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We know that $\Bbb{Q}\hookrightarrow \Bbb{R}$, by a $\Bbb{Q}$-algebra homomorphism from $\Bbb{R}\to\Bbb{R}$ we mean a ring homomorphism from $\Bbb{R}\to\Bbb{R}$ which is the identity on $\Bbb{Q}$.

I first thought in this way- We know that $[\Bbb{R}:\Bbb{Q}]=\infty$, and the set $\{1,\sqrt{2}\}$ can be extended to a basis $B$ of $\Bbb{R}$ over $\Bbb{Q}$. Then I define $f:\Bbb{R}\to\Bbb{R}$ such that $f(1)=\sqrt{2}$, $f(\sqrt{2})=1$ and $f(x)=x\ \forall x\in B\setminus \{1,\sqrt{2}\}$. Now extend $f$ linearly. But this way $f$ will be a group homomorphism but may not preserve the product (i.e. may not be a ring homomorphism). Even this $f$ is onto as well.

Can anyone suggest a way out? Thanks for your help in advance.

MathBS
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    Related: https://math.stackexchange.com/questions/3518835/are-there-discontinuous-ring-endomorphisms-on-real-numbers, https://math.stackexchange.com/questions/1562845/number-of-ring-homomorphism-from-the-ring-of-real-numbers-to-itself – Ravi Fernando Apr 05 '21 at 17:52

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No. In fact, the only $\mathbb Q$-algebra homomorphism $\mathbb R \to \mathbb R$ is the identity map. Proof: Suppose $f$ is such a homomorphism. If $x > 0$, then $x = y^2$ for some $y \neq 0$, so $f(x) = f(y^2) = f(y)^2 > 0$. Therefore $f$ sends all positives to positives. It follows that $f$ preserves the ordering on $\mathbb R$. But $f$ must act as the identity on all rationals, and this forces it to act as the identity everywhere.

Ravi Fernando
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  • Ok, I understand your point. $f$ is increasing and as $f$ is infact a group homomorphism. $f$ is continuous everywhere, so $f(x)=\alpha x\ $ for some $\alpha\in \Bbb{R}$. As $f$ is identity on $\Bbb{Q}$, $\alpha=1$. – MathBS Apr 05 '21 at 17:52
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    See also https://math.stackexchange.com/a/449411/589 – lhf Apr 05 '21 at 18:19
  • @lhf Right, your outline there still works without assuming surjectivity. – Ravi Fernando Apr 05 '21 at 21:42