Hilbert Nullstellensatz

For an algebraically closed field $k$ and an ideal $I\subseteq k[x_1,\dots ,x_n]$, the strong Hilbert Nullstellensatz isIts weak form says that an ideal with empty affine zero set is the unit ideal. Equivalently, if finitely many polynomials have no common zero, a polynomial combination of them equals one.

For a finitely generated algebra over an algebraically closed field $k$, radical ideals correspond, with inclusion reversed, to closed subsets of the maximal spectrum. If $f\notin \sqrt{I}$, the nonzero algebra $(A/I)[1/f]$ is still finitely generated over $k$. Its maximal residue field is $k$ by the Zariski lemma, so contracting a maximal ideal gives a maximal ideal of $A$ containing $I$ and avoiding $f$. Hence the intersection of all maximal ideals containing $I$ is $\sqrt{I}$. This proves the inverse correspondence, including the empty closed set and unit ideal.

For an algebraically closed field $k$, every maximal ideal of $k[x_1,\dots ,x_n]$ has the formfor a unique point $a\in k^n$. Equivalently, every proper ideal has a common zero.

If a field $L$ is finitely generated as an algebra over a subfield $k$, then $L$ is a finite algebraic extension of $k$. Applying this to a residue field of a polynomial ring proves the Weak Hilbert Nullstellensatz.

For an ideal $I\subseteq k[x_1,\dots ,x_n]$ over an algebraically closed field,

Let $I,J\subseteq \mathbb{Z}[x_1,\dots ,x_n]$. If $V_{\mathbb{C}}(I)\subseteq V_{\mathbb{C}}(J)$, thenfor all but finitely many primes $p$. The Nullstellensatz puts a power of each generator of $J$ in $I\mathbb{Q}[x_1,\dots ,x_n]$; clearing denominators leaves only finitely many exceptional characteristics.

To prove the Strong Hilbert Nullstellensatz from its weak form, adjoin a variable $y$ and the equation $1-yf$. If $f$ vanishes on $V(I)$, the ideal $(I,1-yf)$ has no common zero and is therefore the unit ideal. Substituting $y=f^{-1}$ and clearing denominators yields $f^N\in I$ for some $N$.

A complex affine algebraic set is either finite or has the cardinality of $\mathbb{C}$. One proof takes a positive-dimensional irreducible component and applies Noether normalization: the resulting finite dominant map to a positive-dimensional affine space has a point above every closed point.

Let $I\subseteq k[x_0,\dots ,x_n]$ be a homogeneous ideal over an algebraically closed field, and letbe the irrelevant ideal. Then $V_{+}(I)=\varnothing$ exactly when either $I=(1)$ or $\sqrt{I}=\mathfrak{m}$. For a proper homogeneous ideal this is also equivalent to ${\mathfrak{m}}^N\subseteq I$ for some $N$.

It is the homogeneous projective version of the Hilbert Nullstellensatz.

For homogeneous fiber equations $f_i$ of fixed degrees $d_i$ in $S=k[y_0,\dots ,y_n]$, put $S_j=0$ for $j<0$. The linear map ${\mu }_N((g_i{)}_i)={\sum }_i{g}_i{f}_i$ has image the degree-$N$ component of their homogeneous ideal. Hence it is surjective exactly when that ideal contains $(y_0,\dots ,y_n{)}^N$: the degree-$N$ monomials generate this power. In a family, the matrix entries are regular functions on the base. Failure of surjectivity is therefore a closed determinantal variety, cut out by the maximal-size matrix minors.

In the standard graded homogeneous coordinate ring $S=k[x_0,\dots ,x_n]$ of ${\mathbb{P}}^n$, the irrelevant ideal isIt vanishes only at the origin of the corresponding affine cone, which does not represent a projective point.

It is the polynomial-ring instance of the irrelevant ideal of a graded ring.

Every open cover of an affine variety has a finite subcover. Refine it by distinguished opens $D(f_i)$. Their complements have no common point, so the Nullstellensatz writes $1$ as a finite combination of the $f_i$; the corresponding finite family of distinguished opens covers.