# Representability of matroids over finite fields

I have several questions regarding representability of matroids.

Question 1. Does there exist a finite matroid that is representable over an infinite field, but is not representable over any finite field?

Question 2. Does there exist a finite matroid that is representable over a field of characteristic $$0$$, but is not representable over any field of positive characteristic?

Question 3. Does there exist a finite matroid that is representable over a field of characteristic $$0$$, but $$\{\mathrm{char}(F): M \text{ is representable over }F\}$$ is a finite set?

I care about Question 1 most, and find other quetions also interesting. I briefly checked Oxley, Matroid theory but did not find an answer.

• I think the answer to all three questions is no, by a spreading out argument. – Peter McNamara Oct 30 '18 at 3:56
• The answer to Q2 and Q3 is no, by considering possibilities of the characteristic set. – Bullet51 Oct 30 '18 at 4:34
• Since linear representability of a given finite matroid over a field $F$ is a property of $F$ expressible in first-order logic, negative answers to Questions 2 and 3 follow from the compactness theorem of first-order logic. Then the completeness of the theory of algebraically closed fields of any fixed characteristic implies that a finite matroid representable over a field $F$ (and thus over its algebraic closure) is also representable over the algebraic closure of the prime field of the same characteristic. A negative answer to Q1 then follows. – Andreas Blass Oct 30 '18 at 15:44

Theorem 1 (Rado, 1957). Let $$M$$ be a matroid representable over a field $$K$$. Then $$M$$ is representable over a simple algebraic extension of the prime field of $$K$$.
Theorem 2 (Rado, 1957). Let $$K$$ be an extension field of $$\mathbb{Q}$$ of degree $$N$$ , and let $$M$$ be a matroid representable over $$K$$. Then there is a positive integer $$c$$ such that given any prime $$p > c$$ there is a positive integer $$k = k(p) ≤ N$$ such that $$M$$ is representable over $$GF(p^k)$$. For infinitely many $$p$$, $$k(p) = 1$$.