Nullstellensatz

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Jacobson Rings

For a ring \(A\) and an ideal \(\mathfrak{a}\), we saw that the following formula holds:

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

(§Properties of Localization, ⁋Corollary 8) In particular, if \(\mathfrak{a}\) is a prime ideal, it is clear that \(\mathfrak{p}=\sqrt{\mathfrak{p}}\) must hold. More generally, we make the following definition.

Definition 1 An ideal \(\mathfrak{a}\) of a ring \(A\) is called a radical ideal if \(\mathfrak{a}=\sqrt{\mathfrak{a}}\) holds.

Thus, the above observation says in short that any prime ideal is radical. The proof of this observation is somewhat trivial, but if we had considered the intersection of maximal ideals containing \(\mathfrak{p}\) instead of prime ideals containing \(\mathfrak{p}\), in a manner similar to §Integral Extension, §§Nakayama’s Lemma, then this observation would not have been so obvious, and in fact it would not hold. For example, any local ring containing a prime ideal that is not maximal, such as \(\mathbb{Z}_{(2)}\), is a counterexample.

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

Then the following holds.

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

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

Proof

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

\[(0)=\sqrt{(0)}=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime} \mathfrak{p}\]

and thus \(a=0\), which is a contradiction.

Conversely, assume the second condition and let us show the first. Fix a prime ideal \(\mathfrak{p}\) of \(A\), and let \(\mathfrak{P}\) be the intersection of all maximal ideals containing \(\mathfrak{p}\); we must show that \(\mathfrak{p}=\mathfrak{P}\). Suppose, for contradiction, that there exists an element \(a\in \mathfrak{P}\setminus \mathfrak{p}\). Then by [Set Theory] §Axiom of Choice, ⁋Theorem 4, there exists a prime ideal \(\mathfrak{q}\) containing \(\mathfrak{p}\) but not containing \(a\) that is maximal with respect to this property. By definition \(a\not\in \mathfrak{q}\), so \(\mathfrak{q}\) is not a maximal ideal, and hence \(A/\mathfrak{q}\) is not a field. However, in \(A[a^{-1}]\), \(\mathfrak{q}\) must be a maximal ideal by definition, and this contradicts the second condition; therefore \(\mathfrak{p}=\mathfrak{P}\).

Nullstellensatz

We can now state the Nullstellensatz as follows.

Theorem 4 Let \(A\) be a Jacobson ring and let \(E\) be a finitely generated \(A\)-algebra. 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, and in particular any prime ideal of \(E\) is generated by an irreducible monic polynomial. From this we see that no prime ideal can be contained in another prime ideal, so any prime ideal of \(E\) is maximal, and since such an ideal cannot contain \(1\in \mathbb{K}\), its intersection with \(A=\mathbb{K}\) must be \((0)\). In this case, \(E/\mathfrak{n}\) becomes a \(\mathbb{K}\)-vector space of dimension equal to the degree of the irreducible polynomial defining \(\mathfrak{n}\). Finally, to show that \((0)\) is an intersection of maximal ideals, we use the argument that \(E=\mathbb{K}[\x]\) has infinitely many irreducible polynomials, and since the degree of a polynomial must be finite, the only polynomial having all of them as factors is \(0\). Here the infinitude of irreducible polynomials in \(E\) follows exactly from Euclid’s proof of the infinitude of primes.
2. Next, for an arbitrary Jacobson ring \(A\) and an \(A\)-algebra \(E\) generated by one element, we show that \(E\) is Jacobson by verifying the second condition of Lemma 3. That is, our goal in this step is to prove the following proposition.

   Let \(A\) be a Jacobson ring and let \(E\) be an \(A\)-algebra generated by one element. If for a fixed prime ideal \(\mathfrak{q}\subseteq E\), there exists 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.

   Now since \(E'=E/\mathfrak{q}\) is also an \(A\)-algebra generated by one element, showing the above proposition is equivalent to showing the following.

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

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

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

   For this, we show that under the above assumptions \(A'\) must be a field and \(E'\) is a finite extension of \(A'\). Since \(E'\) in the above proposition is an \(A'\)-algebra generated by one element, we can write \(E'=A'[\x]/\mathfrak{q}\). First let us show that \(\mathfrak{q}\neq 0\). Suppose, for contradiction, that \(\mathfrak{q}=0\) and that there exists a suitable \(x\in E'/(0)=A'[\x]\) such that \(E'[x^{-1}]=A'[\x][x^{-1}]\) is a field. Let \(K'=\Frac(A')\); then by this assumption \(K'[\x][x^{-1}]\) 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 \(p(x)\in \mathfrak{q}\) satisfies the following equation in \(E'\):

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

   where \(\alpha\) is the generator of \(E'\) as an \(A'\)-algebra. Then from the above equation we see that \(E'[p_n^{-1}]\) is an integral \(A'[p_n^{-1}]\)-algebra. On the other hand, the \(x\) defined above must also satisfy a suitable polynomial equation

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

   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 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 final case now follows by induction 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 formula

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

then from the isomorphism

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

given by evaluation, we see 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 \(\mathfrak{n}\) of \(E\), \(E/\mathfrak{n}\) is an algebraic extension of \(\mathbb{K}/(\mathfrak{n}\cap \mathbb{K})=\mathbb{K}\), and if \(\mathbb{K}\) is algebraically closed then such an extension can only be itself, so we must have \(E/\mathfrak{n}\cong \mathbb{K}\). On the other hand, letting \(a_i\) be the element of \(\mathbb{K}\) to which each \(\x_i\) is mapped under the canonical surjection \(E \rightarrow E/\mathfrak{n}\cong \mathbb{K}\), we have \(\mathfrak{m}_a\subseteq \mathfrak{n}\), and the desired result follows from the maximality of \(\mathfrak{m}_a\).

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

Lemma 5 Let \(\mathbb{K}\) be a field. Then \(\mathfrak{m}_a=(\x_1-a_1,\ldots, \x_n-a_n)\) is a maximal ideal of \(\mathbb{K}[\x_1,\ldots, \x_n]\). Also, if \(\mathbb{K}\) is algebraically closed, there is a one-to-one correspondence between maximal ideals of \(\mathbb{K}[\x_1,\ldots,\x_n]/(f_1,\ldots, f_r)\) and the solutions \((x_1,\ldots, x_n)\) of the system of equations

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

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

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

and the function \(I\) that takes a subset \(S\) of \(\mathbb{K}^n\) and returns the subset

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

of \(\mathbb{K}[\x_1,\ldots, \x_n]\).

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

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

holds.

Proof

From Lemma 5, we see that the elements of \(Z(\mathfrak{a})\) correspond one-to-one with the maximal ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) containing \(\mathfrak{a}\). Therefore \(I(Z(\mathfrak{a}))\) is the intersection of the maximal ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) containing \(\mathfrak{a}\), and since \(\mathbb{K}[\x_1,\ldots, \x_n]\) is Jacobson by Theorem 4, this equals the intersection of the prime ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) containing \(\mathfrak{a}\), which is exactly 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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