Nullstellensatz

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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 $\mathfrak{a}=\sqrt{\mathfrak{a}}$.

Thus, the above observation amounts to the statement that every prime ideal is radical. The proof of this is somewhat trivial, but if we had instead considered the intersection of maximal ideals containing $\mathfrak{p}$, in a manner similar to §Integral Extensions, §§Nakayama’s Lemma, then this observation would no longer be obvious, and in fact it would fail. For instance, any local ring containing a prime ideal that is not maximal, such as $\mathbb{Z}_{(2)}$, provides 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.

We then have the following.

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

$A$ is a Jacobson ring.
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 immediate from the definition that any quotient $A/\mathfrak{p}$ is also Jacobson. Meanwhile, by §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. Now 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 $a$. 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 $0$. Hence every 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 prove 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 §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma), 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 $A[a^{-1}]$ the ideal $\mathfrak{q}$ must be maximal by construction, 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.

First, consider the case $A=\mathbb{K}$ and $E=\mathbb{K}[\x]$. Then $E$ is a principal ideal domain, and in particular every prime ideal of $E$ is generated by an irreducible monic polynomial. From this we see that no prime ideal can be contained in another, so every prime ideal of $E$ is maximal; moreover, such an ideal cannot contain $1\in \mathbb{K}$, so its intersection with $A=\mathbb{K}$ must be $(0)$. In this case, $E/\mathfrak{n}$ becomes a $\mathbb{K}$-vector space whose dimension equals the degree of the irreducible polynomial defining $\mathfrak{n}$. Finally, to show that $(0)$ is an intersection of maximal ideals, we argue that $E=\mathbb{K}[\x]$ has infinitely many irreducible polynomials, and since any polynomial has finite degree, the only polynomial divisible by all of them is $0$. The infinitude of irreducible polynomials in $E$ follows exactly from Euclid’s proof of the infinitude of primes.
Next, let $A$ be an arbitrary Jacobson ring and let $E$ be an $A$-algebra generated by one element; we show that $E$ is Jacobson by verifying the second condition of Lemma 3 (Rabinowitch). That is, our goal in this step is to prove the following statement.

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.

Since $E'=E/\mathfrak{q}$ is also an $A$-algebra generated by one element, proving the above is equivalent to proving 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 passing to this 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.

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.

To this end, we show that under these hypotheses $A'$ must be a field and $E'$ is a finite extension of $A'$. Since $E'$ is an $A'$-algebra generated by one element, we can write $E'=A'[\x]/\mathfrak{q}$. First we show that $\mathfrak{q}\neq 0$. Suppose, for contradiction, that $\mathfrak{q}=0$ and that there exists $x\in E'/(0)=A'[\x]$ 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 step, 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 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 this equation we see that $E'[p_n^{-1}]$ is an integral $A'[p_n^{-1}]$-algebra. On the other hand, the $x$ above must also satisfy some 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 (Rabinowitch) implies that $A'$ is a field. Therefore $E'$ is an integral $A'$-algebra, and again by §Integral Extensions and Ideals, ⁋Corollary 3, we conclude that $E'$ is a field.
The final case now follows by induction using the second result.

In particular, consider the case $A=\mathbb{K}$ 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

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

then from the evaluation isomorphism

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

we see that $\mathfrak{m}_a$ is a maximal ideal.

Moreover, if $\mathbb{K}$ is algebraically closed, then every maximal ideal of $E$ is of this form. Indeed, for any maximal ideal $\mathfrak{n}$ of $E$, the quotient $E/\mathfrak{n}$ 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 trivial, so we must have $E/\mathfrak{n}\cong \mathbb{K}$. Letting $a_i$ denote the image of each $\x_i$ under the canonical surjection $E \rightarrow E/\mathfrak{n}\cong \mathbb{K}$, we obtain $\mathfrak{m}_a\subseteq \mathfrak{n}$, 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 $\mathbb{K}[\x_1,\ldots, \x_n]$. Furthermore, 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 solutions $(x_1,\ldots, x_n)$ of the system

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

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

References

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

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Jacobson Rings

Nullstellensatz

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