Elementary / Interesting proofs of the Nullstellensatz

Question

Is there an easy proof of the Nullstellensatz that avoids the standard Noether-normalization techniques?

One proof I know proves first the 'weak' Nullstellensatz which ensures that maximal ideals correspond to points (using normalization), and then the stronger version by introducing another variable and the Rabinovich trick.

Answer 1

There is a cheap proof of the weak Nullstellensatz in Artin's Algebra which goes like this: suppose $m$ is a maximal ideal of $F[x_1, \dots x_n]$ where $F$ is an uncountable algebraically closed field. Then $F[x_1, \dots x_n]/m$ is a field extension of $F$ which is either $F$ itself or transcendental. In the first case, $m$ is of the form $(x_i - a_i)$ for some $a_i$. In the second case, a transcendental extension of $F$ must have uncountable dimension, since if $t$ is transcendental then $\{ \frac{1}{t - a} : a \in k \}$ is linearly independent. But $F[x_1, \dots x_n]/m$ has at most countable dimension over $F$. Hence the second case does not occur.

Edit, 2/28/10: See also question #15611, in which Brian Conrad gives an argument that one can reduce the general case to the uncountable case.

Answer 2

I have been thinking about the question "What is the best -- i.e., some combination of shortest, most natural, easiest -- proof of the Nullstellensatz?" recently on the eve of a commutative algebra course.

In my notes up to this point I had been following Kaplansky's treatment of Goldman domains and Hilbert-Jacobson rings. This places the problem in a more general context and allows for an attractively thorough analysis. At the end one comes out with the following results:

(1) The polynomial ring $k[t_1,\ldots,t_n]$ is a Jacobson ring -- i.e., every radical ideal is the intersection of the maximal ideals containing it.

(2) (Zariski's Lemma): If $\mathfrak{m}$ is a maximal ideal of $k[t_1,\ldots,t_n]$, then $k[t_1,\ldots,t_n]/\mathfrak{m}$ is a finite degree field extension of $k$.

(That Hilbert's Nullstellensatz follows from (1) and (2) is an easy, standard argument that I won't discuss here.)

But it is well-known that to prove the Nullstellensatz one needs only (2), because then (1) follows by a short and easy argument that everyone seems to like: Rabinowitsch's Trick. So perhaps this is a sign that developing the theory of (Hilbert-)Jacobson rings to prove the Nullstellensatz is overkill.

So the question seems to be: what is the best proof of Zariski's Lemma?

After looking around at various proofs, here is what I think the answer is now: it is an easy consequence of the following result.

Theorem (Artin-Tate Lemma): Let $R \subset T \subset S$ be a tower of rings such that
(i) $R$ is Noetherian,
(ii) $S$ is finitely generated as an $R$-algebra, and
(iii) $S$ is finitely generated as a $T$-module.
Then $T$ is finitely generated as an $R$-algebra.

Proof: Let $x_1,\ldots,x_n$ be a set of generators for $S$ as an $R$-algebra, and let $\omega_1,\ldots,\omega_m$ be a set of generators for $S$ as a $T$-module. For all $1 \leq i \leq n$, we may write \begin{equation} \label{ARTINTATEEQ1} x_i = \sum_j a_{ij} \omega_j, \ a_{ij} \in T. \end{equation} Similarly, for all $1 \leq i,j \leq m$, we may write \begin{equation} \label{ARTINTATEEQ2} \omega_i \omega_j = \sum_{k} b_{ijk} \omega_k, \ b_{ijk} \in T. \end{equation} Let $T_0$ be the $R$-subalgebra of $T$ generated by the $a_{ij}$ and $b_{ijk}$. Since $T_0$ is a finitely generated algebra over the Noetherian ring $R$, it is itself a Noetherian ring by the Hilbert Basis Theorem. \ \indent Now each element of $S$ may be expressed as a polynomial in the $x_i$'s with $R$-coefficients. Making substitutions using the two equations above shows that $S$ is a finitely generated $T_0$-module. Since $T_0$ is Noetherian, the submodule $T$ is also finitely generated as a $T_0$-module. This immediately implies that $T$ is finitely generated as a $T_0$-algebra and then in turn that $T$ is finitely generated as an $R$-algebra, qed!

Proof that Artin-Tate implies Zariski's Lemma:

It suffices to prove the following: let $K/k$ be a field extension which is finitely generated as a $k$-algebra. Then $K/k$ is algebraic. Indeed, suppose otherwise: let $x_1,\ldots,x_n$ be a transcendence basis for $K/k$ (where $n \geq 1$ since $K/k$ is transcendental), put $k(x) = k(x_1,\ldots,x_n)$ and consider the tower of rings

$k \subset k(x) \subset K$.

To be sure, we recall the definition of a transcendence basis: the elements $x_i$ are algebraically independent over $k$ and $K/k(x)$ is algebraic. But since $K$ is a finitely generated $k$-algebra, it is certainly a finitely generated $k(x)$-algebra and thus $K/k(x)$ is a finite degree field extension. Thus the Artin-Tate Lemma applies to our tower: we conclude that $k(x)/k$ is a finitely generated $k$-algebra. But this is absurd. It implies the much weaker statement that $k(x) = k(x_1,\ldots,x_{n-1})(x_n)$ is finitely generated as a $k(x_1,\ldots,x_{n-1})[x_n]$-algebra, or weaker yet, that there exists some field $F$ such that $F(t)$ is finitely generated as an $F[t]$-algebra: i.e., there exist finitely many rational functions $\{r_i(t) = \frac{p_i(t)}{q_i(t)} \}_{i=1}^N$ such that every rational function is a polynomial in the $r_i$'s with $k$-coefficients. But $F[t]$ is a PID with infinitely many nonassociate nonzero prime elements (e.g. adapt Euclid's argument of the infinitude of the primes), so we may choose a nonzero prime element $q$ which does not divide $q_i(t)$ for any $i$. It is then clear that $\frac{1}{q}$ cannot be a polynomial in the $r_i(t)$'s: for instance, evaluation at a root of $q$ in $\overline{F}$ leads to a contradiction.

Note that this is almost all exactly as in Artin-Tate's paper, except for the endgame above, which has been made a little more explicit and simplified: their conclusion seems to depend upon unique factorization in $k[t_1,\ldots,t_n]$, which does not come until later on in my notes.

Further comments:

(i) The proof is essentially a reduction to Noether's normalization in the case of field extensions, which becomes the familiar result about existence of transcendence bases. Thus it is not so far away from the most traditional proof of the Nullstellensatz. But I think the Artin-Tate Lemma is easier than Noether Normalization.

(ii) Speaking of Noether: the proof of the Artin-Tate Lemma is embedded in the standard textbook proof that if $R$ is a finitely generated $k$-algebra and $G$ is a finite group acting on $R$ by ring automorphisms, then $R^G$ is a finitely generated $k$-algebra. In fact I had already typed this proof up elsewhere in my notes. Realizing that the Artin-Tate Lemma is something I was implicitly proving in the course of another result anyway was part of what convinced me that this was an efficient route to the Nullstellensatz. Note that the paper of Artin and Tate doesn't make any connection with Noether's theorem and conversely the textbooks on invariant theory that prove Noether's theorem don't seem to mention Artin-Tate. (However, googling -- Artin-Tate Lemma, Noether -- finds several research papers which allude to the connection in a way which suggests it is common knowledge among the cognoscenti.)

Added: It turns out this is the proof of Zariski's Lemma given in Chapter 7 of Atiyah-Macdonald. I had missed this because (i) they give another (nice) proof using valuation rings in Chapter 5 and (ii) they do not attribute the Artin-Tate Lemma to Artin and Tate, although their treatment of it is even closer to the Artin-Tate paper than mine is above. (In the introduction of their book, they state cheerfully that they have not attributed any results. I think this is a drawback of their otherwise excellent text.)

Answer 3

Well, easy is a relative term, but Terry Tao has a blog post about a computational proof of the Nullstellensatz he discovered which requires little more than high school algebra. The proof itself may be longer than the one(s) you are used to, but I'm willing to bet the tools needed are easier. The post is available here.

Answer 4

There is the snappy model-theoretic proof of the Nullstellensatz, which goes by first proving that ACF_p, the theory of algebraically closed fields of some fixed characteristic p, is model complete. Once you know this it is easy:

Let k be an algebraically closed field, and let m be a maximal ideal in k[x₁,...,x_n]. Let L be the algebraic closure of k[x₁,...,x_n]/m.

We are trying to prove that there exists $a_1,...,a_n \in k$ which is a common zero of some set of generators for m. Note that this assertion is a first-order sentence using the language of fields and symbols in k. So model completeness of ACF_p tells us that this sentence is true in k if and only if it is true in some (and then every) algebraically closed extension field of k. But we know that this sentence is true in the algebraically closed extension L by construction, so we are done.

Answer 5

I found this "geodesic" proof by Munshi, written by May (I like the very last paragraph (:).

Answer 6

From Serge Lang's Algebra:

Theorem 9.1.1

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Let $\phi:k\to L$ be an embedding of k into an algebraically closed field $L$. Then there exists an extension of $\phi$ to a homomorphism $\bar{\phi}: k[x] \to L$.

Note: The $x_i$ are not indeterminates.

Corollary 9.1.2 (Zariski)

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. If $k[x]$ is a field, then $k[x]$ is algebraic over $k$.

Corollary 9.1.3

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Fix a finite family of elements $(y_i)_{i=1}^m$ of $k[x]$. If $k[x]$ is an integral domain, there exists a homomorphism $\psi:k[x]\to k^a$, where $k^a$ is the algebraic closure of $k$ such that $\psi(y_i)\neq 0$ for all $1\leq i\leq m$.

Theorem 9.1.4 (Weak Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Then either $\mathfrak{q}$ is the unit ideal, or $\mathfrak{q}$ has a zero in $k^a$.

Theorem 9.1.5 (Hilbert's Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Let $f\in k[X]$ be a polynomial vanishing on every zero of $\mathfrak{q}$ in $k^a$. Then there exists $m>0$ such that $f^m\in \mathfrak{q}$.

---

The proof of 9.1.5 follows from the Rabinowitsch trick and 9.1.4, which in turn follows directly from 9.1.2, which is a straighforward application of 9.1.1.

There is an advantage to this proof because it allows us not only to extend the definition of a variety to non-algebraically closed fields, but also to define an algebraic space, which is defined as a functor that takes field extensions of the basefield $k$ to the zero set of the ideal in that extension, with some nice properties. (This is not standard terminology, but I believe that every functor arising this way is in fact an algebraic space.)

In Eisenbud's commutative algebra book, the Nullstellensatz is generalized further from fields to Jacobson rings, which are rings for which any prime ideal is an intersection of some family of maximal ideals (This is a theorem of Bourbaki).

None of these proofs uses Noether normalization.

Corollary 9.1.2 is a lemma of Zariski that he introduced to prove the Nullstellensatz. I believe Lang's proof of Theorem 9.1.1 is similar to Zariski's proof of 9.1.2.

This is Zariski's paper where he introduced the method used by Lang. If you read the introduction, it's really interesting, because all of the previous proofs had been somewhat nontrivial. This was somewhat groundbreaking for proofs of the Nullstellensatz.

Additionally this lemma of Zariski is a special case of Zariski's main theorem for commutative rings. The Nullstellensatz follows with basically no effort. However, Zariski's main theorem is highly nontrivial.

Answer 7

I once wrote up a proof (in the more general context of Jacobson rings) which uses basic commutative algebra techniques and avoids Noether normalization; it is available here.

This proof is very similar to the proof of Chevalley's theorem stating that the image of a constructible set, under a finite type morphism of Noetherian schemes, is again constructible. Indeed this result also implies the Nullstellensatz; see e.g. the discussion and proofs in section II.2 of the unpublished book of Mumford and Oda, where Chevalley's theorem is referred to as Chevalley's Nullstellensatz.

Answer 8

I have a feeling that some modern proofs hiding from the students the intuition of what is really going on... I am now teaching an undergraduate course "Introduction to algebraic geom. and comm. alg.". I am speaking a lot about Grobner bases and resultant -- they look natural and algorithmic. For the course I am going to use a proof of weak Nullstellensatz using the Groebner bases only.

Let $k$ be a field, $a\in k$. Let ev$_{a}:k[x_1,x_2,\dots,x_n]\to k[x_2,\dots,x_n]$ denote the evaluation homomorphism ev$_{a}:f(x_1,x_2,\dots,x_n)\to f(a,x_2,...,x_n)$. The proof is based on the following lemmas.

Lemma1:

Let $k$ be an algebraically closed field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\langle p \rangle$ and $p\in k[x]\setminus k$. Then there exists $a\in k$, $p(a)=0$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$

Lemma2 (It is valid for any field):

Let $k$ be a field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\{0\}$. Then there exist non-zero polynomial $q\in k[x_1]$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for any $a\in k$, $q(a)\neq 0$.

Corollary:

Let $k$ be an infinite field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\{0\}$. Then ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for some $a\in k$.

The weak Nullstellensatz follows by induction...

Let me show how Groebner bases can be used to prove, say, Lemma 2. Let $I\subset k[x_1,\dots,x_n]$ be an ideal. Let $\langle I\rangle_{k(x_1)}$ denote the ideal generated by $I$ in $k(x_1)[x_2,\dots,x_n]$. Lemma 2 follows from the
statements:

1: $\langle I \rangle _{k(x_1)}=k(x_1)[x_2,\dots,x_n]$ iff $I\cap k[x_1]\neq \{0\}$.

2: Let $\Gamma=\{g_1,\dots,g_2\}$ be a Groebener basis for $\langle I\rangle_{k(x_1)}$. Suppose, that leading coefficients of $g_i$ are $1$. Let $q\in k[x_1]$ be the common denominator of all $g_i$
(i.e. $qg_i\in k[x_1,...x_n]$ for any $i$). Then ev$_{a}(\Gamma)$ is a Groebner basis for ev$_{a}(I)$ for any $a\in k$, $q(a)\neq 0$.

To show (2) it is suffices to note that during reduction of $f\in k[x_1,x_2,\dots,x_n]$ by $\Gamma$ all denominators divide a power of $q$.

Answer 9

Richard Swan's proof (which is based on Munshi's ideas): https://www.math.uchicago.edu/~swan/nss.pdf

Answer 10

You can see the following in arxiv

A one-line undergraduate proof of Zariski's lemma and Hilbert's nullstellensatz

Answer 11

I like Daniel Allcock's proof (which still uses the Rabinowitsch trick).

Answer 12

Here is another proof I posted on math.SE.

Answer 13

To me, the reason for the veracity of the Nullstellensatz is that if $R$ is a Hilbert–Jacobson ring, then so is $R[x]$. Coincidentally, I think I've discovered a short way of proving this theorem, which is one of the Goldman–Krull theorems. It uses only division with remainder in its general form.

Let $P \le R[x]$ be prime. By dividing by $R \cap P$, we may assume that $R \cap P = \{0\}$ and that $R$ is integral; indeed, quotients of Hilbert–Jacobson rings are again Hilbert–Jacobson by the correspondence theorem. Let $f \notin P$ of minimal degree such that $f$ is contained in all maximal ideals containing $P$. Then $\deg f > 0$, because otherwise $f \in R$ and $m + \langle x \rangle$ is maximal and separates $f$ from $P$, where $m$ is a maximal ideal of $R$ which does not contain $f$ by assumption, since $\{0\}$ is prime if $R$ is integral.

If $P \neq 0$, let $g$ be of minimal degree in $P$. We write $a$ for the leading coefficient of $f$ and $b$ for the leading coefficient of $g$. If $\deg g \le \deg f$, we divide $f$ by $g$ with remainder and obtain $b^m f = qg + r$, where $\deg r < \deg g \le \deg f$, so that $r$ has the same properties as $f$ but is of lesser degree, a contradiction. Therefore, $\deg g > \deg f$, and now we divide $g$ by $f$ with remainder and obtain $a^k g = hf + s$ with $\deg s < \deg f$ and obtain $hf \notin P$ for degree reasons, thus $s$ has the same properties as $f$ but is of lesser degree, a contradiction ($s \neq 0$ because else $h \in P$, which has to have smaller degree).

If now $P = \{0\}$, choose a maximal ideal of $R[x]_f$, which is not a field, because either $R$ is no field, and $a$ is not divisible by all nonzero elements of $R$, although this would follow from their invertibility in $R[x]_f$, or $R$ is a field and $f$ is divisible (for the same reason) by all the prime elements of the Euclidean domain $R[x]$, which is likewise impossible by Euclid's theorem. Its preimage in $R[x]$ is then a nonzero prime ideal not containing $f$, and now we are in the previous situation.

EDIT: I thought it might be prudent to add my own proof of the second important theorem, which is that a maximal ideal of $R[x]$ contracts to a maximal ideal in $R$ whenever $R$ is Hilbert–Jacobson.

Indeed, let $M \le R[x]$ be maximal. As before we reduce to the case where $R$ is an integral domain and $M \cap R = \emptyset$. Let $f \in m$ be of minimal degree, and suppose that $R$ is not a field. Then there exists a non-invertible $a \in R \setminus \{0\}$. Still, $a$ is invertible in $R[x]/M$, and we get an equation $ap(x) - 1 \in M$ of lowest degree. Thus, the leading coefficient of $f$ (let's denote it by $b$) is not invertible, because otherwise division with remainder would yield an equation $ar(x) - 1 \in M$ of lower degree. But since $R$ is Hilbert–Jacobson, there is a maximal ideal $m \le R$ that does not contain $b$. Thus, the image of $M$ in $(R/m)[x]$ is nonzero (and it does not contain $1$ because any $g \in M$ satisfies an equation $b^k g = f h$ by division with remainder and degree reasons and $b$ is invertible in $R/m$, so that the image of $M$ is in fact the principal ideal generated by $f$), whence by the correspondence theorem $M$ was not maximal to begin with.

Comment: The first proof is a proper simplification of the one given by Goldman and Krull, whereas the second is of comparable length. From the first theorem it is evident that $\operatorname{Spec}_\max(k[x_1, \ldots, x_n])$ contains all the geometrical information, and from the second it may be deduced that $\operatorname{Spec}_\max(k[x_1, \ldots, x_n])$ consists precisely of the points of $k^n$. Thus, the first of the two theorems proves something which is stronger than the classical Nullstellensatz: It proves that the class for which it is sufficient to consider $\operatorname{Spec}_\max$, which are the Hilbert–Jacobson rings, is closed under taking polynomial rings, so that rings of the type $R[x_1, \ldots, x_n]$ with $R$ Hilbert–Jacobson are Hilbert–Jacobson. Is that not remarkable?

Answer 14

The following elementary proof of a generalization of Zariski's Lemma is based on that of Th. VI, 3.15 in the excellent Lombardi and Quitté, where the case that $A$ is von Neumann regular is treated.

If $A$ is a commutive ring of Krull dimension zero, and $B$ an $A$-algebra of finite type such that $\text{dim}(B) = 0$ as well, then $B$ is finitely generated as an $A$-module.

Proof: $B = C/I$ where $I$ is an ideal of $C = A[X_1,\cdots,X_n]$, for some $n$. Write $B = A[x_1,\cdots,x_n]$, and induce on $n$. The case $n = 0$ is trivial.

Since $B$ is zero-dimensional, $x_n^m(1-x_nb) = 0$ for some $m \geq 0$ and $b \in B$ (see below). We can assume $m \geq 1$. Writing $b$ as a polynomial in $x_1,\cdots,x_n$, this gives an $f \in C$ in which the coefficient of $X_n^m$ equals 1, with $f(x_1,\cdots,x_n) = 0$ in $B$ (that is, $f \in I$).

Let $a \in A$ be a coefficient of $f$. Then $a^k(1-az) = 0$ in $A$ for a $k \geq 0$ and a $z \in A$. It follows that $e := a^kz^k$ is idempotent, and $a^kA = eA$. We have $A = A_e \times A_{1-e}$, and $a$ becomes a unit in $A_e$ and nilpotent in $A_{1-e}$, properties stable under localization. If $a_1,\cdots,a_r$ are the coefficients of $f$, in any order, and $e_1,\cdots,e_r$ are the idempotents so obtained, then $A = \prod_{D \subseteq \{1,\cdots,r\}} A_{e_D}$ where $e_D$ denotes the product of the $e_i$ for $i \in D$ and the $1-e_i$ for $i \notin D$. And each $a_i$ is a unit or a nilpotent (or both, perhaps) in every component $A_{e_D}$.

Given $D \subseteq \{1,\cdots,r\}$, put $e := e_D$. We may write $f = g - h$ for $g,h \in C$ with all coefficients of $g$ units in $A_e$ and those of $h$ nilpotent. As $X_n^m$ has coefficient 1 in $f$ and $m \geq 1$, $g$ is non-constant in $C_{e}$. We now apply a trick familiar in this neck of the woods. Taking $d \in \mathbb{N}$ larger than the degree of $g$ in each $X_j$ ($1 \leq j \leq n$), we have $C = A[Y_1,\cdots,Y_{n-1},X_n]$, where $Y_j = X_j - X_n^{d^j}$ for $1 \leq j \leq n-1$. Under this change of variables, the transform $\tilde{g}$ of $g$ will consist of a monomial $X_n^N$ with $N \geq 1$, having the same coefficient as the unique monomial of $g$ giving rise to it, a unit of $A_e$ therefore, plus monomials in $Y_1,\cdots,Y_{n-1},X_n$ that are of degree less than $N$ in $X_n$. And the coefficients of $\tilde{h}$ are again nilpotent in $A_e$.

Let $y_j$ be the image of $Y_j$ in $B$. Then $\tilde{g}(y_1,\cdots,y_{n-1},x_n) = \tilde{h}(y_1,\cdots,y_{n-1},x_n)$ is nilpotent in $B_e$. By raising $\tilde{g}$ to a suitable power, we find that the element $x_n$ of $B_e$, and therefore $B_e$ itself, is integral over $R_e$, where $R := A[y_1,\cdots,y_{n-1}] \subseteq B$. Now $\text{dim}(R_e) \leq 0$ because $\text{dim}(B_e) \leq 0$ and integral extensions have Lying Over and Going Up. (Or, to keep things completely elementary, for any $u \in R_e$ we have $u^s = u^{s+1}v$ for an $s \geq 0$ and a $v \in B_e$. Thus $u^s = u^{s+i}v^i$ for all $i \geq 0$, and if $v^t + r_{t-1}v^{t-1} + \cdots + r_0 = 0$ with the $r_i \in R_e$, multiplying the equation with $u^{s+t}$ gives $u^s(1+uw) = 0$ for $w := r_{t-1} + r_{t-2}u + \cdots + r_0u^{t-1} \in R_e$.)

By the induction hypothesis, $R_{e}$ is a finitely generated $A_{e}$-module, and, by transitivity, so is $B_e = B_{e_D}$. Hence, gluing all the $D$'s together, $B$ is finitely generated as a module over $A$, $\square$.

[For completeness: $\text{dim}(A) \leq 0 \iff \forall_{a \in A}\,0 \in S_a$, where $S_a = \{a^n(1-ax)\,\mid\, n \geq 0,\, x \in A\}$. For $S_a$ is multiplicatively closed, so if $0 \notin S_a$, the localization $S_a^{-1}A$ is not the trivial ring, so it contains a prime ideal of the form $\mathfrak{p}S_a^{-1}A$ with $\mathfrak{p} \in \text{spec}(A) = \text{max}(A)$ and $\mathfrak{p} \cap S_a = \varnothing$. Then $a \notin \mathfrak{p}$, so $aA + \mathfrak{p} = A$, so $ax+y = 1$ for some $x \in A$ and $y \in \mathfrak{p}$. But then $y = 1-ax \in \mathfrak{p} \cap S_a$, contradiction. Conversely, if $\mathfrak{p} \subsetneq \mathfrak{q}$ in $\text{spec}(A)$, and $a \in \mathfrak{q}-\mathfrak{p}$, and $a^n(1-ax) = 0$, then $1-ax \in \mathfrak{p}$, so that both $a$ and $1-ax$ are in $\mathfrak{q}$, another contradiction.]