Dear Minhyong,

I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures).

Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.)

In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$.

Added: The comments below have forced me to think a little harder about my position. Here is an attempt to refine it:

Any countably generated extension of $\mathbb Q$ can be embedded into either $\mathbb C$ or $\overline{\mathbb Q}\_{\ell}$, and when I invoke, or seen invoked, an isomorphism between the latter two fields, I think of it as a short-hand for something like the following: in the given proof, a countably generated subfield of $\mathbb C$ will appear (e.g. the field generated by the Hecke eigenvalues of a Maass form). Having fixed the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, we have in particular fixed an embedding of this field into $\overline{\mathbb Q}\_{\ell}$, and hence have chosen an extension of the $\ell$-adic absolute value to this field. (Of course, one could switch the roles of $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$ here.)

By virtue of fixing the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, one is ensuring that any such extensions are compatible, if along the way we encounter different subfields of $\mathbb C$, and that is one big advantage, when writing an argument, of fixing such an isomorphism once and for all. But in practice I don't know that one encounters anything more serious than one single countably generated subfield that contains all the complex numbers appearing in the proof. And hence one doesn't use anything like the full strength of the isomorphism.

I guess this does put me in Deligne's camp: the isomorphism is convenient, but one could get by with something much weaker, just involving countably generated subfields of $\mathbb C$.

I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures).

Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.)

In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$.

Added: The comments below have forced me to think a little harder about my position. Here is an attempt to refine it:

Any countably generated extension of $\mathbb Q$ can be embedded into either $\mathbb C$ or $\overline{\mathbb Q}_{\ell}$, and when I invoke, or seen invoked, an isomorphism between the latter two fields, I think of it as a short-hand for something like the following: in the given proof, a countably generated subfield of $\mathbb C$ will appear (e.g. the field generated by the Hecke eigenvalues of a Maass form). Having fixed the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$, we have in particular fixed an embedding of this field into $\overline{\mathbb Q}_{\ell}$, and hence have chosen an extension of the $\ell$-adic absolute value to this field. (Of course, one could switch the roles of $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$ here.)

By virtue of fixing the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$, one is ensuring that any such extensions are compatible, if along the way we encounter different subfields of $\mathbb C$, and that is one big advantage, when writing an argument, of fixing such an isomorphism once and for all. But in practice I don't know that one encounters anything more serious than one single countably generated subfield that contains all the complex numbers appearing in the proof. And hence one doesn't use anything like the full strength of the isomorphism.

I guess this does put me in Deligne's camp: the isomorphism is convenient, but one could get by with something much weaker, just involving countably generated subfields of $\mathbb C$.

Dear Minhyong,

I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures).

Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.)

In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$.

Dear Minhyong,

I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures).

Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.)

In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$.

Added: The comments below have forced me to think a little harder about my position. Here is an attempt to refine it:

Any countably generated extension of $\mathbb Q$ can be embedded into either $\mathbb C$ or $\overline{\mathbb Q}\_{\ell}$, and when I invoke, or seen invoked, an isomorphism between the latter two fields, I think of it as a short-hand for something like the following: in the given proof, a countably generated subfield of $\mathbb C$ will appear (e.g. the field generated by the Hecke eigenvalues of a Maass form). Having fixed the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, we have in particular fixed an embedding of this field into $\overline{\mathbb Q}\_{\ell}$, and hence have chosen an extension of the $\ell$-adic absolute value to this field. (Of course, one could switch the roles of $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$ here.)

By virtue of fixing the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, one is ensuring that any such extensions are compatible, if along the way we encounter different subfields of $\mathbb C$, and that is one big advantage, when writing an argument, of fixing such an isomorphism once and for all. But in practice I don't know that one encounters anything more serious than one single countably generated subfield that contains all the complex numbers appearing in the proof. And hence one doesn't use anything like the full strength of the isomorphism.

I guess this does put me in Deligne's camp: the isomorphism is convenient, but one could get by with something much weaker, just involving countably generated subfields of $\mathbb C$.