Hilbert's Nullstellensatz and Splitting Fields

Fulton Algebraic Curves

We aim to prove the following theorems:

(Hilbert’s Nullstellensatz) Let $k$ be an algebraically closed field. For any ideal $I\subset k[X_1,\dots ,X_n]$, $I(V(I))=\mathrm{Rad}(I)$.

(Weak Nullstellensatz) Let $k$ be an algebraically closed field. For any proper ideal $I\subset k[X_1,\dots ,X_n]$, $V(I)$ is nonempty.

Hilbert’s Nullstellensatz follows from the Weak Nullstellensatz by the following theorem:

Theorem 1. Hilbert’s Nullstellensatz follows from the Weak Nullstellensatz.

Proof) For notational convenience, take $L=k[X_1,\dots ,X_n]$. Let $I$ be an arbitrary ideal in $L$. As $L$ is Noetherian, $I$ is finitely generated; let $I=(f_1,f_2,\dots ,f_l)$. It suffices to show that $I(V(I))\subset \mathrm{Rad}(I)$; in other words, it suffices to show that if some polynomial $G\in L$ vanishes whenever $f_1,\dots ,f_l$ all do, then there is some $m>0$ and $a^l\in L$ such that $G^m=f_i{a}^i$.

Consider the ideal $J=(f_1(X_1,\dots ,X_n),f_2(X_1,\dots ,X_n),\dots ,f_l(X_1,\dots ,X_n),G(X_1,\dots ,X_n)X_{n+1}-1)$ in $k[X_1,\dots ,X_{n+1}]=L[X_{n+1}]$. $V(J)$ must be empty as whenever $f_1,\dots ,f_l$ all vanish, by the definition of $G$, $G{X}_{n+1}-1=-1\ne 0$. By the Weak Nullstellensatz, this implies $J$ is the entire field, and $1\in J$. Therefore, there are polynomials $b^1,\dots ,b^{l+1}\in L[X_{n+1}]$ such that $b^i{f}_i(X_1,\dots ,X_n)+b^{l+1}(G{X}_{n+1}-1)=1$. We now interpret this identity in a larger ring $L[X_{n+1},1/X_{n+1}]$ where we let $Y=1/X_{n+1}$ be the multiplicative inverse of $X_{n+1}$. One can multiply the identity by a sufficiently large power of $Y$ to obtain a new identity $c^i{f}_i(X_1,\dots ,X_n)+c^{l+1}(G-Y)=Y^N$ where $c_i\in L[Y]$; we now restrict our attention back to $L[Y]$ as the identity resides within this subring. Consider the evaluation map $\varphi :L[Y]\to L$ where $\varphi (f)=f(G)$. Then, $c^i{f}_i(X_1,\dots ,X_n)=\varphi (c^i{f}_i(X_1,\dots ,X_n)+c^{l+1}(G-Y))=\varphi (Y^N)=G^N$ in $L$. We get $G\in \mathrm{Rad}(I)$, as desired.

Thus, we focus our efforts on proving the Weak Nullstellensatz. Make the following observations:

Theorem 2 (Determinant in Arbitrary Fields). Let $F$ be a field. For any $f_{11},f_{12},\dots ,f_{nn}\in F$, there are elements $g^1,\dots ,g^n\in F$ such that $f_{ji}{g}^i=0$ for all $1\le j\le n$ iff $\det F\ne 0$ where $F$ is the matrix with $f_{ji}$ in the $i$th column and $j$th row.

Proof) One can check that the determinant function in fields satisfies its normal axioms, namely that is row-multilinear, row exchanges change the sign of the determinant, and the identity matrix has determinant $1$. Thus, the standard proof in linear algebra easily transfers over to fields.

Theorem 3. Let $R$ be a subring of a domain $S$, $v\in S$. Then the following are equivalent. (1) $v$ is integral over $R$. (2) $R[v]$ is module-finite over $R$. (3) There is a subring $R^{\prime }$ of $S$ containing $R[v]$ such that $R^{\prime }$ is module-finite over $R$.

Proof) (1) implies (2), as if there exists $r_0,\dots ,r_{l-1}\in R$ such that $r_i{v}^i+v^l=0$ in $S$, then ${\sum }_{i=0}^{i=l-1}R{v}^i=R[v]$. Therefore, $R[v]$ is module-finite over $R$. (2) clearly implies (3), as seen when one takes $R^{\prime }=R[v]$. We now show that (3) implies (1).

Assume $R^{\prime }={\sum }_{i=1}^{k}R{w}^i$ is a subring containing $R[v]$, where all the $w^i$s are nonzero. Then, for any $w^i$, there are coefficients $r_{ij}\in R$ for $1\le j\le k$ such that $v{w}^i=r_{ij}{w}^j$; this can be rewritten as $(r_{ij}-{\delta }_i^{j}v)w^j=0$. We find that $w^1,w^2,\dots ,w^k$ is a solution to $Mx=0$ where $M$ is the matrix such that $M_{ij}=r_{ij}-{\delta }_i^{j}v$. As $R^{\prime }$ is the subring of an integral domain, it itself is an integral domain and thus has a field of fractions $K$. Applying Theorem 2 in $K$, we get that $\det M=0$. In the matrix $M$, $v$ only appears on the diagonals. Thus, computing the determinant gives an expression $v^k+v^{k-1}{n}_{k-1}+\dots +n_0=0$ where $n_i\in R$. Therefore, $v$ is integral over $R$, and we are done.

Corollaries) Given Theorem 3 the following observations can be made (1) Say $R$ is a subring of $S$ and $S$ is a subring of a domain $T$. If $S$ is integral over $R$ and $T$ is integral over $S$, then $T$ is integral over $R$. (2) If a domain $S$ is ring-finite over a ring $R$, $S$ is module-finite over $R$ iff $S$ is integral over $R$. (3) An algebraically closed field has no module-finite field extensions except itself. (4) The set of elements of the (domain) $S$ integral over a subring $R$ is a subring of $S$.

Theorem 4. If a field $L$ is ring-finite over a subfield $K$, then $L$ is module-finite over $K$.

Proof) Let $L=K[e_1,\dots ,e_n]$; we use induction on $n$. First consider the case where $L=K[e_1]=K(e_1)$; we must show $L$ is module-finite over $K$. Consider the evaluation ring homomorphism $\varphi :K[X]\to K[e_1]$ where $f\mapsto f(e_1)$; as the image $K[e_1]$ is a field and $K[x]$ is a PID, $\ker \varphi$ is of the form $(g)$ where $g\in K[X]$ is irreducible. We can assume $g$ is monomial. As $g$ satisfies $g(e_1)=0$, $e_1$ is integral over $K$ and $L$ is module-finite over $K$.

Now assume the $n-1$ case holds and consider the $n$ case. Note that $L=K[e_1,\dots ,e_n]=K(e_1)[e_2,\dots ,e_n]$. By induction, if we take $W=K(e_1)$, $L={\sum }_{i=1}^{m}W{l}_i$. If $e_1$ is integral over $K$, we are done by corollary 1 of Theorem 3; assume otherwise. Each $l_i$ satisfies some equation of the form $l_i^{d_i}+c_{i,d_i-1}{l}_i^{d_i-1}+c_{i,0}=0$ where $c_{ij}\in W$. Thus, for each $l_i$, there is some polynomial $g_i\in K[e_1]$ such that $g_i{c}_{i,j}\in K[e_1]$ for all $j$; then, $g_i{l}_i=k_i$ satisfies $k_i^{d_l}+(g_i{c}_{i,d_i-1})k_i^{d_l-1}+\dots +(g_i^{d_i}{c}_{i,0})=0$, and therefore $k_i$ is integral over $K[e_1]$. As the elements integral over $K[e_1]$ is a subring of $L$ (by corollary 4 of Theorem 3), if we take $G=g_1{g}_2\dots g_m\in K[e_1]$ we have that $G{l}_i$ is integral over $K[e_1]$ for all $i$. This implies that for any $l\in L$, $Gl$ is integral over $K[e_1]$.

We focus our attention on $K(e_1)\subset L$. Our previous argument supplies some $G\in K[e_1]$ such that for any $l\in K(e_1)$, $Gl$ is integral over $K[e_1]$. Note that any polynomial integral over $K[e_1]$ must be in $K(e_1)$. Thus, $G$ must be such that, for any $l\in K(e_1)$, $Gl\in K[e_1]$. This is obviously impossible, completing the proof.

Theorem 5. Let $k$ be an algebraically closed field. For any proper ideal $I\subset k[X_1,\dots ,X_n]$, $V(I)$ is nonempty.

Proof) As $k[X_1,\dots ,X_n]$ is Noetherian, we may assume $I$ is maximal. We prove that $I$ must be of the form $(X_1-a_1,X_2-a_2,\dots ,X_n-a_n)$ for some $(a_1,\dots ,a_n)\in {\mathbb{A}}^n(k)$, in which case $V(I)=(a_1,\dots ,a_n)$.

Note that $J=k[X_1,\dots ,X_n]/I$ must be a field with $k$ as a subfield. $J$ must also be ring-finite over $k$, and by Theorem 4 must be module-finite. Thus, $J$ must be integral over $k$. However, as $k$ is algebraically closed, this is only possible if $J=k$. Thus, each $X_i$ has some $a_i\in K$ such that $X_i-a_i\in I$, implying $(X_1-a_1,X_2-a_2,\dots ,X_n-a_n)\subset I$. As the LHS is a maximal ideal, $I=(X_1-a_1,X_2-a_2,\dots ,X_n-a_n)$, as desired.

Remark: The Nullstellensatz provides a correspondence between algebraic sets and radical ideals in algebraically closed fields.

We end with some brief theorems on splitting fields.

Theorem 6. Let $K$ be a field and $F\in K[X]$ be monic. Show that there is a field $L$ containing $K$ such that $F={\prod }_{i=1}^n(X-x_i)\in L[X]$. $L$ is called the splitting field of $F$.

Proof) As $K[X]$ is a PID, it suffices to show this assuming $F$ is irreducible and monic. We use induction on the degree of $F$; the $\mathrm{deg}F=1$ case is trivial. Consider the maximal ideal $(F)$ of $K[X]$ and the field $L=K[X]/(F)$; if we denote the equivalence class of $f$ by $\bar{f}$, one can observe that $\bar{X}$ is a root of $F$ in $L$. Thus, in $L[Y]$, $F$ can be further factorized at least once by $(Y-\bar{X})$, representing it as a product of polynomials at least one degree lower in the field extension $L$ of $F$. The inductive assumption completes the proof.

Theorem 7. Suppose $K$ is a field of characteristic zero, $F$ is an irreducible monic polynomial in $K[X]$ of degree $n>0$. Let $L$ be a splitting field of $F$, so $F={\prod }_{i=1}^n(X-x_i)$, $x_i\in L$. Show that the $X_i$ are distinct.

Proof) First, suppose $W$ is a field extension of $K$ such that there is some $y\in W$ where $F(y)=0$. The ring homomorphism $\varphi :K[X]\to W$ where $f\mapsto f(y)$ induces an isomorphism between $K[X]/\ker \varphi$ and $K[y]=K(y)$, and thus $\ker \varphi$ is a maximal ideal, forcing $\ker \varphi =(F)$. Thus, for any $G\in K[X]$ such that $G(y)=0$, $F|G$.

Now, FTSOC assume $(X-x_1{)}^2|F$ in $L[X]$. Consider the derivative $G=F_{x_1}$; in $L[X]$, $X-x_1$ still divides $G$, so therefore $G(x_1)=0$. The above argument then implies that $F|g=F_{x_1}$, which is impossible. We are done.