If a map of $\mathrm{GL}_n$-representations is an isomorphism over every field, is it an isomorphism of $\mathrm{GL}_n$-representations?

Question

Consider the group scheme $G=\mathrm{GL}_n$ over $\mathbb{Z}$. Let $M$, $N$ be representations of $G$ and suppose we have a map $f:M\to N$ of $G$-representations such that $f$ is an isomorphism over every field, regardless of characteristic. Does it follow that $f$ is an isomorphism of $G$-representations? Slightly weaker, is $f$ an isomorphism over every ring?

Answer 1

Note that

1. The action of $G$ could be trivial, and

2. Whatever the action of $G$ is, $f$ is an isomorphism of $G$-representations iff it's an isomorphism, since its inverse, if it exists, is automatically a morphism of $G$-representations.

So $G$ is irrelevant to the question, meaning the answer depends entirely on what you mean by "$G$-representation" here. I assume at the very least that $M, N$ are commutative group schemes over $\mathbb{Z}$. In the nicest possible case we have $M \cong \mathbb{A}^m, N \cong \mathbb{A}^n$ (as group schemes), since this would produce linear representations in the usual sense over a field, in which case your question is:

If $f : \mathbb{A}^m \to \mathbb{A}^n$ is a morphism of group schemes which is an isomorphism over every field, must $f$ be an isomorphism?

The answer is yes. First note that $f$ is given by a matrix in $M_{n \times m}(\mathbb{Z})$ (this follows from a computation of $\text{Hom}(\mathbb{A}^1, \mathbb{A}^1) \cong \mathbb{Z}$; if we were working over an arbitrary base ring $k$ more complicated things could happen, e.g. if $\text{char}(k) = p$ we'd also have Frobenii). Once we know that $f$ is an isomorphism over any field it follows that $n = m$ and that $\det(f) \in \mathbb{Z}$ is nonzero. Once we know that $f$ is an isomorphism over $\mathbb{F}_p$ (or any other field of characteristic $p$) for every prime $p$ it follows that $\det(f) = \pm 1$, so $f \in GL_n(\mathbb{Z})$ is invertible over $\mathbb{Z}$.

(Less obviously and more interestingly, if $f : \mathbb{A}^m \to \mathbb{A}^n$ is a morphism of schemes, not necessarily additive, then it's an isomorphism iff it's an isomorphism over every field. See the discussion here, although it's a bit messy and I haven't corrected an incorrect proof.)

If you allow $M, N$ to be more general group schemes then the answer depends on what they are. I don't know, for example, if every commutative group scheme over $\mathbb{Z}$ is reduced. I assume you don't care about this level of generality, though.

Edit: Arguably most of the work in this argument is in the calculation of $\text{Hom}(\mathbb{A}^1, \mathbb{A}^1) \cong \mathbb{Z}$ here's some more detail on that. This homset consists of all polynomials $f \in \mathbb{Z}[x]$ such that

$$f(x + y) = f(x) + f(y)$$

identically in $\mathbb{Z}[x, y]$. The most concrete way to do this calculation is to just write $f(x) = \sum f_k x^k$ and expand out powers; the LHS has cross terms $x^i y^j$ but the RHS doesn't, unless $\deg f \le 1$ (here we need to use that $\mathbb{Z}$ has characteristic $0$). Then since $f(0) = f(0) + f(0)$ we get $f(0) = 0$, so the conclusion is that $f(x) = kx$ for some $k \in \mathbb{Z}$, as expected.

More abstractly, this condition implies that the roots of $f(x)$ over $\overline{\mathbb{Q}}$ are closed under addition, since if $f(\alpha) = f(\beta) = 0$ then $f(\alpha + \beta) = f(\alpha) + f(\beta) = 0$. But there are finitely many such roots, and $\overline{\mathbb{Q}}$ has characteristic $0$, so this is not possible unless $0$ is the only root. So $f(x) = kx^n$ for some $n \ge 1$ and expanding powers again gives $n = 1$ (here we again need to use that $\mathbb{Z}$ has characteristic $0$).

Another abstract argument is to take $y = \varepsilon \in \mathbb{Z}[\varepsilon]/\varepsilon^2$, or equivalently to quotient by $y^2$; this gives

$$f(x + \varepsilon) = f(x) + \varepsilon f'(x) = f(x) + f(\varepsilon) = f(x) + f(0) + \varepsilon f'(0)$$

where $f'(x)$ is the formal derivative; this gives that $f(0) = 0$ and $f'(x) = f'(0)$ is a constant, so $f'(x) = k$ for some $k \in \mathbb{Z}$, which gives $f(x) = kx$ (here we again need to use that $\mathbb{Z}$ has characteristic zero).

See this answer for what happens if we work over a field of characteristic $p$ instead of $\mathbb{Z}$.