# Four proofs of a theorem by Vámos

Let $\chi(M)$ be the characteristic set of the matroid $M$; that is, $\chi(M)=\{\operatorname{char}(\mathbb{F})\colon M\ \text{is representable over the field}\ \mathbb{F}\},$ where $\operatorname{char}(\mathbb{F})$ is the characteristic of $\mathbb{F}$. It is well known that $0\in\chi(M)$ if and only if $\chi(M)$ contains infinitely many primes. One direction is due to Rado [4]. The following theorem contains the other direction.

Theorem 1. If $\chi(M)$ contains infinitely many primes, then it contains zero.

One unusual aspect of Theorem 1 is that there are four proofs of it in the literature. All of them are attractive, and they call upon different (but related) sets of tools: Zorn’s lemma, compactness in first-order logic, ultraproducts, and Gröbner bases. In this post I will review all four proofs.

To start with, let $r$ be $r(M)$, and assume that $E(M)=\{1,\ldots, n\}$. Let $X$ be the set of variables $\{x_{i,j}\}_{1\leq i \leq r,\ 1\leq j \leq n}$, and let $R$ be the polynomial ring $\mathbb{Z}[X]$. We will let $A$ be the $r\times n$ matrix where the entry in row $i$ and column $j$ is $x_{i,j}$. Each $r$-element subset of $E(M)$ corresponds to an $r\times r$ submatrix of $A$, and thereby to a homogeneous polynomial in $R$, namely the determinant of that submatrix. Let $\mathcal{B}$ be the set of polynomials that correspond to bases of $M$, and let $\mathcal{N}$ contain the polynomials that correspond to dependent $r$-element subsets in $M$. Thus a representation of $M$ is a ring homomorphism from $R$ to a field that sends every polynomial in $\mathcal{N}$, but no polynomial in $\mathcal{B}$, to zero.

The first proof of Theorem 1 is due to Peter Vámos. He published two articles in the proceedings entitled Möbius Algebras, produced by the University of Waterloo in 1971. The proof of Theorem 1 has sometimes been cited as coming from [6], but in fact the correct attribution is to [5]. His proof hinges on the following lemma. Recall that an ideal, $P$, is prime, if $ab\in P$ implies that $a\in P$ or $b\in P$.

Lemma 1. Let $I$ be an ideal in a commutative ring, $S$, and let $T$ be a multiplicatively closed set of $S$ that is disjoint from $I$. There exists a prime ideal, $P$, such that $I\subseteq P$, and $P\cap T=\emptyset$.

Proof. We consider the partial order of ideals in $S$ that contain $I$ and that are disjoint from $T$. Assume that $J_{1}\subset J_{2}\subset J_{3}\subset \cdots$ is a chain in the order. It is easy to confirm that the $J_{1}\cup J_{2}\cup J_{3}\cup \cdots$ is itself an ideal in the order. This means that every such chain of ideals has a maximal element, so we can apply Zorn’s Lemma. Therefore the partial order has a maximal element, $P$. It remains to show that $P$ is a prime ideal. Assume otherwise, so that the product $ab$ belongs to $P$, even though $a\notin P$ and $b\notin P$. The maximality of $P$ means that the ideal generated by $P$ and $a$ is not disjoint from $T$, so there is some element, $ca+p$ in $T\cap \langle P, a\rangle$. (Here $p\in P$, and $c$ is an element of $S$.) Similarly, there are elements $q\in P$ and $d\in S$ such that $db+q$ belongs to $T$. Since $T$ is multiplicatively closed, we see that $(ca+p)(db+q)=(cd)(ab)+(ca)q+(db)p+pq$ is in $T$. But this element is also in $P$, since $ab\in P$. Now we have a contradiction to the fact that $P$ and $T$ are disjoint. $\square$

Proof 1. Let $p_{1},p_{2},p_{3},\ldots$ be distinct primes such that $M$ is representable over a field of characteristic $p_{i}$ for each $i$. Let $\phi_{i}$ be a homomorphism from $R$ to a field of characteristic $p_{i}$ that takes all the polynomials in $\mathcal{N}$, and no polynomial in $\mathcal{B}$, to zero. Then the kernel, $\ker(\phi_{i})$, of $\phi_{i}$ is a prime ideal of $R$ containing $\mathcal{N}$, but disjoint from $\mathcal{B}$. Moreover, the integer $p_{i}$ is contained in $\ker(\phi_{i})$. In fact, the only integers in $\ker(\phi_{i})$ are the multiples of $p_{i}$, since if $\ker(\phi_{i})$ contains a non-multiple of $p_{i}$, then it contains the greatest common divisor of that non-multiple and $p_{i}$ (namely $1$). This would imply that $\ker(\phi_{i})$ is equal to $R$, which is impossible.

Assume that there is an element in both the ideal $\langle \mathcal{N}\rangle$ and the multiplicatively closed set generated by $\mathcal{B}\cup\mathbb{Z}$. Such an element is of the form $kf$, where $k$ is an integer, and $f$ is a product of polynomials from $\mathcal{B}$. Since $kf$ is in $\langle \mathcal{N}\rangle$, it is in each $\ker(\phi_{i})$, even though $f$ is not. Thus $k$ is in each ideal $\ker(\phi_{i})$. Thus $k$ is a multiple of each of the distinct primes $p_{1},p_{2},p_{3},\ldots$, and we have a contradiction. Now, by Lemma $1$, there is a prime ideal, $P$, of $R$ which contains $\langle \mathcal{N}\rangle$, and which avoids $\mathcal{B}\cup\mathbb{Z}$. Because $P$ is prime, the quotient ring, $R/P$, is an integral domain. Let $F$ be the field of fractions of $R/P$. The natural homomorphism from $R$ to $F$ gives us a representation of $M$. As $P$ does not contain any integer, it follows that $R/P$ has characteristic zero, and we are done. $\square$

The next proof of Theorem 1 makes use of the Compactness Theorem in first-order logic. It is due to Samuel Wagstaff and appeared two years after Vámos’s proof, in 1973 [7]. Wagstaff does not supply many details — perhaps the account in his PhD thesis (upon which the article based) is more explicit. But I believe that my version below captures his key ideas.

Proof 2. We construct a first-order language of vector spaces. We have a countable supply of variables, $x_{1},x_{2},x_{3},\ldots$, and unary predicates $\operatorname{Scal}$ and $\operatorname{Vec}$. In our interpretation, these predicates will indicate that a variable stands for a scalar or a vector. We also have constants, $0_{S}$, $1_{S}$, and $0_{V}$, which are interpreted as the identities of the vector space. The functions $+_{S}$, $\times_{S}$, $+_{V}$, and $\times_{V}$, are interpreted as the binary operations of the field of scalars, as addition of vectors, and as scalar multiplication. The only binary relation we need is equality. Now, in this language, we can express the axioms for fields and for vector spaces. To illustrate, the compatibility of scalar and vector multiplication would be expressed with the sentence \begin{multline*}\forall x_{1}\forall x_{2}\forall x_{3} (\operatorname{Scal}(x_{1})\land \operatorname{Scal}(x_{2})\land \operatorname{Vec}(x_{3}) \to\\ ((x_{1}\times_{S}x_{2})\times_{V}x_{3}=x_{1}\times_{V}(x_{2}\times_{V}x_{3}))).\end{multline*} We can assert the representability of $M$ by requiring that there exist $x_{1},x_{2},\ldots, x_{n}$ such that $\operatorname{Vec}(x_{i})$ holds for each $i$. For each subset $X\subseteq E(M)=\{1,\ldots, n\}$, we include a sentence asserting that the corresponding subset of $\{x_{1},x_{2},\ldots, x_{n}\}$ is linearly independent if $X$ is independent in $M$, and otherwise asserting that it is linearly dependent. Let us illustrate how this can be accomplished by writing the formula which asserts that $\{x_{1},x_{2},x_{3}\}$ is linearly independent. Let $y_{1}, y_{2}, y_{3}$ be variables that are not used elsewhere in our sentence. The formula we need is as follows: \begin{multline*}\forall y_{1}\forall y_{2}\forall y_{3} (\operatorname{Scal}(y_{1})\land \operatorname{Scal}(y_{2})\land \operatorname{Scal}(y_{3}) \land (y_{1}\ne 0_{S} \lor y_{2}\ne 0_{S} \lor y_{3}\ne 0_{S})\to\\ (y_{1}\times_{V}x_{1}) +_{V} (y_{2}\times_{V} x_{2}) +_{V} (y_{3}\times_{V}x_{3})\ne 0_{V}).\end{multline*} We simply perform this trick for each subset of $E(M)$.

Now we have a finite collection, $\mathcal{T}$, of sentences which can be satisfied if and only if $M$ is representable over a field. For each prime, $p$, let $S_{p}$ be the sentence asserting that the sum of $p$ copies of $1_{S}$ is equal to $0_{S}$. Thus $M$ is representable over a field of characteristic $p$ if and only if $\mathcal{T}\cup\{S_{p}\}$ can be satisfied. Assume that $M$ is not representable over a field with characteristic zero. Then the collection $\mathcal{T}\cup\{\neg S_{p}\colon p\ \text{is a prime}\}$ is inconsistent: it cannot be satisfied by any model. The Compactness Theorem tells us that there is a finite subcollection which is inconsistent. Therefore we can assume that $\mathcal{T}\cup\{\neg S_{p} \colon p\ \text{is a prime and}\ p\leq N\}$ has no model, for some integer $N$. This implies that there can be no representation of $M$ over a field with characteristic greater than $N$, so we have proved the contrapositive of Theorem 1. $\square$

The next proof appeared in 1989, and comes courtesy of Imre Leader [3]. I think it might be my favourite out of the four. It uses the notion of an ultraproduct.

Proof 3. Let $F_{1},F_{2},F_{3},\ldots$ be an infinite sequence of fields such that $M$ is representable over each $F_{i}$ and $\operatorname{char}(F_{i+1})>\operatorname{char}(F_{i})$. Let $\mathcal{U}$ be a non-principal ultrafilter of the positive integers. This means that $\mathcal{U}$ is a collection of subsets of positive integers, and $\mathcal{U}$ satisfies: (1) any superset of a member of $\mathcal{U}$ is also in $\mathcal{U}$; (2) if $U,V\in\mathcal{U}$, then $U\cap V\in \mathcal{U}$; and (3) if $U$ is any subset of $\mathbb{Z}^{+}$, then exactly one of $U$ or $\mathbb{Z}^{+}-U$ is in $\mathcal{U}$. Non-principal ultrafilters cannot be described explicitly, but we know that they exist by an application of Zorn’s Lemma. If $U$ is a finite subset of $\mathbb{Z}^{+}$, then $\mathbb{Z}^{+}-U$ is in $\mathcal{U}$. We consider sequences $(x_{1},x_{2},x_{3},\ldots)$, where each $x_{i}$ is an element of $F_{i}$. We declare $(x_{1},x_{2},x_{3},\ldots)$ and $(y_{1},y_{2},y_{3},\ldots)$ to be equivalent if $\{i\in\mathbb{Z}^{+}\colon x_{i}=y_{i}\}$ is a member of $\mathcal{U}$. We perform componentwise addition and multiplication on sequences. It is not too difficult to see that these operations respect our notion of equivalence, and that the equivalence classes of sequences form a field, $F$, under componentwise addition and multiplication. Moreover, the characteristic of $F$ is zero. We consider a sequence of representations of $M$ over the fields $F_{1},F_{2},F_{3},\ldots$. This leads to a representation of $M$ over $F$ in the obvious way: we simply “stack” the infinite sequence of matrices, and produce a matrix whose entries are infinite sequences of numbers. $\square$

Finally, we come full circle! In 2003, Vámos (along with Rosemary Baines) published another proof of Theorem 1 [2]. The proof is actually an easy consequence of the algorithm that Stefan mentioned in a previous post, and the main ingredient is the use of Gröbner bases. I will tweak the proof slightly, for the sake of simplicity.

Proof 4. Let $f$ be the product of all polynomials in $\mathcal{B}$. We introduce the new, “dummy”, variable, $y$. From now on, we will operate in the ring $\mathbb{Z}[X\cup\{y\}]$. Note that an evaluation of the variables in $X\cup\{y\}$ will make $yf-1$ zero if and only if it makes each polynomial in $\mathcal{B}$ non-zero. Let $I$ be the ideal of $\mathbb{Z}[X\cup\{y\}]$ that is generated by $\mathcal{N}\cup\{yf-1\}$. Thus a representation of $M$ is really a ring homomorphism from $\mathbb{Z}[X\cup\{y\}]$ to a field that takes every polynomial in $I$ to zero.

Let $G$ be a strong Gröbner basis of $I$ (see [1]). Thus $I$ is the ideal generated by $G$. In addition, we have an ordering on products of variables (the lexicographic order will suffice). The leading term of a polynomial is the highest non-zero term under this ordering. For $G$ to be a strong Gröbner basis, we require that if $h$ is in $I$, then there is a polynomial $g\in G$ such that the leading term of $g$ divides the leading term of $h$. Assume that $G$ contains an integer, $N$. If $\phi$ is a homomorphism from $\mathbb{Z}[X\cup\{y\}]$ to a field, $F$, such that $\ker(\phi)$ contains $I$, then $N$ is taken to zero by $\phi$. This means that the characteristic of $F$ is at most $N$, so $M$ is not representable over infinitely many prime characteristics. Therefore we will now assume that $G$ contains no integer.

Let $I_{\mathbb{Q}}$ be the ideal of $\mathbb{Q}[X\cup\{y\}]$ generated by $\mathcal{N}\cup\{yf-1\}$. Assume that $1\in I_{\mathbb{Q}}$. Then $1$ can be expressed as a polynomial combination of the polynomials in $\mathcal{N}\cup\{yf-1\}$. By multiplying this combination through by the denominators of the coefficients, we see that $I$ contains an integer, $N$. Thus $G$ contains a polynomial whose leading term divides the leading term of $N$, which is $N$ itself. But in the lexicographic ordering, a constant polynomial is below all non-constant polynomials, from which we deduce that $G$ contains a constant polynomial, contradicting our conclusion from the previous paragraph. Therefore $1\notin I_{\mathbb{Q}}$, which means that $I_{\mathbb{Q}}$ is contained in a maximal ideal, $P$. Note that $P$ contains no constant polynomial. Now $\mathbb{Q}[X\cup\{y\}]/P$ is a field, and the natural homomorphism from $\mathbb{Q}[X\cup\{y\}]$ to this field leads to a representation of $M$ over a field of characteristic zero. $\square$

References.

[1] W. W. Adams and P. Loustaunau. An introduction to Gröbner bases. Graduate Studies in Mathematics, 3. American Mathematical Society, Providence, RI, 1994.

[2] R. Baines and P. Vámos. An algorithm to compute the set of characteristics of a system of polynomial equations over the integers. J. Symbolic Comput. 35 (2003), no. 3, 269–279.

[3] I. Leader. A short proof of a theorem of Vámos on matroid representations. Discrete Math. 75 (1989), no. 1-3, 315–317.

[4] R. Rado. Note on independence functions. Proc. London Math. Soc. (3) 7 (1957), 300–320.

[5] P. Vámos. A necessary and sufficient condition for a matroid to be linear. Möbius algebras (Proc. Conf., Univ. Waterloo, Waterloo, Ont., 1971), pp. 162–169. Univ. Waterloo, Waterloo, Ont., 1971.

[6] P. Vámos. Linearity of matroids over division rings. Notes by G. Roulet. Möbius algebras (Proc. Conf., Univ. Waterloo, Waterloo, Ont., 1971), pp. 170–174. Univ. Waterloo, Waterloo, Ont., 1971.

[7] S. S. Wagstaff Jr. Infinite matroids. Trans. Amer. Math. Soc. 175 (1973), 141–153.

## 7 thoughts on “Four proofs of a theorem by Vámos”

1. What a nice post, Dillon! I only knew the Groebner basis argument.

I was a bit confused by proof #2 at first, but I think you made a typo. Don’t you mean that if M has no representation in char. 0, then T + { *not* S_p : p prime } is inconsistent?

• Thanks Rudi, I’ve corrected the typo.

2. Lemma 1 (and hence Proof 1) depends on Zorn’s Lemma (which is equivalent to the Axiom of Choice). We can also use Zorn’s Lemma to prove the Compactness Theorem (which underlies Proof 2), and the existence of ultrafilters (which are necessary for Proof 3). However, it is possible to prove Theorem 1 without the Axiom of Choice. The Compactness Theorem and the existence of ultrafilters are equivalent to each other, and they are independent of the Axiom of Choice. I’m not sure about Proof 4. It relies on the existence of Groebner bases, which in turn depends on the Hilbert Basis Theorem. I don’t know what axioms are required for that theorem.

I think there may be an opportunity for some reverse mathematics here: which axiom scheme is necessary to prove Theorem 1?

• I don’t think you need any power tools from logic to prove that Groebner bases exist: it suffices that the Buchberger algorithm finishes on any input, which is true for ‘combinatorial’ reasons.

But these other proofs (especially 2) are really elegant.

Proof #3 — which I don’t really understand — looks a lot like ‘the product partial field of infinitely many finite fields has a homomorphism to a field of char 0’.

• That’s interesting – I don’t think I’ve seen a ‘combinatorial’ proof that the Buchberger algorithm terminates. Adams and Loustaunau rely on Hilbert’s Basis Theorem. Can you link to some details?

I think your interpretation of the ultraproduct is exactly right.

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