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In mathematical logic class, I've learned that the only isomorphism on $\mathbb R$ is only the identity since isomorphism must preserve the order relation of reals.

But in my abstract algebra textbook, problem says:

Prove that there exists an isomorphism of fields $f: \mathbb R\rightarrow \mathbb R$ that maps $\pi$ to $-\pi$.

Does it make sense?

Darae-Uri
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  • The only isomorphism on $\mathbb{R}$ with what structure? As a field $(\mathbb{R},+,\cdot)$ the only automorphism is the identity; but if you consider $(\mathbb{R},+)$ as a group you could do such a thing, I believe. – sTertooy Sep 29 '16 at 16:05
  • The problem states that $f$ is an isomorphism of fields so that $\mathbb R$ is a field with addition and multiplication. – Darae-Uri Sep 29 '16 at 16:06
  • Does your book require that homomorphisms of rings preserve the multiplicative identity? – user350031 Sep 29 '16 at 16:10
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    See http://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map why it has to be the identity. – sTertooy Sep 29 '16 at 16:10
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    @user350031 No. In my textbook, it is not necessarily true, even if both have identities. – Darae-Uri Sep 29 '16 at 16:12
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    Maybe this is where the problem lies :) – user350031 Sep 29 '16 at 16:14
  • @user350031 No: any automorphism of a ring with multiplicative identity must preserve that identity, even if we don't demand a priori that ring homomorphisms preserve the multiplicative identity. (If $f(1)\not=1$, then there is some $a$ such that $f(1)a\not=a$; since $f$ is an automorphism, $a=f(b)$ for some $b$, but then $f(1b)\not=f(1)f(b)$.) – Noah Schweber Sep 29 '16 at 16:16
  • @NoahSchweber: Ah yes, of course, apologies! (Proof: We have $\phi(1)\phi(a)=\phi(a)$ for all $a$ and now surjectivity finishes the proof). – user350031 Sep 29 '16 at 16:22
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    I would like to look at this book. Do you mind sharing the title and page number of the problem? – Keith Kearnes Sep 29 '16 at 18:00

1 Answers1

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If your text phrases the problem in that way, then it is indeed incorrect: since the ordering on $\mathbb{R}$ is definable from the field structure ($a\le b\iff\exists c(a+c^2=b)$), any field automorphism must preserve the ordering, and from this (as you well know) it's not hard to show that there are no nontrivial field automorphisms at all.

Noah Schweber
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