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According to Wikipedia https://en.wikipedia.org/wiki/Algebraic_matroid "For fields of characteristic zero (such as the real numbers) linear and algebraic matroids coincide"

I cannot find the two references it provides. There is a theorem that if you are algebraic over F then you are linear over an extension of F. In the case of characteristic zero, or more precisely, the complex numbers, is it true without going to an extension?

I checked the examples at the end of Oxley but none seem to have this property.

EDIT: According to the review in Mathscinet, this is proved in Ingleton, A. W. Representation of matroids. 1971 Combinatorial Mathematics and its Applications (Proc. Conf., Oxford, 1969) pp. 149–167 Academic Press, London

but I cannot access that reference, I just want to double check, or maybe see if there is another reference.

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Algebraic over $\mathbb{C}$ implies realizable over $\mathbb{C}$. If your matroid is algebraic over $\mathbb{C}$, then it is realizable over some $\mathbb{C} \subset K$, so it is realizable over the algebraic closure $\overline{K}$, which explicitly means that a list of equalities and inequalities over $\overline{K}$ has a solution. But then these equations and inequalities have a solution over $\mathbb{C}$ as well.

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    $\begingroup$ We could also argue that we can assume the transcendence degree of $K$ over $\mathbb{C}$ is finite (cardinality at most the cardinality of $\mathbb{C}$ will do), in which case $\bar{K}$ is isomorphic to $\mathbb{C}$. However, I believe that this isomorphism requires the axiom of choice, unlike David's argument. $\endgroup$
    – Richard Stanley
    Nov 9, 2016 at 20:54

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