The Nullstellensatz

Jacobson Rings

We have seen that for a ring \(A\) and an arbitrary ideal \(\mathfrak{a}\), the formula

\[\sqrt{\mathfrak{a}}=\bigcap_\text{\scriptsize$\mathfrak{p}$ prime containing $\mathfrak{a}$} \mathfrak{p}\]

holds. (§Properties of Localization, ⁋Corollary 8) In particular, if \(\mathfrak{p}\) is a prime ideal, then \(\mathfrak{p}=\sqrt{\mathfrak{p}}\) should naturally hold. More generally, we define the following.

Definition 1 For a ring \(A\), an arbitrary ideal \(\mathfrak{a}\) is called a radical ideal if \(\mathfrak{a}=\sqrt{\mathfrak{a}}\) holds.

Therefore, in short, the observation above says that every prime ideal is radical. The proof of this observation is more or less trivial; however, had we considered, instead of the intersection of the prime ideals containing \(\mathfrak{p}\), the intersection of the maximal ideals containing \(\mathfrak{p}\) in a manner similar to §Integral Extensions, §§Nakayama’s Lemma, this observation would not have been so trivial, and in fact it does not even hold. For instance, any local ring containing a prime ideal that is not maximal, such as \(\mathbb{Z}_{(2)}\), gives a counterexample.

Definition 2 A ring \(A\) is called a Jacobson ring if every prime ideal is an intersection of maximal ideals.

The following then holds.

Lemma 3 (Rabinowitch) For a ring \(A\), the following are equivalent.

1. \(A\) is a Jacobson ring.
2. For a prime ideal of \(A\), \(\mathfrak{p}\), if \((A/\mathfrak{p})[a^{-1}]\) is a field for some \(a\in A/\mathfrak{p}\), then \(A/\mathfrak{p}\) is a field.

Proof

First, suppose that \(A\) is Jacobson. Then it is also clear by definition that its quotient \(A/ \mathfrak{p}\) is Jacobson. Meanwhile, by [Algebraic Structures] §Field of Fractions, ⁋Proposition 9, \(A/\mathfrak{p}\) is an integral domain, and since \((0)\) is a prime ideal in an integral domain, we can express \((0)\) as an intersection of maximal ideals. Now, by §Localization, ⁋Proposition 8, there is a one-to-one correspondence between the prime ideals of \((A/\mathfrak{p})[a^{-1}]\) and the prime ideals of \(A/\mathfrak{p}\) not containing \(a\); since by assumption the only prime ideal of \((A/\mathfrak{p})[a^{-1}]\) is \(0\), the only prime ideal of \(A/\mathfrak{p}\) not containing \(a\) is also \(0\). In other words, every nonzero prime ideal of \(A/\mathfrak{p}\) must always contain \(a\). But if such a prime ideal exists, then \((0)\) is not a maximal ideal, so every maximal ideal of \(A/\mathfrak{p}\) is nonzero and therefore contains \(a\). Meanwhile,

\[(0)=\bigcap_\text{\scriptsize$\mathfrak{m}$ maximal} \mathfrak{m}\]

and hence \(a=0\), a contradiction.

Conversely, let us assume the second condition and prove the first. That is, we fix a prime ideal of \(A\), \(\mathfrak{p}\), and letting the intersection of all maximal ideals containing \(\mathfrak{p}\) be \(\mathfrak{P}\), we must show that \(\mathfrak{p}=\mathfrak{P}\). Suppose, contrary to the conclusion, that there exists an element \(a\in \mathfrak{P}\setminus \mathfrak{p}\). Then by [Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma), among the prime ideals containing \(\mathfrak{p}\) but not containing \(a\), there exists a maximal prime ideal \(\mathfrak{q}\). Since \(a\not\in \mathfrak{q}\) by definition, \(\mathfrak{q}\) is not a maximal ideal, and therefore \(A/\mathfrak{q}\) is not a field. However, in \(A[a^{-1}]\), \(\mathfrak{q}\) must be a maximal ideal by definition, which contradicts the second condition; hence we must have \(\mathfrak{p}=\mathfrak{P}\).

The Nullstellensatz

Now the Nullstellensatz can be stated as follows.

Theorem 4 Let a Jacobson ring \(A\) and a finitely generated \(A\)-algebra \(E\) be given. Then \(E\) is also a Jacobson ring. Moreover, if \(\mathfrak{n}\) is a maximal ideal of \(E\), then \(\mathfrak{m}=\mathfrak{n}\cap A\) is a maximal ideal of \(A\), and \(E/\mathfrak{n}\) is a finite field extension of \(A/\mathfrak{m}\).

Proof

We divide the proof into three steps.

1. First, consider the case where \(A=\mathbb{K}\) and \(E=\mathbb{K}[\x]\). Then \(E\) is a principal ideal domain; in particular, every nonzero prime ideal of \(E\) is generated by an irreducible monic polynomial. From this, since no nonzero prime ideal can be contained in another prime ideal, every nonzero prime ideal of \(E\) is maximal; such an ideal cannot contain \(1\in \mathbb{K}\), so its intersection with \(A=\mathbb{K}\) must be \((0)\). In this case, \(E/\mathfrak{n}\), whose dimension equals the degree of the irreducible polynomial defining \(\mathfrak{n}\), is a \(\mathbb{K}\)-vector space. Finally, to show that \((0)\) is the intersection of maximal ideals, it suffices to use the argument that \(E=\mathbb{K}[\x]\) has infinitely many irreducible polynomials, and since the degree of a polynomial is necessarily finite, the only polynomial having all of them as factors is \(0\). Here, the infinitude of irreducible polynomials in \(E\) follows by directly mimicking Euclid’s proof of the infinitude of primes.
2. For the next step, consider an arbitrary Jacobson ring \(A\) and an \(A\)-algebra \(E\) generated by a single element; to show that \(E\) is Jacobson, we verify that the second condition of Lemma 3 (Rabinowitch) holds. In other words, our goal in this step is to prove the following proposition.

   Let a Jacobson ring \(A\) be given, and let an \(A\)-algebra \(E\) generated by a single element be given. If, for a fixed prime ideal \(\mathfrak{q}\subseteq E\), \(E/\mathfrak{q}\) contains a nonzero \(x\in E/\mathfrak{q}\) such that \((E/\mathfrak{q})[x^{-1}]\) is a field, then \(E/\mathfrak{q}\) is also a field.

   Since \(E'=E/\mathfrak{q}\) is again an \(A\)-algebra generated by a single element, the proposition above amounts to showing the following.

   Let a Jacobson ring \(A\) be given, and suppose that an \(A\)-algebra \(E'\) generated by a single element is an integral domain. If \(E'\) contains a nonzero \(x\in E'\) such that \(E'[x^{-1}]\) is a field, then \(E'\) is also a field.

   In taking this quotient, \(A\) is replaced by \(A'=A/(A\cap \mathfrak{q})\), which is again a Jacobson ring; consequently, what we must ultimately show is the following proposition.

   Suppose that an integral domain \(A'\) is Jacobson, and that an \(A'\)-algebra \(E'\) generated by a single element is an integral domain containing \(A'\). If \(E'\) contains a nonzero \(x\in E'\) such that \(E'[x^{-1}]\) is a field, then \(E'\) is also a field.

   To this end, we show that under the above assumptions \(A'\) must be a field, and \(E'\) is a finite extension of \(A'\). Since in the above proposition \(E'\) is an \(A'\)-algebra generated by a single element, we may write \(E'=A'[\x]/\mathfrak{q}\). We first show that \(\mathfrak{q}\neq 0\). Suppose for contradiction that \(\mathfrak{q}=0\), and assume that there exists some \(x\in E'/(0)=A'[\x]\) such that \(E'[x^{-1}]=A'[\x][x^{-1}]\) is a field. Letting \(K'=\Frac(A')\), this assumption implies that \(K'[\x][x^{-1}]\) is also a field. But \(K'[\x]\) is Jacobson by the first result, so \(K'[\x]\) must be a field, which is a contradiction. Therefore \(\mathfrak{q}\neq 0\), and \(E'[x^{-1}]=K'[\x]/\mathfrak{q}K'[\x]\) is a finite-dimensional extension of \(K'\).
   Now suppose that \(p(\x)\in \mathfrak{q}\) satisfies in \(E'\) the equation

   \[p(\alpha)=p_n\alpha^n+\cdots+p_0=0\]

   where \(\alpha\) is the generator of \(E'\) as an \(A'\)-algebra. Then from the above equation, \(E'[p_n^{-1}]\) is an integral \(A'[p_n^{-1}]\)-algebra. Meanwhile, the element \(x\) defined above must also satisfy some polynomial

   \[q(x)=q_mx^m+\cdots+q_0=0\]

   and since \(E'\) is an integral domain, we may assume without loss of generality that \(q_0\neq 0\). Then from the monic polynomial

   \[\left(\frac{1}{x}\right)^m+\frac{q_1}{q_0}\left(\frac{1}{x}\right)^{m-1}+\cdots+\frac{q_m}{q_0}=0\]

   we see that \(E'[x^{-1}]\) is an integral \(A'[(p_nq_0)^{-1}]\)-algebra. Now by §Integral Extensions and Ideals, ⁋Corollary 3, \(A'[(p_nq_0)^{-1}]\) is a field, and since \(A'\) is Jacobson by assumption, Lemma 3 (Rabinowitch) implies that \(A'\) is a field. Therefore \(E'\) is an integral \(A'\)-algebra, and again by §Integral Extensions and Ideals, ⁋Corollary 3 we see that \(E'\) is a field.

3. The general case follows by induction on the number of generators using the second result.

In particular, consider the case where \(A=\mathbb{K}\) and \(E=\mathbb{K}[\x_1,\ldots, \x_n]\). Then, for any

\[a=(a_1,\ldots, a_n)\in \mathbb{K}^n\]

if we define the ideal \(\mathfrak{m}_a\) by the equation

\[\mathfrak{m}_a=(\x_1-a_1,\ldots, \x_n-a_n)\]

the isomorphism given by evaluation

\[\ev_a:\mathbb{K}[\x_1,\ldots, \x_n]/\mathfrak{m}_a\rightarrow \mathbb{K}\]

shows that \(\mathfrak{m}_a\) is a maximal ideal.

Moreover, if \(\mathbb{K}\) is an algebraically closed field, then every maximal ideal of \(E\) is of this form. First, for any maximal ideal of \(E\), say \(\mathfrak{n}\), \(E/\mathfrak{n}\) is an algebraic extension of \(\mathbb{K}/(\mathfrak{n}\cap \mathbb{K})=\mathbb{K}\); but if \(\mathbb{K}\) is algebraically closed, the only such extension is itself, so we must have \(E/\mathfrak{n}\cong \mathbb{K}\). Meanwhile, under the canonical surjection \(E \rightarrow E/\mathfrak{n}\cong \mathbb{K}\), if we let the image of each \(\x_i\) in \(\mathbb{K}\) be \(a_i\), then \(\mathfrak{m}_a\subseteq \mathfrak{n}\), and now from the maximality of \(\mathfrak{m}_a\) we obtain the desired result.

Therefore, from §Basic Notions, ⁋Proposition 11, we obtain the following.

Lemma 5 Let a field \(\mathbb{K}\) be given. Then \(\mathfrak{m}_a=(\x_1-a_1,\ldots, \x_n-a_n)\) is a maximal ideal of \(\mathbb{K}[\x_1,\ldots, \x_n]\). Moreover, if \(\mathbb{K}\) is algebraically closed, then between the maximal ideals of \(\mathbb{K}[\x_1,\ldots,\x_n]/(f_1,\ldots, f_r)\) and, satisfying the equation

\[f_1(x_1,\ldots, x_n)=\cdots=f_r(x_1,\ldots, x_n)=0\]

the \((x_1,\ldots, x_n)\), there is a one-to-one correspondence.

A somewhat more traditional version of the Nullstellensatz is also obtained from this. To state it, consider the function that takes an ideal of \(\mathbb{K}[\x_1,\ldots, \x_n]\), \(\mathfrak{a}\), and yields the subset of \(\mathbb{K}^n\), \(Z(\mathfrak{a})\),

\[Z(\mathfrak{a})=\{(a_1,\ldots, a_n)\in \mathbb{K}^n\mid \text{$f(a_1,\ldots, a_n)=0$ for all $f\in \mathfrak{a}$}\}\]

and the function that takes a subset of \(\mathbb{K}^n\), \(S\), and yields the subset of \(\mathbb{K}[\x_1,\ldots, \x_n]\)

\[I(S)=\{f\in \mathbb{K}[\x_1,\ldots, \x_n]\mid\text{$f(a_1,\ldots, a_n)=0$ for all $(a_1,\ldots, a_n)\in S$}\}\]

This is the function \(I\).

Proposition 6 Let an algebraically closed field \(\mathbb{K}\) and an ideal \(\mathfrak{a}\subseteq \mathbb{K}[\x_1,\ldots, \x_n]\) be given. Then

\[I(Z(\mathfrak{a}))=\sqrt{\mathfrak{a}}\]

holds.

Proof

From Lemma 5, we know that the elements of \(Z(\mathfrak{a})\) are in one-to-one correspondence with the maximal ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) containing \(\mathfrak{a}\). Hence \(I(Z(\mathfrak{a}))\) is the intersection of the maximal ideals containing \(\mathfrak{a}\) in \(\mathbb{K}[\x_1,\ldots, \x_n]\), and since \(\mathbb{K}[\x_1,\ldots, \x_n]\) is Jacobson by Theorem 4, this equals the intersection of the prime ideals containing \(\mathfrak{a}\) in \(\mathbb{K}[\x_1,\ldots, \x_n]\), which is precisely the right-hand side.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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