of $k$, and since finite extensions are algebraic it embeds into $K$, the algebraic closure of $k$. In other words $\mathfrak{m}$ is the kernel of a $k$-algebra homomorphism $A \to K$. So it seems like we can associate the maximal ideals with points in $K^n$, but unlike when $k$ was algebraically closed there can be several homomorphisms with the same kernel, so this association is not unique. For example if we post-compose with any $k$-algebra automorphism of $K$ we would get a different homomorphism with the same kernel. In fact we will see that this is the only thing you can do.

Theorem: If $\varphi, \psi : A \to K$ are $k$-algebra homomorphisms from a polynomial algebra to the algebraic closure of $k$ with the same kernel then there is a $k$-algebra automorphism of $\alpha: K \to K$ such that $\varphi = \alpha \psi$.

proof: Let $E$ be the image of $\psi$ and $E'$ the image of $\varphi$. Since they are both isomorphic to $A/\text{ker}\varphi$ there is an isomorphism $E \to E'$. Let $\tilde{\alpha}$ be the composition $E \to E' \hookrightarrow K$ of the isomorphism with the inclusion of $E'$. Then I claim $\tilde{\alpha}$ extends to a $k$-algebra automorphism of $K$. The proof uses Zorn’s lemma.

Let $P := \left\lbrace (F, \gamma) \mid E \subseteq F,\; \gamma:F \to K,\; \gamma|_E = \tilde{\alpha} \right\rbrace$ be the set of extensions of $\tilde{\alpha}$ to field extensions of $E$. Define a partial order on $P$ by $(F, \gamma) \leq (F', \gamma')$ if $F \subseteq F'$ and $\gamma'|_F = \gamma$. $P$ is clearly non-empty since it contains $E$, further if $\big((F_i, \gamma_i)\big)_{i\in I}$ is a chain in $P$ then $(F, \gamma)$ is an upper bound with $F = \bigcup\limits_{i\in I} F_i$ and $\gamma(x) = \gamma_i(x)$ whenever $x$ is in $F_i$. Now applying Zorn’s lemma we get that $P$ has a maximal element $(F, \alpha)$. I claim that $F = K$ and that $\alpha$ is an automorphism.

Let $a$ be any element of $K$. Since $F \subseteq K$ is an algebraic extension $a$ has a minimal polynomial $p(x)$ with coefficients in $F$. We have that $F(a) \simeq F[x]/(p(x))$. Let $F'$ denote the image of $\alpha$. Since $\alpha$ is a homomorphism of fields it is an isomorphism onto its image and thus induces an isomorphism of rings $\hat{\alpha}:F[x] \to F'[x]$. Let $a'$ be a root of $\hat{\alpha}(p(x))$, then $F'(a') \simeq F'[x]/\hat{\alpha}(p(x)) \simeq F[x]/(p(x)) \simeq F(a)$. Thus we can extend $\alpha$ to $F(a)$ by mapping $a$ to $a'$. Since $(F, \alpha)$ is maximal this implies that $F=K$. And since $F'\simeq F = K$ is algebraically closed and $K$ doesn’t have any proper algebraic subfields $\alpha$ is surjective. Therefor $\alpha$ is an automorphism such that $\varphi = \alpha \psi$. □

So what does this mean? If we let $G$ be the group of $k$-algebra automorphisms of $K = \bar{k}$, and we let $G$ act diagonally on $K^n$. Then the maximal ideals of $k[x_1, \cdots, x_n]$ are in bijection with the orbits of $K^n$. These orbits are called Galois orbits (even though there is no assumption that the extension $k \to K$ is Galois).

Geometric pictures

Let’s use this knowledge to understand $\Spec \mathbb{R}[x]$. The real automorphisms of $\C$ are the identity and complex conjugation, so $\Spec \mathbb{R}[x]$ looks like $\C$ modulo conjugation. It’s the complex plane and fold it along $\mathbb{R}$, leaving us with the upper half plane $\C^\geq$.

If we add more variables and look at for example $\Spec \mathbb{R}[x, y]$ we get $\C^> \times \C \cup \mathbb{R} \times \C^\geq$, where $\C^>$ is the strict upper half plane (i.e. not including $\mathbb{R}$). Since I have not yet learned to draw in four dimensions I leave you with this illustration of the imaginary component.

The maximal ideal corresponding to a point $(z=a+bi, w=c+di) \in \C^> \times \C$ is $\big((x-z)(x-\bar{z}), (xd - yb - (ad-bc))\big)$, while the maximal ideal corresponding to $(r, w) \in \mathbb{R} \times \C^\geq$ is $\big((x-r), (y-w)\big)$ if $w$ is real and $\big((x-r), (y-w)(y-\bar{w})\big)$ if it is not. So the maximal ideals of $\mathbb{R}[x, y]$ are those generated by an irreducible polynomial in one of the variables and a linear polynomial in both/the other variable.

• Algerbaic Geometry
• Polynomial rings