Affine schemes and Nullstellensatz
(These are the notes I made for a talk at Junior Logic Seminar at Oxford.)
The category of affine schemes is the category opposite to the category of commutative rings.
Let $A$ be a ring. One associates to it a topological space called $\mathrm{Spec} A$:
$$\begin{array}{l}\mathrm{Spec} A = \{ \textrm{prime ideals of } A \} \textrm{ with closed sets } \\ \quad \quad \quad X_\mathfrak{a} = \{\mathfrak{p} \subseteq \mathfrak{a} \mid \mathfrak{p} \in \mathrm{Spec} A\} \textrm{ for prime } \mathfrak{a}\in A\end{array}$$
Clearly, the points that correspond to maximal ideals are closed. If $\mathfrak{m} \subset A$ is a maximal ideal, $A/\mathfrak{m}$ is called the residue field of $\mathfrak{m}$. In classical algebraic geometry one deals with schemes over a field. Affine scheme over a field $ k$ is $ \mathrm{Spec} A$, where $ A=k[x_1,\ldots,x_n]/\mathfrak{a}$.
Proposition. Let $ A$ be the coordinate algebra of an affine scheme over $ k$. Let $ \mathrm{Max}_k\,A$ be the subset of maximal ideals with the residue field $ k$. Then $ \mathrm{Max}_k\,A\cong V(\mathfrak{a})=\{\bar{a}\in k^n \mid\forall f(\bar{x}) \in\mathfrak{a} f(\bar{a})=0\}$. Note that in general one cannot recover $ A$ from $ \mathrm{Spec} A$: consider $ k[x]/(x^2)$.
Theorem. Let $ K \supset k$ be a field extension, and $ K$ is finitely generated over $ k$ with generators $x_1, \ldots, x_n \in K$. Then $K$ is algebraic over $k$.
Proof. Regroup $x_1, \ldots, x_n$ so that $ K_0=k(x_1, \ldots, x_r)$ is a transcendental extension of $ k$ and $ K_{i-r}=K_0\langle x_{r+1}, \ldots, x_{i-1} \rangle$, $ x_i$ algebraic over $ K_{i-r}$ for $ i > r$. Let $ P_i(x)$-s be minimal polynomials of $ x_i$-s. They are polynomials over $ K_0$, i.e. have rational functions of $ x_1,\ldots,x_r $ as their coefficients, so there is a least common denominator of these, a polynomial $ g \in k[x_1,\ldots, x_r]$, such that $ P_i$ has coefficients in the ring $B=k \langle x_1,\ldots, x_r, g^{-1} \rangle$. Any element of the field $K$ has the form
$$ \sum b_\alpha\displaystyle\mathop{\prod}_{r+1 \leq e_i \leq d_i} x^{e_i}$$
where $\alpha$ is a multi-index, and $d_i$ are degrees of polynomials $P_i$, and $b_\alpha \in B$. From this one can see that if $I$ is a proper ideal of $ B$ (as a ring), then $ IK$ is a proper ideal of the field $ K$. Since $ I$ can only be trivial, $ B$ is a field. In other words it contains every rational function of $ x_1, \ldots, x_r, g^{-1}$, which cannot be true. ☐
Corollary. If $ A$ is finitely generated over an algebraically closed field then $ \mathrm{Max}_k A=\mathrm{Max} A$. ☐
Corollary. [Nullstellensatz] Let $ k$ be an a.c.f. and $ \mathfrak{a} \subset k[x_1,\ldots,x_n]$ be an ideal. Then $V(\mathfrak{a})$ is empty iff $ \mathfrak{a}=k[x_1,\ldots,x_n]$.
Model-theoretical proof of Nullstellensatz.
Theorem. [Nullstellensatz, weaker form. Together with Hilbert's basis theorem implies the version in the previous section] Let $ k$ be an algebraically closed field, $ I \subset k[x_1, \ldots,x_n]$ a proper ideal, generated by polynomials $ f_1, \ldots,f_m$. Then these polynomials have at least one common zero in $k^n$. Proof. First prove that
Claim. $ACF_p$ has quantifier elimination.
Suffices to construct a family of partial isomorphisms $ I$ between finitely generated substructures of two sufficiently saturated models $ ACF_p \models k_1, k_2$. $ I$ is non-empty: there is at least the partial isomorphism that identifies prime fields of $ k_1$ and $ k_2$. Construct the rest by induction. Let $ f$ be a partial isomorphism between substructures $ R_1 \subset k_1$, $ R_2 \subset k_2$ (remember, the language is just the language of rings) related by the family of partial isomorphisms. Let $ L_1$, $ L_2$ be the fraction fields of $ R_1$, $ R_2$. Pick an element $ a \in k_1-L_1$. There are two cases to consider:
- $ a$ is algebraic over $ L_1$ ; then pick the minimal polynomial of $ a$ and extend $ f$ by putting $ f(a)=b$ where $ b$ is a solution of the minimal polynomial of $ a$ in $ k_2$;
- $ a$ is transcendental over $ L_1$; one can choose any transcendental (over $ L_2$) element $ b$ of $ k_2$ and extend $ f$: $ f(a)=b$. The question is if there is any such element. Since any polynomial only has finitely many roots, the set of formulae $ \{ P(\bar{x}) \neq 0 \mid P \in L_2[\bar{x}]\}$ is finitely satisfiable (and is a type over a finite set, since $ L_2$ is finitely generated). By saturatedness (in fact $ \aleph_0$-saturatedness would saffice), it has a solution in $ L_2$.
Now take the smallest prime ideal $ J$ that contains $ I$. It is not the whole of $ k[\bar{x}]$ unless $ I$ is. Let $ L$ be the field of fractions of the quotient $ k[x_1, \ldots, x_n]/J$, then $ k$ canonically identifies with a substructure of $ K$. Clearly, $ f_i(\bar{x}) = 0$ in $ K$, and all the more in the algebraic closure of $ K$, $ \bar{K}$. By quantifier elimination, $ k$ is an elementary substructure of $ \bar{K}$ and so has a solution to $ f_1,\ldots,f_n$. ☐