# When is there a subring of the complex numbers surjecting onto a given field of prime characteristic?

To make use of the Lie algebra action of $$\mathsf{gl}_2(\mathbb{C})$$ to establish a isomorphism in modular representation theory, I would like an answer to this question:

Let $$K$$ be a field of prime characteristic. When is there a subring $$R$$ of the complex numbers and a maximal ideal $$M$$ of $$R$$ such that $$R/M \cong K$$?

Clearly no such ring $$R$$ exists if $$K$$ has strictly more than $$|\mathbb{C}|$$ independent transcendental elements. Is this the only obstruction? Is there a reasonably explicit way to construct a suitable $$R$$ when $$K$$ is the algebraic closure of $$\mathbb{F}_p$$?

As a follow-up (which at first I thought I needed, but I now see I can get around by working with $$\mathrm{GL}_2(\mathbb{C})$$ rather than $$\mathrm{SL}_2(\mathbb{C})$$), note that if $$R/M \cong K$$ then the induced map $$\mathrm{GL}_d(R) \rightarrow \mathrm{GL}_d(K)$$, defined on a $$d \times d$$ matrix with entries in $$R$$ by applying $$R \twoheadrightarrow R/M \cong K$$ to each entry, is a surjective group homomorphism.

Is is true in general that the restriction of the group homomorphism $$\mathrm{GL}_d(R) \rightarrow \mathrm{GL}_d(K)$$ to $$\mathrm{SL}_d(R)$$ is surjective onto $$\mathrm{SL}_d(K)$$, or are there further obstructions?

• If $X$ is a transcendence base for $\mathbb C$ over $\mathbb Q$, then the subring it generates is a free commutative ring over $X$. That subring has a surjection onto any ring of cardinality at most $|X|=|\mathbb C|$. – Keith Kearnes Feb 26 '20 at 15:15
• Follow-up question: yes (exercise). Hint: elementary matrices. – YCor Feb 26 '20 at 16:26
• Clearly this is very easy for some people. But it genuinely comes up in my research and I think the down-voters are being rather harsh. – Mark Wildon Feb 26 '20 at 16:29
• @MarkWildon no, it's that it generates $SL_d(K)$ for $K$ a field, and the arrow is in the right direction. The map $SL_d(R)\to SL_d(K)$ is already surjective in restriction to the elementary subgroup $E_d(R)$. – YCor Feb 26 '20 at 18:40
• An algebraic closure of $\mathbb{F}_p$ is given by $\bar{\mathbb{F}}_p = R/M$ where $R$ is the ring of algebraic integers and $M$ any maximal ideal containing $p$. – Todd Leason Feb 28 '20 at 5:04

The complex numbers have transcendence degree the continuum over $$\mathbb Q$$ so contain a copy of the field of rational functions in continuum many variables over $$\mathbb Q$$. This in turn contains the ring of polynomials in continuum many variables over $$\mathbb Z$$, which surjects onto any ring of cardinality at most the continuum.
• Thank you. This answers my first question. Maybe it's too much to ask for a more explicit construction, but I would be interested to know when the subring can be taken inside the ring of algebraic integers of $\mathbb{C}$. – Mark Wildon Feb 26 '20 at 16:22
• @MarkWildon I think the construction of a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$ (or one of $\mathbb{R}$ over $\mathbb{Q}$) is not easy (up to my untutored knowledge, only 3 explicit algebraically independent elements are known !). – Duchamp Gérard H. E. Feb 26 '20 at 17:43
• The subring can be taken inside the algebraic integers iff $K$ is an algebraic extension of $\mathbb{F}_p$. For, let $R$ be the ring of algebraic integers in $\mathbb{C}$. Since any subring $S$ of $R$ is an algebraic extension of $\mathbb{Z}$, the quotient of $S$ by a maximal ideal $m$ is an algebraic extension of $\mathbb{F}_p$ where $(p)=\mathbb{Z}\cap m$. ... – Todd Leason Feb 28 '20 at 5:29
• ... Conversely, let $K$ be an algebraic extension of $\mathbb{F}_p$. Let $M$ be a maimal ideal of $R$ containing $p$. Then $R/M$ is an algebraic closure of $\mathbb{F}_p$. Hence we can assume $K \le R/M$. Let $\pi: R \to R/M$ and let $S = \pi^{-1}(K)$. Then $L=S/(M \cap S)$. qed – Todd Leason Feb 28 '20 at 5:36