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$\def\bbR{\mathbb{R}} \def\bbQ{\mathbb{Q}}$The comment from Vladimir Sotirov in March 2022 in this answer could be interpreted as suggesting the possibility that every ring homomorphism between $\bbR$-algebras must be $\bbR$-linear. But is this true? From the comment, it seems that this would be a consequence of the fact that the unique ring endomorphism of $\bbR$ is the identity (here is a proof of the fact).

Here are my thoughts: If $A,B$ are non-zero real algebras and $f:A\to B$ is a ring homomorphism, then, from the fact that $f(1)=1$, we can conclude that $f$ is $\bbQ$-linear. On the other hand, let $\alpha:\bbR\to A$ and $\beta:\bbR\to B$ be the structure morphisms. If we knew that $f(\alpha(\bbR))\subset \beta(\bbR)$, then the composite ring homomorphism $\bbR\xrightarrow{\alpha}\alpha(\bbR)\xrightarrow{f}\beta(\bbR)\xrightarrow{\beta^{-1}}\bbR$ would be the identity, and so, $f$ would be $\bbR$-linear. But is $f(\alpha(\bbR))\subset \beta(\bbR)$ always true? How one does one see this?

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  • $\begingroup$ I did not suggest that every ring homomorphism between $\mathbb R$-algebras is $\mathbb R$-linear. I only claimed it for $1$-dimensional $\mathbb R$-algberas (i.e. between rings of real numbers). The consequence of that was that the components of morphisms of certain locally ringed spaces (those arising from real algebras of smooth functions) are automatically real-linear if they are ring homomorphisms. The argument crucially used that those homomorphisms are components of a morphism between those locally ringed spaces. $\endgroup$
    – Vladimir Sotirov
    Commented Mar 21, 2023 at 19:11
  • $\begingroup$ @VladimirSotirov I just edited the phrasing in the first sentence. Thank you so much for the new comment in the other post detailing your argument ^^ $\endgroup$
    – Elías Guisado Villalgordo
    Commented Mar 22, 2023 at 8:41

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No, this is indeed not correct. For instance, there are two $\mathbb{R}$-algebra automorphisms of $\mathbb{C}$ (the identity and conjugation), but under the axiom of choice there are infinitely many field automorphisms of $\mathbb{C}$ (which are not continuous, and are just generally pathological).

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