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Let $V$ be a finite dimensional $K$ vector space and $T \in L(V)$. Moreover, let $\overline K$ be an algebraic closure of $K$, $\overline V = \overline K \otimes V,$ and $\overline T \in L(\overline V)$ be the canonical extension of $T$. Then, without defining trace or determinant, one can show that $\overline T$ has $\dim V$ many eigenvalues $\lambda_1, ..., \lambda_n \in \overline K$, with possible repetitions. One can now define the trace and determinant of $T$ as the sum and product (respectively) of the $\lambda_i$. However, it's not obvious that these quantities are in $K$ as opposed to $\overline K$. $^{(*)}$ How can we show this, using Galois theory?

This exact question was discussed in the comments to this answer to a previous question of mine. However, those comments remained at a very rough and sketchy level (and I'm currently re-learning Galois theory and want to understand this better), so I'd like a full, detailed explanation.

$^{(*)}$ A closely related fact is, we can define the characteristic polynomial of $T$ as $p_T(X) = (X - \lambda_1)\dots(X - \lambda_n) \in \overline K[X],$ and we should then be able to show that in fact $p_T(X) \in K[X]$. Perhaps the best strategy is to prove this fact, since then the desired claim about trace and determinant follows immediately.

WillG
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1 Answers1

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(I want to add the remark that Galois theory only works for separable extensions. If $K$ is not perfect, then $\overline{K}/K$ is not Galois. Thus we need a further argument in the imperfect case. Below, I will only treat the case that $K$ is perfect.)

The key fact is this:

If $\lambda \in \overline{K}$ is an eigenvalue of $T$ and $\sigma \in G:=\mathrm{Gal}(\overline{K}/K)$, then $\sigma(\lambda)$ is also an eigenvalue of $T$ with the same multiplicity.

Proof $G=\mathrm{Gal}(\overline{K}/K)$ acts on $\overline{K} \otimes_K V$ by acting on the first component, i.e. by having act $g \in G$ act via $g \otimes \mathrm{id}$. On elementary tensors, this action is given by $(g \otimes \mathrm{id})(\lambda \otimes v)=g(\lambda) \otimes v$. Similarly $\overline{T}=\mathrm{id} \otimes T$ acts on $\overline{K} \otimes_K V$ by acting on the second factor. It follows that these actions commute, i.e. we have for all $g \in G$: $(g \otimes \mathrm{id}) \circ (\mathrm{id} \otimes T)=(\mathrm{id} \otimes T) \circ (g \otimes \mathrm{id})$. Both sides equal $g \otimes T$ which acts on elementary tensors as $(g \otimes T)(\lambda \otimes v)=g(\lambda) \otimes T(v)$.

Note that $g \otimes \mathrm{id}$ is not $\overline{K}$-linear. But it is $K$-linear and satifies $(g \otimes \mathrm{id})(\lambda v)=g(\lambda) (g \otimes \mathrm{id})(v)$, as one can check by writing $v$ as a sum of elementary tensors.

Now let $\lambda \in \overline{K}$. Then if for some $v \in \overline{V}$ we have $\overline{T}(v)=\lambda v$, apply $g \otimes \mathrm{id}$ to both sides and use that this commutes with $\overline{T}$ to get $\overline{T}((g\otimes \mathrm{id})(v))=(g \otimes \mathrm{id})(\lambda v)=g(\lambda) (g \otimes \mathrm{id})(v)$. Thus $(g \otimes \mathrm{id})(v)$ is an eigenvector for the eigenvalue $g(\lambda)$. Thus $G$ permutes the eigenvalues. To match the multiplities, we need to show that $G$ also permutes the generalized eigenspaces.

But the argument is essentially the same.

If for some $\lambda \in \overline{K}, v \in \overline{V}$ and $k \in \Bbb N$, we have $(\overline{T}-\lambda \mathrm{Id})^k(v)=0$, then apply $(g \otimes \mathrm{id})$ to both sides and obtain $(\overline{T}-g(\lambda) \mathrm{Id})^k((g\otimes \mathrm{id})(v))=0$. This establishes the key fact.

Now using this, we get that if $\lambda_1, \ldots, \lambda_k$ are the distinct eigenvalues with multiplicities $n_1, \ldots, n_k$, we get.that their sum and product are each invariant by the action of $G$ (because $G$ permutes them and preserves multiplicities) and hence (if $K$ is perfect), they lie in $K$. By the same token, $p_T(X) \in K[X]$.

Lukas Heger
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  • In the last paragraph, I think you're invoking Galois theory to assert that the fixed field of $\text{Gal}(\overline{K}/K)$ is $K$ (which is why you restrict to perfect fields). My concern is that $\text{Gal}(\overline{K}/K)$ is typically infinite, whereas the standard treatment of the Galois correspondence only considers finite Galois extensions (it seems the theory can be extended to the infinite case, but that it's much messier). – WillG Jun 13 '25 at 01:08
  • Can you explain why this works even if $\text{Gal}(\overline{K}/K)$ is infinite? I suspect maybe we can restrict to a finite sub-extension of $\overline{K}/K$, but haven't fully thought it through. – WillG Jun 13 '25 at 01:09
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    @WillG as long as $K$ is perfect, the fixed field of $\mathrm{Gal}(\overline{K}/K)$ is indeed $K$, even in the infinite case. Note that for any $\alpha \in \overline{K}$, $\alpha$ is contained in a finite Galois extension $L/K$ (one can take a splitting field on the minimal polynomial of $\alpha$ for example). Now one can show that $L$ is stable under the action of $\mathrm{Gal}(\overline{K}/K)$ and the map $\mathrm{Gal}(\overline{K}/K) \to \mathrm{Gal}(L/K)$ is surjective (because we can always extend automorphisms here). – Lukas Heger Jun 13 '25 at 01:13
  • Thus if $\alpha$ is in the fixed field of $\mathrm{Gal}(\overline{K}/K)$, then it's in the fixed field of $\mathrm{Gal}(L/K)$ for some finite Galois extension $L/K$ and thus $\alpha \in K$. This can be found in any treatment of infinite Galois theory. – Lukas Heger Jun 13 '25 at 01:14
  • If you're asking for a full proof of the Galois correspondence in the infinite case, I can give you a reference, but writing out the full proof here seems a bit much. The key idea is what I sketched above: an infinite Galois extension is a union of finite Galois subextensions, i.e. every element is contained in a finite Galois subextension. Then you can use the finite case of the Galois correspondence. – Lukas Heger Jun 13 '25 at 01:19
  • Ah ok, I think I just need to review this. I'm not really interested in the full correspondence in the infinite case (yet)—I think the sketch you gave is sufficient for my purposes, I just need to think about it. Thanks! – WillG Jun 13 '25 at 01:43