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I attended a talk which generalized matroid realizability over a field to matroid realizability over division rings, and showed that the question of realizability is undecidable. However, they used a word problem arising from the division ring.

Is it known whether the question of "Is a matroid M realizable over any field F" is computable?

It seems to me that it shouldn't be, because they you would have a prover of all theorems akin to the Pappus theorem, but that's just my intuition.

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    $\begingroup$ Here's a somewhat related paper where Sturmfels shows that realizability over $\mathbb{Q}$ of matroids is equivalent to Hilbert's 10th problem over $\mathbb{Q}$: projecteuclid.org/download/pdf_1/euclid.bams/1183553968 $\endgroup$ Commented Oct 13, 2020 at 4:49
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    $\begingroup$ Of course I understand you're asking about realizability over any field. Possibly Mnev's Universality Theorem is relevant here. $\endgroup$ Commented Oct 13, 2020 at 4:49
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    $\begingroup$ Yes, it is computable. I think Oxley's book has a proof (but regardless of whether or not it does, it is true). $\endgroup$ Commented Oct 13, 2020 at 14:15
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    $\begingroup$ Found it: theorem 6.8.9. $\endgroup$ Commented Oct 13, 2020 at 14:24

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Contra my suspicions, the Internet is telling me that Vámos proved in "A necessary and sufficient condition for a matroid to be linear" (citation below) that it is decidable if a matroid is representable over a field. See quote on pg. 3 of Solving Rota’s Conjecture which says "The first exception is the algorithmic problem of determining when a given matroid is representable over an unspecified field, which was proved to be decidable by Vámos [28]."

Vamos, P., A necessary and sufficient condition for a matroid to be linear, Möbius Algebras, Conf. Proc. Waterloo 1971, 162-169 (1975). ZBL0374.05017.

EDIT:

Since the Vámos reference seems hard to track down, let me quote from the MathReviews [MR0349447] by J. E. Graver, which gives more details about the result:

The author gives a necessary and sufficient condition for a matroid to be $f$-linear, i.e., representable over some field. Let $E$ be a set and $\scr{E}$ a collection of subsets of $E$ satisfying the conditions for the independent sets of a matroid on $E$. We denote this matroid by the pair $(E,\mathscr{E})$. Fix $B\in \mathscr{E}$, a basis for the matroid. For each $x\in E$ define $S(a)$ as follows: $S(a)=\varnothing$ if $\{a\}\notin \mathscr{E}$; $S(a)=\{a\}$ if $a\in B$; otherwise $S(a)$ is the unique subset of $B$ such that $S(a)∪\{a\}$ is a circuit. Let $\Gamma=\{(a,b):a\in E,b\in S(a)\}$, and consider $R=Z[X_{\gamma}]$, the ring of polynomials over the integers in variables $\{X_\gamma: \gamma \in \Gamma\}$. A finite subset $A \subseteq E$ is said to be regular if $|A|=|\bigcup_{a \in A}S(a)|$. For each regular set $A$ a polynomial in $R$ is constructed as follows: Let $r_{(a,b)}=X_{(a,b)}$ if $a \in A$ and $b \in S(a)$, and $r_{(a,b)}=0$ if $a\in A$ and $b\in A−S(a)$. Ordering the elements of $A$, constructing the $|A|\times |A|$ matrix $[r_{(a,b)}]$ and taking the determinant yield a polynomial $P(A)$, unique up to sign. However, the ideal $I$ generated by $\{P(A):\textrm{$A$ is regular}, A\notin \mathscr{E}\}$ and the multiplicatively closed set $T$ generated by $\{P(A):\textrm{$A$ is regular}, A\in \mathscr{E}\}$ are uniquely determined. We may now state the main result. Theorem: The matroid $E, \mathscr{E}$ is $f$-linear if and only if $T\cap I=\varnothing$.

Note that in the review there is a confusing typo where it describes $T$ as generated by $\{P(A):\textrm{$A$ is regular}, A\notin \mathscr{E}\}$ rather than $\{P(A):\textrm{$A$ is regular}, A\in \mathscr{E}\}$. Note also, as pointed out in the comments by Geva Yashfe, that this result appears as Theorem 6.8.9 in the 2nd edition of Oxley's Matroid Theory text, which is probably a more accessible source. To deduce decidability from this algebraic crieterion, Oxley appeals to Gröbner basis theory.

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  • $\begingroup$ Nice find! I'm currently trying to figure out a way to access that paper. $\endgroup$ Commented Oct 13, 2020 at 7:15
  • $\begingroup$ However, that problem is still different from my problem, I think? Specifying a field for the algorithm is different from allowing ANY field. $\endgroup$ Commented Oct 13, 2020 at 7:16
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    $\begingroup$ @bottledcaps: I'm pretty sure it's the same problem. You're asking: is there any F such that my M is realizable, right? $\endgroup$ Commented Oct 13, 2020 at 12:15
  • $\begingroup$ I agree that this reference (what looks like conference proceedings) is pretty hard to track down though. $\endgroup$ Commented Oct 13, 2020 at 13:44
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    $\begingroup$ @SamHopkins There is a mistake in the statement above, and possibly in the review. It should be $T = \{P(A) : A\text{ is regular, } A \in \mathscr{E}\}$. (Instead of $\notin$.) The point is: you want to invert the determinants which should be invertible, demand the determinants of dependent subsets' column-sets vanish, and check whether the resulting quotient is not the zero ring (in which case there is a maximal ideal $\mathfrak{m}$ containing $I$ in a localization inverting $T$, and this gives a representation in the field $\mathbb{Z}[x_{i,j}]/\mathfrak{m}$). $\endgroup$ Commented Oct 13, 2020 at 14:31
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Sam Hopkins has answered what I think is your stated question. But in the comments, it seems that you're also interested in the following question: Given a matroid $M$ and field $F$, is it decidable whether $M$ is $F$-representable?

In Oxley's book Matroid Theory (2nd edition), he discusses this problem on pages 226–227. If $F$ is a finite field then decidability is trivial since the rank of the matroid tells you the dimension of the vector space to try. For infinite fields the situation is more subtle; for starters, one has to address the question of what it means to be "given" $F$. Gröbner basis algorithms may solve the problem if $F$ is algebraically closed. On the other hand, Sturmfels (On the decidability of Diophantine problems in combinatorial geometry, Bull. Amer. Math. Soc. 17 (1987), 121–124) has shown that for $F=\mathbb{Q}$, the question is equivalent to Hilbert's 10th problem for $\mathbb{Q}$, which remains an open problem at the time of this writing.

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  • $\begingroup$ For $\mathbb{R}$ it is apparently also decidable as mentioned here as a result of the real nullstellsatz: mathoverflow.net/a/48185/25028. I wonder if there is any field $F$ (however artificial) for which realizability is known to be undecidable? $\endgroup$ Commented Oct 13, 2020 at 23:09
  • $\begingroup$ Ahh, I realize I am thinking of the problem: fix a field $F$; is there an algorithm which decides if a matroid $M$ is $F$-representable? As you say, if $F$ is part of the input you have to be careful about what it means to be given $F$. But I think the field-by-field algorithm question is totally clear. $\endgroup$ Commented Oct 13, 2020 at 23:31
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    $\begingroup$ @SamHopkins This is equivalent to solvability of polynomial equations in $F$ with coefficients in $\mathbb{Z}$. There are fields for which there are strong undecidability results (for instance, equations with coefficients in $\mathbb{Z}[t_1,t_2]$ cannot be solved algorithmically over $\mathbb{C}(t_1,t_2)$), but I'm not aware of any $F$ (unless very artificial) for which it is known. $\endgroup$ Commented Oct 14, 2020 at 10:15
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A matroid being representable is equivalent to a certain locally closed subscheme of the Grassmannian, defined by the non-vanishing of the Plücker coordinates corresponding to bases and the vanishing of the Plücker coordinates coordinates corresponding to non-bases, being nonempty. This can be done (using Gröbner basis algorithms) in a few different ways.

In the paper Singular matroid realization spaces, Dan Corey and Dante Luber have written code in OSCAR which does this, apparently quite efficiently. Remarkably, they were able to determine exactly which matroids of rank 3 on 11 elements are realizable over $\mathbb{C}$.

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  • $\begingroup$ The link to the GitHub repository for their code that they post in the paper (being this one) is broken... $\endgroup$ Commented Sep 9, 2023 at 4:00
  • $\begingroup$ This answers (and IMHO is the “right” answer to) the question of algorithmically deciding whether a matroid, or indeed any configuration of linear subspaces, is representable over a field of given characteristic. I think one needs to add a little explanation as to why it's decidable whether a system of algebraic equations (with coefficients in $\mathbb{Z}$) has a solution over any field (which is true, but the reason momentarily escapes me). $\endgroup$
    – Gro-Tsen
    Commented Sep 9, 2023 at 7:13
  • $\begingroup$ There are Gröbner basis algorithms that work over $\mathbb{Z}$, e.g., can decide is some polynomials generate the unit ideal over $\mathbb{Z}$. $\endgroup$ Commented Sep 9, 2023 at 15:43
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Being decidable is a very weak property - one can also ask about the exact complexity class of this problem. In fact, post-Mnёv theorem this is extremely well understood. I recommend Matoušek's survey which explains why over $\Bbb R$ the problem is $\exists\Bbb R$-complete. Here "realizability" in the special case of rank $3$ oriented matroids becomes a question of "stretchability" of pseudoline arrangements.

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