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Analytic number theory is often concerned with the asymptotic behaviour of various arithmetic functions: functions ${f: {\bf N} \rightarrow {\bf R}}$ or ${f: {\bf N} \rightarrow {\bf C}}$ from the natural numbers ${{\bf N} = \{1,2,\dots\}}$ to the real numbers ${{\bf R}}$ or complex numbers ${{\bf C}}$. In this post, we will focus on the purely algebraic properties of these functions, and for reasons that will become clear later, it will be convenient to generalise the notion of an arithmetic function to functions ${f: {\bf N} \rightarrow R}$ taking values in some abstract commutative ring ${R}$. In this setting, we can add or multiply two arithmetic functions ${f,g: {\bf N} \rightarrow R}$ to obtain further arithmetic functions ${f+g, fg: {\bf N} \rightarrow R}$, and we can also form the Dirichlet convolution ${f*g: {\bf N} \rightarrow R}$ by the usual formula

$\displaystyle f*g(n) := \sum_{d|n} f(d) g(\frac{n}{d}).$

Regardless of what commutative ring ${R}$ is in used here, we observe that Dirichlet convolution is commutative, associative, and bilinear over ${R}$.

An important class of arithmetic functions in analytic number theory are the multiplicative functions, that is to say the arithmetic functions ${f: {\bf N} \rightarrow {\bf R}}$ such that ${f(1)=1}$ and

$\displaystyle f(nm) = f(n) f(m)$

for all coprime ${n,m \in {\bf N}}$. A subclass of these functions are the completely multiplicative functions, in which the restriction that ${n,m}$ be coprime is dropped. Basic examples of completely multiplicative functions (in the classical setting ${R={\bf C}}$) include

• the Kronecker delta ${\delta}$, defined by setting ${\delta(n)=1}$ for ${n=1}$ and ${\delta(n)=0}$ otherwise;
• the constant function ${1: n \mapsto 1}$ and the linear function ${n \mapsto n}$ (which by abuse of notation we denote by ${n}$);
• more generally monomials ${n \mapsto n^s}$ for any fixed complex number ${s}$ (in particular, the “Archimedean characters” ${n \mapsto n^{it}}$ for any fixed ${t \in {\bf R}}$), which by abuse of notation we denote by ${n^s}$;
• Dirichlet characters ${\chi}$;
• the Liouville function ${\lambda}$;
• the indicator function of the ${z}$-smooth numbers (numbers whose prime factors are all at most ${z}$), for some given ${z}$; and
• the indicator function of the ${z}$-rough numbers (numbers whose prime factors are all greater than ${z}$), for some given ${z}$.

Examples of multiplicative functions that are not completely multiplicative include

• the Möbius function ${\mu}$;
• the divisor function ${\tau}$ (also referred to as ${d}$);
• more generally, the higher order divisor functions ${\tau_k(n) = \sum_{d_1,\dots,d_k: d_1 \dots d_k = n} 1}$ for ${k \geq 1}$;
• the Euler totient function ${\phi}$;
• the number of roots ${n \mapsto \# \{ a \in {\bf Z}/n{\bf Z}: P(a) = 0\}}$ of a given polynomial ${P}$ defined over ${{\bf Z}}$;
• more generally, the point counting function ${n \mapsto V[{\bf Z}/n{\bf Z}]}$ of a given algebraic variety ${V}$ defined over ${{\bf Z}}$ (closely tied to the Hasse-Weil zeta function of ${V}$);
• the function ${r: n \mapsto r(n)}$ that counts the number of representations of ${n}$ as the sum of two squares;
• more generally, the function that maps a natural number ${n}$ to the number of ideals in a given number field ${K}$ of absolute norm ${n}$ (closely tied to the Dedekind zeta function of ${K}$).

These multiplicative functions interact well with the multiplication and convolution operations: if ${f,g: {\bf N} \rightarrow R}$ are multiplicative, then so are ${fg}$ and ${f * g}$, and if ${\psi}$ is completely multiplicative, then we also have

$\displaystyle \psi (f*g) = (\psi f) * (\psi g). \ \ \ \ \ (1)$

Finally, the product of completely multiplicative functions is again completely multiplicative. On the other hand, the sum of two multiplicative functions will never be multiplicative (just look at what happens at ${n=1}$), and the convolution of two completely multiplicative functions will usually just be multiplicative rather than completley multiplicative.

The specific multiplicative functions listed above are also related to each other by various important identities, for instance

$\displaystyle \delta * f = f; \quad \mu * 1 = \delta; \quad 1 * 1 = \tau; \quad \phi * 1 = n$

where ${f}$ is an arbitrary arithmetic function.

On the other hand, analytic number theory also is very interested in certain arithmetic functions that are not exactly multiplicative (and certainly not completely multiplicative). One particularly important such function is the von Mangoldt function ${\Lambda}$. This function is certainly not multiplicative, but is clearly closely related to such functions via such identities as ${\Lambda = \mu * L}$ and ${L = \Lambda * 1}$, where ${L: n\mapsto \log n}$ is the natural logarithm function. The purpose of this post is to point out that functions such as the von Mangoldt function lie in a class closely related to multiplicative functions, which I will call the derived multiplicative functions. More precisely:

Definition 1 A derived multiplicative function ${f: {\bf N} \rightarrow R}$ is an arithmetic function that can be expressed as the formal derivative

$\displaystyle f(n) = \frac{d}{d\varepsilon} F_\varepsilon(n) |_{\varepsilon=0}$

at the origin of a family ${f_\varepsilon: {\bf N}\rightarrow R}$ of multiplicative functions ${F_\varepsilon: {\bf N} \rightarrow R}$ parameterised by a formal parameter ${\varepsilon}$. Equivalently, ${f: {\bf N} \rightarrow R}$ is a derived multiplicative function if it is the ${\varepsilon}$ coefficient of a multiplicative function in the extension ${R[\varepsilon]/(\varepsilon^2)}$ of ${R}$ by a nilpotent infinitesimal ${\varepsilon}$; in other words, there exists an arithmetic function ${F: {\bf N} \rightarrow R}$ such that the arithmetic function ${F + \varepsilon f: {\bf N} \rightarrow R[\varepsilon]/(\varepsilon^2)}$ is multiplicative, or equivalently that ${F}$ is multiplicative and one has the Leibniz rule

$\displaystyle f(nm) = f(n) F(m) + F(n) f(m) \ \ \ \ \ (2)$

for all coprime ${n,m \in {\bf N}}$.

More generally, for any ${k\geq 0}$, a ${k}$-derived multiplicative function ${f: {\bf N} \rightarrow R}$ is an arithmetic function that can be expressed as the formal derivative

$\displaystyle f(n) = \frac{d^k}{d\varepsilon_1 \dots d\varepsilon_k} F_{\varepsilon_1,\dots,\varepsilon_k}(n) |_{\varepsilon_1,\dots,\varepsilon_k=0}$

at the origin of a family ${f_{\varepsilon_1,\dots,\varepsilon_k}: {\bf N} \rightarrow R}$ of multiplicative functions ${F_{\varepsilon_1,\dots,\varepsilon_k}: {\bf N} \rightarrow R}$ parameterised by formal parameters ${\varepsilon_1,\dots,\varepsilon_k}$. Equivalently, ${f}$ is the ${\varepsilon_1 \dots \varepsilon_k}$ coefficient of a multiplicative function in the extension ${R[\varepsilon_1,\dots,\varepsilon_k]/(\varepsilon_1^2,\dots,\varepsilon_k^2)}$ of ${R}$ by ${k}$ nilpotent infinitesimals ${\varepsilon_1,\dots,\varepsilon_k}$.

We define the notion of a ${k}$-derived completely multiplicative function similarly by replacing “multiplicative” with “completely multiplicative” in the above discussion.

There are Leibniz rules similar to (2) but they are harder to state; for instance, a doubly derived multiplicative function ${f: {\bf N} \rightarrow R}$ comes with singly derived multiplicative functions ${F_1, F_2: {\bf N} \rightarrow R}$ and a multiplicative function ${G: {\bf N} \rightarrow R}$ such that

$\displaystyle f(nm) = f(n) G(m) + F_1(n) F_2(m) + F_2(n) F_1(m) + G(n) f(m)$

for all coprime ${n,m \in {\bf N}}$.

One can then check that the von Mangoldt function ${\Lambda}$ is a derived multiplicative function, because ${\delta + \varepsilon \Lambda}$ is multiplicative in the ring ${{\bf C}[\varepsilon]/(\varepsilon^2)}$ with one infinitesimal ${\varepsilon}$. Similarly, the logarithm function ${L}$ is derived completely multiplicative because ${\exp( \varepsilon L ) := 1 + \varepsilon L}$ is completely multiplicative in ${{\bf C}[\varepsilon]/(\varepsilon^2)}$. More generally, any additive function ${\omega: {\bf N} \rightarrow R}$ is derived multiplicative because it is the top order coefficient of ${\exp(\varepsilon \omega) := 1 + \varepsilon \omega}$.

Remark 1 One can also phrase these concepts in terms of the formal Dirichlet series ${F(s) = \sum_n \frac{f(n)}{n^s}}$ associated to an arithmetic function ${f}$. A function ${f}$ is multiplicative if ${F}$ admits a (formal) Euler product; ${f}$ is derived multiplicative if ${F}$ is the (formal) first derivative of an Euler product with respect to some parameter (not necessarily ${s}$, although this is certainly an option); and so forth.

Using the definition of a ${k}$-derived multiplicative function as the top order coefficient of a multiplicative function of a ring with ${k}$ infinitesimals, it is easy to see that the product or convolution of a ${k}$-derived multiplicative function ${f: {\bf N} \rightarrow R}$ and a ${l}$-derived multiplicative function ${g: {\bf N} \rightarrow R}$ is necessarily a ${k+l}$-derived multiplicative function (again taking values in ${R}$). Thus, for instance, the higher-order von Mangoldt functions ${\Lambda_k := \mu * L^k}$ are ${k}$-derived multiplicative functions, because ${L^k}$ is a ${k}$-derived completely multiplicative function. More explicitly, ${L^k}$ is the top order coeffiicent of the completely multiplicative function ${\prod_{i=1}^k \exp(\varepsilon_i L)}$, and ${\Lambda_k}$ is the top order coefficient of the multiplicative function ${\mu * \prod_{i=1}^k \exp(\varepsilon_i L)}$, with both functions taking values in the ring ${C[\varepsilon_1,\dots,\varepsilon_k]/(\varepsilon_1^2,\dots,\varepsilon_k^2)}$ of complex numbers with ${k}$ infinitesimals ${\varepsilon_1,\dots,\varepsilon_k}$ attached.

It then turns out that most (if not all) of the basic identities used by analytic number theorists concerning derived multiplicative functions, can in fact be viewed as coefficients of identities involving purely multiplicative functions, with the latter identities being provable primarily from multiplicative identities, such as (1). This phenomenon is analogous to the one in linear algebra discussed in this previous blog post, in which many of the trace identities used there are derivatives of determinant identities. For instance, the Leibniz rule

$\displaystyle L (f * g) = (Lf)*g + f*(Lg)$

for any arithmetic functions ${f,g}$ can be viewed as the top order term in

$\displaystyle \exp(\varepsilon L) (f*g) = (\exp(\varepsilon L) f) * (\exp(\varepsilon L) g)$

in the ring with one infinitesimal ${\varepsilon}$, and then we see that the Leibniz rule is a special case (or a derivative) of (1), since ${\exp(\varepsilon L)}$ is completely multiplicative. Similarly, the formulae

$\displaystyle \Lambda = \mu * L; \quad L = \Lambda * 1$

are top order terms of

$\displaystyle (\delta + \varepsilon \Lambda) = \mu * \exp(\varepsilon L); \quad \exp(\varepsilon L) = (\delta + \varepsilon \Lambda) * 1,$

and the variant formula ${\Lambda = - (L\mu) * 1}$ is the top order term of

$\displaystyle (\delta + \varepsilon \Lambda) = (\exp(-\varepsilon L)\mu) * 1,$

which can then be deduced from the previous identities by noting that the completely multiplicative function ${\exp(-\varepsilon L)}$ inverts ${\exp(\varepsilon L)}$ multiplicatively, and also noting that ${L}$ annihilates ${\mu*1=\delta}$. The Selberg symmetry formula

$\displaystyle \Lambda_2 = \Lambda*\Lambda + \Lambda L, \ \ \ \ \ (3)$

which plays a key role in the Erdös-Selberg elementary proof of the prime number theorem (as discussed in this previous blog post), is the top order term of the identity

$\displaystyle \delta + \varepsilon_1 \Lambda + \varepsilon_2 \Lambda + \varepsilon_1\varepsilon_2 \Lambda_2 = (\exp(\varepsilon_2 L) (\delta + \varepsilon_1 \Lambda)) * (\delta + \varepsilon_2 \Lambda)$

involving the multiplicative functions ${\delta + \varepsilon_1 \Lambda + \varepsilon_2 \Lambda + \varepsilon_1\varepsilon_2 \Lambda_2}$, ${\exp(\varepsilon_2 L)}$, ${\delta+\varepsilon_1 \Lambda}$, ${\delta+\varepsilon_2 \Lambda}$ with two infinitesimals ${\varepsilon_1,\varepsilon_2}$, and this identity can be proven while staying purely within the realm of multiplicative functions, by using the identities

$\displaystyle \delta + \varepsilon_1 \Lambda + \varepsilon_2 \Lambda + \varepsilon_1\varepsilon_2 \Lambda_2 = \mu * (\exp(\varepsilon_1 L) \exp(\varepsilon_2 L))$

$\displaystyle \exp(\varepsilon_1 L) = 1 * (\delta + \varepsilon_1 \Lambda)$

$\displaystyle \delta + \varepsilon_2 \Lambda = \mu * \exp(\varepsilon_2 L)$

and (1). Similarly for higher identities such as

$\displaystyle \Lambda_3 = \Lambda L^2 + 3 \Lambda L * \Lambda + \Lambda * \Lambda * \Lambda$

which arise from expanding out ${\mu * (\exp(\varepsilon_1 L) \exp(\varepsilon_2 L) \exp(\varepsilon_3 L))}$ using (1) and the above identities; we leave this as an exercise to the interested reader.

An analogous phenomenon arises for identities that are not purely multiplicative in nature due to the presence of truncations, such as the Vaughan identity

$\displaystyle \Lambda_{> V} = \mu_{\leq U} * L - \mu_{\leq U} * \Lambda_{\leq V} * 1 + \mu_{>U} * \Lambda_{>V} * 1 \ \ \ \ \ (4)$

for any ${U,V \geq 1}$, where ${f_{>V} = f 1_{>V}}$ is the restriction of a multiplicative function ${f}$ to the natural numbers greater than ${V}$, and similarly for ${f_{\leq V}}$, ${f_{>U}}$, ${f_{\leq U}}$. In this particular case, (4) is the top order coefficient of the identity

$\displaystyle (\delta + \varepsilon \Lambda)_{>V} = \mu_{\leq U} * \exp(\varepsilon L) - \mu_{\leq U} * (\delta + \varepsilon \Lambda)_{\leq V} * 1$

$\displaystyle + \mu_{>U} * (\delta+\varepsilon \Lambda)_{>V} * 1$

which can be easily derived from the identities ${\delta = \mu_{\leq U} * 1 + \mu_{>U} * 1}$ and ${\exp(\varepsilon L) = (\delta + \varepsilon \Lambda)_{>V} * 1 + (\delta + \varepsilon \Lambda)_{\leq V} + 1}$. Similarly for the Heath-Brown identity

$\displaystyle \Lambda = \sum_{j=1}^K (-1)^{j-1} \binom{K}{j} \mu_{\leq U}^{*j} * 1^{*j-1} * L \ \ \ \ \ (5)$

valid for natural numbers up to ${U^K}$, where ${U \geq 1}$ and ${K \geq 1}$ are arbitrary parameters and ${f^{*j}}$ denotes the ${j}$-fold convolution of ${f}$, and discussed in this previous blog post; this is the top order coefficient of

$\displaystyle \delta + \varepsilon \Lambda = \sum_{j=1}^K (-1)^{j-1} \binom{K}{j} \mu_{\leq U}^{*j} * 1^{*j-1} * \exp( \varepsilon L )$

and arises by first observing that

$\displaystyle (\mu - \mu_{\leq U})^{*K} * 1^{*K-1} * \exp(\varepsilon L) = \mu_{>U}^{*K} * 1^{*K-1} * \exp( \varepsilon L )$

vanishes up to ${U^K}$, and then expanding the left-hand side using the binomial formula and the identity ${\mu^{*K} * 1^{*K-1} * \exp(\varepsilon L) = \delta + \varepsilon \Lambda}$.

One consequence of this phenomenon is that identities involving derived multiplicative functions tend to have a dimensional consistency property: all terms in the identity have the same order of derivation in them. For instance, all the terms in the Selberg symmetry formula (3) are doubly derived functions, all the terms in the Vaughan identity (4) or the Heath-Brown identity (5) are singly derived functions, and so forth. One can then use dimensional analysis to help ensure that one has written down a key identity involving such functions correctly, much as is done in physics.

In addition to the dimensional analysis arising from the order of derivation, there is another dimensional analysis coming from the value of multiplicative functions at primes ${p}$ (which is more or less equivalent to the order of pole of the Dirichlet series at ${s=1}$). Let us say that a multiplicative function ${f: {\bf N} \rightarrow R}$ has a pole of order ${j}$ if one has ${f(p)=j}$ on the average for primes ${p}$, where we will be a bit vague as to what “on the average” means as it usually does not matter in applications. Thus for instance, ${1}$ or ${\exp(\varepsilon L)}$ has a pole of order ${1}$ (a simple pole), ${\delta}$ or ${\delta + \varepsilon \Lambda}$ has a pole of order ${0}$ (i.e. neither a zero or a pole), Dirichlet characters also have a pole of order ${0}$ (although this is slightly nontrivial, requiring Dirichlet’s theorem), ${\mu}$ has a pole of order ${-1}$ (a simple zero), ${\tau}$ has a pole of order ${2}$, and so forth. Note that the convolution of a multiplicative function with a pole of order ${j}$ with a multiplicative function with a pole of order ${j'}$ will be a multiplicative function with a pole of order ${j+j'}$. If there is no oscillation in the primes ${p}$ (e.g. if ${f(p)=j}$ for all primes ${p}$, rather than on the average), it is also true that the product of a multiplicative function with a pole of order ${j}$ with a multiplicative function with a pole of order ${j'}$ will be a multiplicative function with a pole of order ${jj'}$. The situation is significantly different though in the presence of oscillation; for instance, if ${\chi}$ is a quadratic character then ${\chi^2}$ has a pole of order ${1}$ even though ${\chi}$ has a pole of order ${0}$.

A ${k}$-derived multiplicative function will then be said to have an underived pole of order ${j}$ if it is the top order coefficient of a multiplicative function with a pole of order ${j}$; in terms of Dirichlet series, this roughly means that the Dirichlet series has a pole of order ${j+k}$ at ${s=1}$. For instance, the singly derived multiplicative function ${\Lambda}$ has an underived pole of order ${0}$, because it is the top order coefficient of ${\delta + \varepsilon \Lambda}$, which has a pole of order ${0}$; similarly ${L}$ has an underived pole of order ${1}$, being the top order coefficient of ${\exp(\varepsilon L)}$. More generally, ${\Lambda_k}$ and ${L^k}$ have underived poles of order ${0}$ and ${1}$ respectively for any ${k}$.

By taking top order coefficients, we then see that the convolution of a ${k}$-derived multiplicative function with underived pole of order ${j}$ and a ${k'}$-derived multiplicative function with underived pole of order ${j'}$ is a ${k+k'}$-derived multiplicative function with underived pole of order ${j+j'}$. If there is no oscillation in the primes, the product of these functions will similarly have an underived pole of order ${jj'}$, for instance ${\Lambda L}$ has an underived pole of order ${0}$. We then have the dimensional consistency property that in any of the standard identities involving derived multiplicative functions, all terms not only have the same derived order, but also the same underived pole order. For instance, in (3), (4), (5) all terms have underived pole order ${0}$ (with any Mobius function terms being counterbalanced by a matching term of ${1}$ or ${L}$). This gives a second way to use dimensional analysis as a consistency check. For instance, any identity that involves a linear combination of ${\mu_{\leq U} * L}$ and ${\Lambda_{>V} * 1}$ is suspect because the underived pole orders do not match (being ${0}$ and ${1}$ respectively), even though the derived orders match (both are ${1}$).

One caveat, though: this latter dimensional consistency breaks down for identities that involve infinitely many terms, such as Linnik’s identity

$\displaystyle \Lambda = \sum_{i=0}^\infty (-1)^{i} L * 1_{>1}^{*i}.$

In this case, one can still rewrite things in terms of multiplicative functions as

$\displaystyle \delta + \varepsilon \Lambda = \sum_{i=0}^\infty (-1)^i \exp(\varepsilon L) * 1_{>1}^{*i},$

so the former dimensional consistency is still maintained.

I thank Andrew Granville, Kannan Soundararajan, and Emmanuel Kowalski for helpful conversations on these topics.

[Note: the idea for this post originated before the recent preprint of Mochizuki on the abc conjecture was released, and is not intended as a commentary on that work, which offers a much more non-trivial perspective on scheme theory. -T.]

In classical algebraic geometry, the central object of study is an algebraic variety ${V}$ over a field ${k}$ (and the theory works best when this field ${k}$ is algebraically closed). One can talk about either affine or projective varieties; for sake of discussion, let us restrict attention to affine varieties. Such varieties can be viewed in at least four different ways:

• (Algebraic geometry) One can view a variety through the set ${V(k)}$ of points (over ${k}$) in that variety.
• (Commutative algebra) One can view a variety through the field of rational functions ${k(V)}$ on that variety, or the subring ${k[V]}$ of polynomial functions in that field.
• (Dual algebraic geometry) One can view a variety through a collection of polynomials ${P_1,\ldots,P_m}$ that cut out that variety.
• (Dual commutative algebra) One can view a variety through the ideal ${I(V)}$ of polynomials that vanish on that variety.

For instance, the unit circle over the reals can be thought of in each of these four different ways:

• (Algebraic geometry) The set of points ${\{ (x,y) \in {\bf R}^2: x^2+y^2 = 1 \}}$.
• (Commutative algebra) The quotient ${{\bf R}[x,y] / (x^2+y^2-1)}$ of the polynomial ring ${{\bf R}[x,y]}$ by the ideal generated by ${x^2+y^2-1}$ (or equivalently, the algebra generated by ${x,y}$ subject to the constraint ${x^2+y^2=1}$), or the fraction field of that quotient.
• (Dual algebraic geometry) The polynomial ${x^2+y^2-1}$.
• (Dual commutative algebra) The ideal ${(x^2+y^2-1)}$ generated by ${x^2+y^2-1}$.

The four viewpoints are almost equivalent to each other (particularly if the underlying field ${k}$ is algebraically closed), as there are obvious ways to pass from one viewpoint to another. For instance, starting with the set of points on a variety, one can form the space of rational functions on that variety, or the ideal of polynomials that vanish on that variety. Given a set of polynomials, one can cut out their zero locus, or form the ideal that they generate. Given an ideal in a polynomial ring, one can quotient out the ring by the ideal and then form the fraction field. Finally, given the ring of polynomials on a variety, one can form its spectrum (the space of prime ideals in the ring) to recover the set of points on that variety (together with the Zariski topology on that variety).

Because of the connections between these viewpoints, there are extensive “dictionaries” (most notably the ideal-variety dictionary) that convert basic concepts in one of these four perspectives into any of the other three. For instance, passing from a variety to a subvariety shrinks the set of points and the function field, but enlarges the set of polynomials needed to cut out the variety, as well as the associated ideal. Taking the intersection or union of two varieties corresponds to adding or multiplying together the two ideals respectively. The dimension of an (irreducible) algebraic variety can be defined as the transcendence degree of the function field, the maximal length of chains of subvarieties, or the Krull dimension of the ring of polynomials. And so on and so forth. Thanks to these dictionaries, it is now commonplace to think of commutative algebras geometrically, or conversely to approach algebraic geometry from the perspective of abstract algebra. There are however some very well known defects to these dictionaries, at least when viewed in the classical setting of algebraic varieties. The main one is that two different ideals (or two inequivalent sets of polynomials) can cut out the same set of points, particularly if the underlying field ${k}$ is not algebraically closed. For instance, if the underlying field is the real line ${{\bf R}}$, then the polynomial equations ${x^2+1=0}$ and ${1=0}$ cut out the same set of points, namely the empty set, but the ideal generated by ${x^2+1}$ in ${{\bf R}[x]}$ is certainly different from the ideal generated by ${1}$. This particular example does not work in an algebraically closed field such as ${{\bf C}}$, but in that case the polynomial equations ${x^2=0}$ and ${x=0}$ also cut out the same set of points (namely the origin), but again ${x^2}$ and ${x}$ generate different ideals in ${{\bf C}[x]}$. Thanks to Hilbert’s nullstellensatz, we can get around this problem (in the case when ${k}$ is algebraically closed) by always passing from an ideal to its radical, but this causes many aspects of the theory of algebraic varieties to become more complicated when the varieties involved develop singularities or multiplicities, as can already be seen with the simple example of Bezout’s theorem.

Nowadays, the standard way to deal with these issues is to replace the notion of an algebraic variety with the more general notion of a scheme. Roughly speaking, the way schemes are defined is to focus on the commutative algebra perspective as the primary one, and to allow the base field ${k}$ to be not algebraically closed, or even to just be a commutative ring instead of a field. (One could even consider non-commutative rings, leading to non-commutative geometry, but we will not discuss this extension of scheme theory further here.) Once one generalises to these more abstract rings, the notion of a rational function becomes more complicated (one has to work locally instead of globally, cutting out the points where the function becomes singular), but as a first approximation one can think of a scheme as basically being the same concept as a commutative ring. (In actuality, due to the need to localise, a scheme is defined as a sheaf of rings rather than a single ring, but these technicalities will not be important for the purposes of this discussion.) All the other concepts from algebraic geometry that might previously have been defined using one of the other three perspectives, are then redefined in terms of this ring (or sheaf of rings) in order to generalise them to schemes.

Thus, for instance, in scheme theory the rings ${{\bf R}[x]/(x^2)}$ and ${{\bf R}[x]/(x)}$ describe different schemes; from the classical perspective, they cut out the same locus, namely the point ${\{0\}}$, but the former scheme makes this point “fatter” than the latter scheme, giving it a degree (or multiplicity) of ${2}$ rather than ${1}$.

Because of this, it seems that the link between the commutative algebra perspective and the algebraic geometry perspective is still not quite perfect in scheme theory, unless one is willing to start “fattening” various varieties to correctly model multiplicity or singularity. But – and this is the trivial remark I wanted to make in this blog post – one can recover a tight connection between the two perspectives as long as one allows the freedom to arbitrarily extend the underlying base ring.

Here’s what I mean by this. Consider classical algebraic geometry over some commutative ring ${R}$ (not necessarily a field). Any set of polynomials ${P_1,\ldots,P_m \in R[x_1,\ldots,x_d]}$ in ${d}$ indeterminate variables ${x_1,\ldots,x_d}$ with coefficients in ${R}$ determines, on the one hand, an ideal

$\displaystyle I := (P_1,\ldots,P_m)$

$\displaystyle = \{P_1Q_1+\ldots+P_mQ_m: Q_1,\ldots,Q_m \in R[x_1,\ldots,x_d]\}$

in ${R[x_1,\ldots,x_d]}$, and also cuts out a zero locus

$\displaystyle V[R] := \{ (y_1,\ldots,y_d) \in R^d: P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \},$

since each of the polynomials ${P_1,\ldots,P_m}$ clearly make sense as maps from ${R^d}$ to ${R}$. Of course, one can also write ${V[R]}$ in terms of ${I}$:

$\displaystyle V[R] := \{ (y_1,\ldots,y_d) \in R^d: P(y_1,\ldots,y_d) = 0 \hbox{ for all } P \in I \}.$

Thus the ideal ${I}$ uniquely determines the zero locus ${V[R]}$, and we will emphasise this by writing ${V[R]}$ as ${V_I[R]}$. As the previous counterexamples illustrate, the converse is not true. However, whenever we have any extension ${R'}$ of the ring ${R}$ (i.e. a commutative ring ${R'}$ that contains ${R}$ as a subring), then we can also view the polynomials ${P_1,\ldots,P_m}$ as maps from ${(R')^d}$ to ${R'}$, and so one can also define the zero locus for all the extensions:

$\displaystyle V[R'] := \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}.$

As before, ${V[R']}$ is determined by the ideal ${I}$:

$\displaystyle V[R'] = V_I[R'] := \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P(y_1,\ldots,y_d) = 0 \hbox{ for all } P \in I \}.$

The trivial remark is then that while a single zero locus ${V_I[R]}$ is insufficient to recover ${I}$, the collection of zero loci ${V_I[R']}$ for all extensions ${R'}$ of ${R}$ (or more precisely, the assignment map ${R' \mapsto V_I[R']}$, known as the functor of points of ${V_I}$) is sufficient to recover ${I}$, as long as at least one zero locus, say ${V_I[R_0]}$, is non-empty. Indeed, suppose we have two ideals ${I, I'}$ of ${R[x_1,\ldots,x_d]}$ that cut out the same non-empty zero locus for all extensions ${R'}$ of ${R}$, thus

$\displaystyle V_I[R'] = V_{I'}[R'] \neq \emptyset$

for all extensions ${R'}$ of ${R}$. We apply this with the extension ${R'}$ of ${R}$ given by ${R' := R_0[x_1,\ldots,x_d]/I}$. Note that the embedding of ${R}$ in ${R_0[x_1,\ldots,x_d]/I}$ is injective, since otherwise ${I}$ would cut out the empty set as the zero locus over ${R_0}$, and so ${R'}$ is indeed an extension of ${R}$. Tautologically, the point ${(x_1 \hbox{ mod } I, \ldots, x_d \hbox{ mod } I)}$ lies in ${V_I[R']}$, and thus necessarily lies in ${V_{I'}[R']}$ as well. Unpacking what this means, we conclude that ${P \in I}$ whenever ${P \in I'}$, that is to say that ${I' \subset I}$. By a symmetric argument, we also have ${I \subset I'}$, and thus ${I=I'}$ as claimed. (As pointed out in comments, this fact (and its proof) is essentially a special case of the Yoneda lemma. The connection is tighter if one allows ${R'}$ to be any ring with a (not necessarily injective) map from ${R}$ into it, rather than an extension of ${R}$, in which case one can also drop the hypothesis that ${V_I[R_0]}$ is non-empty for at least one ${R_0}$. For instance, ${V_{(2)}[R'] = V_{(3)}[R'] = \emptyset}$ for every extension ${R'}$ of the integers, but if one also allows quotients such as ${{\bf Z}/(2)}$ or ${{\bf Z}/(3)}$ instead, then ${V_{(2)}[R']}$ and ${V_{(3)}[R']}$ are no longer necessarily equal.)

Thus, as long as one thinks of a variety or scheme as cutting out points not just in the original base ring or field, but in all extensions of that base ring or field, one recovers an exact correspondence between the algebraic geometry perspective and the commutative algebra perspective. This is similar to the classical algebraic geometry position of viewing an algebraic variety as being defined simultaneously over all fields that contain the coefficients of the defining polynomials, but the crucial difference between scheme theory and classical algebraic geometry is that one also allows definition over commutative rings, and not just fields. In particular, one needs to allow extensions to rings that may contain nilpotent elements, otherwise one cannot distinguish an ideal from its radical.

There are of course many ways to extend a field into a ring, but as an analyst, one way to do so that appeals particularly to me is to introduce an epsilon parameter and work modulo errors of ${O(\varepsilon)}$. To formalise this algebraically, let’s say for sake of concreteness that the base field is the real line ${{\bf R}}$. Consider the ring ${\tilde R}$ of real-valued quantities ${x = x_\varepsilon}$ that depend on a parameter ${\varepsilon \geq 0}$ (i.e. functions from ${{\bf R}^+}$ to ${{\bf R}}$), which are locally bounded in the sense that ${x}$ is bounded whenever ${\varepsilon}$ is bounded. (One can, if one wishes, impose some further continuity or smoothness hypotheses on how ${x}$ depends on ${\varepsilon}$, but this turns out not to be relevant for the following discussion. Algebraists often prefer to use the ring of Puiseux series here in place of ${\tilde R}$, and a nonstandard analyst might instead use the hyperreals, but again this will not make too much difference for our purposes.) Inside this commutative ring, we can form the ideal ${I}$ of quantities ${x = x_\varepsilon}$ that are of size ${O(\varepsilon)}$ as ${\varepsilon \rightarrow 0}$, i.e. there exists a quantity ${C>0}$ independent of ${\varepsilon}$ such that ${|x| \leq C\varepsilon}$ for all sufficiently small ${\varepsilon}$. This can easily be seen to indeed be an ideal in ${\tilde R}$. We then form the quotient ring ${R' := \tilde R/I := \{ x \hbox{ mod } I: x \in \tilde R \}}$. Note that ${x = y \hbox{ mod } I}$ is equivalent to the assertion that ${x = y + O(\varepsilon)}$, so we are encoding the analyst’s notion of “equal up to errors of ${O(\varepsilon)}$” into algebraic terms.

Clearly, ${R'}$ is a commutative ring extending ${{\bf R}}$. Hence, any algebraic variety

$\displaystyle V[{\bf R}] = \{ (y_1,\ldots,y_d) \in {\bf R}^d: P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}$

defined over the reals ${{\bf R}}$ (so the polynomials ${P_1,\ldots,P_m}$ have coefficients in ${{\bf R}}$), also is defined over ${R'}$:

$\displaystyle V[R'] = \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}.$

In language that more closely resembles analysis, we have

$\displaystyle V[R'] = \{ (y_1,\ldots,y_d) \in \tilde R^d: P_1(y_1,\ldots,y_d), \ldots, P_m(y_1,\ldots,y_d) = O(\varepsilon) \}$

$\displaystyle \hbox{ mod } I^d.$

Thus we see that ${V[R']}$ is in some sense an “${\varepsilon}$-thickening” of ${V[{\bf R}]}$, and is thus one way to give rigorous meaning to the intuition that schemes can “thicken” varieties. For instance, the scheme associated to the ideal ${(x)}$, when interpreted over ${R'}$, becomes an ${O(\varepsilon)}$ neighbourhood of the origin

$\displaystyle V_{(x)}[R'] = \{ y \in \tilde R: y = O(\varepsilon) \} \hbox{ mod } I,$

but the scheme associated to the smaller ideal ${(x^2)}$, when interpreted over ${R'}$, becomes an ${O(\varepsilon^{1/2})}$-neighbourhood of the origin, thus being a much “fatter” point:

$\displaystyle V_{(x^2)}[R'] = \{ y \in \tilde R: y^2 = O(\varepsilon) \} \hbox{ mod } I$

$\displaystyle = \{ y \in \tilde R: y = O(\varepsilon^{1/2} ) \} \hbox{ mod } I.$

Once one introduces the analyst’s epsilon, one can see quite clearly that ${V_{(x^2)}[R']}$ is coming from a larger scheme than ${V_{(x)}[R']}$, with fewer polynomials vanishing on it; in particular, the polynomial ${x}$ vanishes to order ${O(\varepsilon)}$ on ${V_{(x)}[R']}$ but does not vanish to order ${O(\varepsilon)}$ on ${V_{(x^2)}[R']}$.

By working with this analyst’s extension of ${{\bf R}}$, one can already get a reasonably good first approximation of what schemes over ${{\bf R}}$ look like, which I found particularly helpful for getting some intuition on these objects. However, since this is only one extension of ${{\bf R}}$, and not a “universal” such extension, it cannot quite distinguish any two schemes from each other, although it does a better job of this than classical algebraic geometry. For instance, consider the scheme cut out by the polynomials ${x^2, y^2}$ in two dimensions. Over ${R'}$, this becomes

$\displaystyle V_{(x^2,y^2)}[R'] = \{ (x,y) \in \tilde R^2: x^2, y^2 = O(\varepsilon) \} \hbox{ mod } I^2$

$\displaystyle = \{ (x,y) \in \tilde R^2: x, y = O(\varepsilon^{1/2}) \} \hbox{ mod } I^2.$

Note that the polynomial ${xy}$ vanishes to order ${O(\varepsilon)}$ on this locus, but ${xy}$ fails to lie in the ideal ${(x^2,y^2)}$. Equivalently, we have ${V_{(x^2,y^2)}[R'] = V_{(x^2,y^2,xy)}[R']}$, despite ${(x^2,y^2)}$ and ${(x^2,y^2,xy)}$ being distinct ideals. Basically, the analogue of the nullstellensatz for ${R'}$ does not completely remove the need for performing a closure operation on the ideal ${I}$; it is less severe than taking the radical, but is instead more like taking a “convex hull” in that one needs to be able to “interpolate” between two polynomials in the ideal (such as ${x^2}$ and ${y^2}$ to arrive at intermediate polynomials (such as ${xy}$) that one then places in the ideal.

One can also view ideals (and hence, schemes), from a model-theoretic perspective. Let ${I}$ be an ideal of a polynomial ring ${R[x_1,\ldots,x_d]}$ generated by some polynomials ${P_1,\ldots,P_m \in R[x_1,\ldots,x_d]}$. Then, clearly, if ${Q}$ is another polynomial in the ideal ${I}$, then we can use the axioms of commutative algebra (which are basically the axioms of high school algebra) to obtain the syntactic deduction

$\displaystyle P_1(x_1,\ldots,x_d) = \ldots = P_m(x_1,\ldots,x_d) = 0 \vdash Q(x_1,\ldots,x_d) = 0$

(since ${Q}$ is just a sum of multiples of ${P_1,\ldots,P_m}$). In particular, we have the semantic deduction

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \implies Q(y_1,\ldots,y_d) = 0 \ \ \ \ \ (1)$

for any assignment of indeterminates ${y_1,\ldots,y_d}$ in ${R}$ (or in any extension ${R'}$ of ${R}$). If we restrict ${y_1,\ldots,y_d}$ to lie in ${R}$ only, then (even if ${R}$ is an algebraically closed field), the converse of the above statement is false; there can exist polynomials ${Q}$ outside of ${I}$ for which (1) holds for all assignments ${y_1,\ldots,y_d}$ in ${R}$. For instance, we have

$\displaystyle y^2 = 0 \implies y = 0$

for all ${y}$ in an algebraically closed field, despite ${x}$ not lying in the ideal ${(x^2)}$. Of course, the nullstellensatz again explains what is going on here; (1) holds whenever ${Q}$ lies in the radical of ${I}$, which can be larger than ${I}$ itself. But if one allows the indeterminates ${y_1,\ldots,y_d}$ to take values in arbitrary extensions ${R'}$ of ${R}$, then the truth of the converse is restored, thus giving a “completeness theorem” relating the syntactic deductions of commutative algebra to the semantic interpretations of such algebras over the extensions ${R'}$. For instance, since

$\displaystyle y^2 = O(\varepsilon) \not \implies y = O(\varepsilon)$

we no longer have a counterexample to the converse coming from ${x}$ and ${(x^2)}$ once we work in ${R'}$ instead of ${{\bf R}}$. On the other hand, we still have

$\displaystyle x^2, y^2 = O(\varepsilon) \implies xy = O(\varepsilon)$

so the extension ${R'}$ is not powerful enough to detect that ${xy}$ does not actually lie in ${(x^2,y^2)}$; a larger ring (which is less easy to assign an analytic interpretation to) is needed to achieve this.

This will be a more frivolous post than usual, in part due to the holiday season.

I recently happened across the following video, which exploits a simple rhetorical trick that I had not seen before:

If nothing else, it’s a convincing (albeit unsubtle) demonstration that the English language is non-commutative (or perhaps non-associative); a linguistic analogue of the swindle, if you will.

Of course, the trick relies heavily on sentence fragments that negate or compare; I wonder if it is possible to achieve a comparable effect without using such fragments.

A related trick which I have seen (though I cannot recall any explicit examples right now; perhaps some readers know of some?) is to set up the verses of a song so that the last verse is identical to the first, but now has a completely distinct meaning (e.g. an ironic interpretation rather than a literal one) due to the context of the preceding verses.  The ultimate challenge would be to set up a Möbius song, in which each iteration of the song completely reverses the meaning of the next iterate (cf. this xkcd strip), but this may be beyond the capability of the English language.

On a related note: when I was a graduate student in Princeton, I recall John Conway (and another author whose name I forget) producing another light-hearted demonstration that the English language was highly non-commutative, by showing that if one takes the free group with 26 generators $a,b,\ldots,z$ and quotients out by all relations given by anagrams (e.g. $cat=act$) then the resulting group was commutative.    Unfortunately I was not able to locate this recreational mathematics paper of Conway (which also treated the French language, if I recall correctly); perhaps one of the readers knows of it?

Jean-Pierre Serre (whose papers are, of course, always worth reading) recently posted a lovely lecture on the arXiv entitled “How to use finite fields for problems concerning infinite fields”. In it, he describes several ways in which algebraic statements over fields of zero characteristic, such as ${{\mathbb C}}$, can be deduced from their positive characteristic counterparts such as ${F_{p^m}}$, despite the fact that there is no non-trivial field homomorphism between the two types of fields. In particular finitary tools, including such basic concepts as cardinality, can now be deployed to establish infinitary results. This leads to some simple and elegant proofs of non-trivial algebraic results which are not easy to establish by other means.

One deduction of this type is based on the idea that positive characteristic fields can partially model zero characteristic fields, and proceeds like this: if a certain algebraic statement failed over (say) ${{\mathbb C}}$, then there should be a “finitary algebraic” obstruction that “witnesses” this failure over ${{\mathbb C}}$. Because this obstruction is both finitary and algebraic, it must also be definable in some (large) finite characteristic, thus leading to a comparable failure over a finite characteristic field. Taking contrapositives, one obtains the claim.

Algebra is definitely not my own field of expertise, but it is interesting to note that similar themes have also come up in my own area of additive combinatorics (and more generally arithmetic combinatorics), because the combinatorics of addition and multiplication on finite sets is definitely of a “finitary algebraic” nature. For instance, a recent paper of Vu, Wood, and Wood establishes a finitary “Freiman-type” homomorphism from (finite subsets of) the complex numbers to large finite fields that allows them to pull back many results in arithmetic combinatorics in finite fields (e.g. the sum-product theorem) to the complex plane. (Van Vu and I also used a similar trick to control the singularity property of random sign matrices by first mapping them into finite fields in which cardinality arguments became available.) And I have a particular fondness for correspondences between finitary and infinitary mathematics; the correspondence Serre discusses is slightly different from the one I discuss for instance in here or here, although there seems to be a common theme of “compactness” (or of model theory) tying these correspondences together.

As one of his examples, Serre cites one of my own favourite results in algebra, discovered independently by Ax and by Grothendieck (and then rediscovered many times since). Here is a special case of that theorem:

Theorem 1 (Ax-Grothendieck theorem, special case) Let ${P: {\mathbb C}^n \rightarrow {\mathbb C}^n}$ be a polynomial map from a complex vector space to itself. If ${P}$ is injective, then ${P}$ is bijective.

The full version of the theorem allows one to replace ${{\mathbb C}^n}$ by an algebraic variety ${X}$ over any algebraically closed field, and for ${P}$ to be an morphism from the algebraic variety ${X}$ to itself, but for simplicity I will just discuss the above special case. This theorem is not at all obvious; it is not too difficult (see Lemma 4 below) to show that the Jacobian of ${P}$ is non-degenerate, but this does not come close to solving the problem since one would then be faced with the notorious Jacobian conjecture. Also, the claim fails if “polynomial” is replaced by “holomorphic”, due to the existence of Fatou-Bieberbach domains.

In this post I would like to give the proof of Theorem 1 based on finite fields as mentioned by Serre, as well as another elegant proof of Rudin that combines algebra with some elementary complex variable methods. (There are several other proofs of this theorem and its generalisations, for instance a topological proof by Borel, which I will not discuss here.)

Update, March 8: Some corrections to the finite field proof. Thanks to Matthias Aschenbrenner also for clarifying the relationship with Tarski’s theorem and some further references.

I had occasion recently to look up the proof of Hilbert’s nullstellensatz, which I haven’t studied since cramming for my algebra qualifying exam as a graduate student. I was a little unsatisfied with the proofs I was able to locate – they were fairly abstract and used a certain amount of algebraic machinery, which I was terribly rusty on – so, as an exercise, I tried to find a more computational proof that avoided as much abstract machinery as possible. I found a proof which used only the extended Euclidean algorithm and high school algebra, together with an induction on dimension and the obvious observation that any non-zero polynomial of one variable on an algebraically closed field has at least one non-root. It probably isn’t new (in particular, it might be related to the standard model-theoretic proof of the nullstellensatz, with the Euclidean algorithm and high school algebra taking the place of quantifier elimination), but I thought I’d share it here anyway.

Throughout this post, F is going to be a fixed algebraically closed field (e.g. the complex numbers ${\Bbb C}$). I’d like to phrase the nullstellensatz in a fairly concrete fashion, in terms of the problem of solving a set of simultaneous polynomial equations $P_1(x) = \ldots = P_m(x) = 0$ in several variables $x = (x_1,\ldots,x_d) \in F^d$ over F, thus $P_1,\ldots,P_m \in F[x]$ are polynomials in d variables. One obvious obstruction to solvability of this system is if the equations one is trying to solve are inconsistent in the sense that they can be used to imply 1=0. In particular, if one can find polynomials $Q_1,\ldots,Q_m \in F[x]$ such that $P_1 Q_1 + \ldots + P_m Q_m = 1$, then clearly one cannot solve $P_1(x)=\ldots=P_m(x)=0$. The weak nullstellensatz asserts that this is, in fact, the only obstruction:

Weak nullstellensatz. Let $P_1,\ldots,P_m \in F[x]$ be polynomials. Then exactly one of the following statements holds:

1. The system of equations $P_1(x)=\ldots=P_m(x)=0$ has a solution $x \in F^d$.
2. There exist polynomials $Q_1,\ldots,Q_m \in F[x]$ such that $P_1 Q_1 + \ldots + P_m Q_m = 1$.

Note that the hypothesis that F is algebraically closed is crucial; for instance, if F is the reals, then the equation $x^2+1=0$ has no solution, but there is no polynomial $Q(x)$ such that $(x^2+1) Q(x) = 1$.

Like many results of the “The only obstructions are the obvious obstructions” type, the power of the nullstellensatz lies in the ability to take a hypothesis about non-existence (in this case, non-existence of solutions to $P_1(x)=\ldots=P_m(x)=0$) and deduce a conclusion about existence (in this case, existence of $Q_1,\ldots,Q_m$ such that $P_1 Q_1 + \ldots + P_m Q_m = 1$). The ability to get “something from nothing” is clearly going to be both non-trivial and useful. In particular, the nullstellensatz offers an important correspondence between algebraic geometry (the conclusion 1 is an assertion that a certain algebraic variety is empty) and commutative algebra (the conclusion 2 is an assertion that a certain ideal is non-proper).

Now suppose one is trying to solve the more complicated system $P_1(x)=\ldots=P_d(x)=0; R(x) \neq 0$ for some polynomials $P_1,\ldots,P_d, R$. Again, any identity of the form $P_1 Q_1 + \ldots + P_m Q_m = 1$ will be an obstruction to solvability, but now more obstructions are possible: any identity of the form $P_1 Q_1 + \ldots + P_m Q_m = R^r$ for some non-negative integer r will also obstruct solvability. The strong nullstellensatz asserts that this is the only obstruction:

Strong nullstellensatz. Let $P_1,\ldots,P_m, R \in F[x]$ be polynomials. Then exactly one of the following statements holds:

1. The system of equations $P_1(x)=\ldots=P_m(x)=0$, $R(x) \neq 0$ has a solution $x \in F^d$.
2. There exist polynomials $Q_1,\ldots,Q_m \in F[x]$ and a non-negative integer r such that $P_1 Q_1 + \ldots + P_m Q_m = R^r$.

Of course, the weak nullstellensatz corresponds to the special case in which R=1. The strong nullstellensatz is usually phrased instead in terms of ideals and radicals, but the above formulation is easily shown to be equivalent to the usual version (modulo Hilbert’s basis theorem).

One could consider generalising the nullstellensatz a little further by considering systems of the form $P_1(x)=\ldots=P_m(x)=0, R_1(x),\ldots,R_n(x) \neq 0$, but this is not a significant generalisation, since all the inequations $R_1(x) \neq 0, \ldots, R_n(x) \neq 0$ can be concatenated into a single inequation $R_1(x) \ldots R_n(x) \neq 0$. The presence of the exponent r in conclusion (2) is a little annoying; to get rid of it, one needs to generalise the notion of an algebraic variety to that of a scheme (which is worth doing for several other reasons too, in particular one can now work over much more general objects than just algebraically closed fields), but that is a whole story in itself (and one that I am not really qualified to tell).

[Update, Nov 26: It turns out that my approach is more complicated than I first thought, and so I had to revise the proof quite a bit to fix a certain gap, in particular making it significantly messier than my first version. On the plus side, I was able to at least eliminate any appeal to Hilbert’s basis theorem, so in particular the proof is now manifestly effective (but with terrible bounds). In any case, I am keeping the argument here in case it has some interest.]