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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?

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    $\begingroup$ 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 $|X|=|\mathbb C|$. $\endgroup$
    – Keith Kearnes
    Commented Feb 26, 2020 at 15:15
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    $\begingroup$ Follow-up question: yes (exercise). Hint: elementary matrices. $\endgroup$
    – YCor
    Commented Feb 26, 2020 at 16:26
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    $\begingroup$ 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. $\endgroup$
    – Mark Wildon
    Commented Feb 26, 2020 at 16:29
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    $\begingroup$ @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 $E_d(R)$. $\endgroup$
    – YCor
    Commented Feb 26, 2020 at 18:40
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    $\begingroup$ 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 $p$. $\endgroup$
    – Todd Leason
    Commented Feb 28, 2020 at 5:04

1 Answer 1

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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.

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  • $\begingroup$ 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 $\mathbb{C}$. $\endgroup$
    – Mark Wildon
    Commented Feb 26, 2020 at 16:22
  • $\begingroup$ @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 !). $\endgroup$
    – Duchamp Gérard H. E.
    Commented Feb 26, 2020 at 17:43
  • $\begingroup$ 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 $(p)=\mathbb{Z}\cap m$. ... $\endgroup$
    – Todd Leason
    Commented Feb 28, 2020 at 5:29
  • $\begingroup$ ... 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 $\endgroup$
    – Todd Leason
    Commented Feb 28, 2020 at 5:36

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