As the title says Let $f: \mathbb{R} \to \mathrm{End}(\mathbb{R},+,-,0)$ be a ring homomorphism, is $f$ necessarily trivial? By trivial I mean the canonical ring homomorphism, which assigns to each $r\in R$ the action $x \mapsto r\cdot x$, because $\mathbb{R}$ is a field we have that any such ring homomorphism has to be injective and for each $r\in\mathbb{R}, r\neq 0$ $f(r)$ has to be a $\mathbb{Q}$-linear automorphism, so in order to generate a counterexample I've tried to play around with a basis of $\mathbb{R}$ over $\mathbb{Q}$, but can't seem to define any permutation of said basis in a compatible way. I've spent way too much time on this problem and can't seem to make any further progress on it. I assume there is something simple that I'm missing.
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The elements $\operatorname{End}(\mathbb R,+)$ are linear transformations of $\mathbb R$ as a vector space over the field $\mathbb Q.$ That is a vector space with an uncountable dimension. Not sure if that helps. – Thomas Andrews Oct 12 '24 at 13:40
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It is not unique. There is an isomorphism of abelian groups $\mathbb{R} \cong \mathbb{R}^2$, inducing an isomorphism of rings $\mathrm{End}_{\mathbf{Ab}}(\mathbb{R}) \cong \mathrm{End}_{\mathbf{Ab}}(\mathbb{R}^2)$. The latter has a subring $\mathrm{End}_{\mathbb{R}-\mathbf{Mod}}(\mathbb{R}^2)$ which is isomorphic to $M_2(\mathbb{R})$. It was shown in Uniqueness of Ring Homomorphisms from $\mathbb{R}$ to $M_2(\mathbb{R})$ that there are lots of ring homomorphisms $\mathbb{R} \to M_2(\mathbb{R})$.
There is also the related question Which commutative rings admit exactly one ring homomorphism from $\mathbb{R}$?
Martin Brandenburg
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The isomorphism $\mathbb{R}\cong \mathbb{R}^2$ can be seen from both of these spaces being $\mathbb{Q}$-vector spaces of the same dimension. – Jakobian Oct 12 '24 at 14:03
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“No” is an answer to the question in the body of the OP, rather than the question in the title? :-) – Jeremy Rickard Oct 12 '24 at 14:03
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