Is matroid realizability computable?

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.

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.