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If a division ring is finite-dimensional over its center then we can apply Skolem-Noether theorem (which asserts that every endomorphism is a conjugation).

Can someone give a counterexample of the Skolem-Noether theorem when the division ring is infinite-dimensional over its center?

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    $\begingroup$ So it's enough to find a division ring and an injective, non-surjective endomorphism. Consider the free group over $(x_n)_{n\ge 0}$, and the endomorphism induced by $x_n\mapsto x_{n+1}$. Extend it to an endomorphism of the group ring (over some given field $k$). This is an Ore domain, so I guess this extends to an injective, non-surjective endomorphism of the right division ring of fractions, whose center is reduced to $k$. But I haven't checked details. $\endgroup$
    – YCor
    Commented Jul 5, 2018 at 10:23

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The following statement is taken from Cohn's 1995 book Skew Fields: Theory of General Division Rings (Cambridge EOM 57), proposition 2.3.4 on page 69, which is attributed to Köthe:

Let $k$ be a commutative field with an automorphism $\alpha$ and put $E = k(\!(t;\alpha)\!)$ [the division ring of $\alpha$-twisted Laurent series over $k$]. Given any automorphism $\beta$ of $k$ such that $\alpha\beta = \beta\alpha$, extend $\beta$ to $E$ by the rule $t^\beta = t$. Then $\beta$ is an inner automorphism of $E$ if and only if $\beta = \alpha^r$ for some $r\in\mathbb{Z}$.

Cohn adds:

For example, if $k = F(s)$, where $F$ is any field of characteristic $0$, and $\alpha\colon s\mapsto s+1$ the $\beta\colon s\mapsto s+\frac{1}{2}$ is an outer automorphism of $E = k(\!(t;\alpha)\!)$.

For completeness of MathOverflow, let me recall the definition of the division ring $E = k(\!(t;\alpha)\!)$ of $\alpha$-twisted (or "skew") Laurent series: its elements are of the form $\sum_{i=i_0}^{+\infty} t^i c_i$ with $c_i \in k$ [here I'm following Cohn's convention of using right-multiplication by $k$], addition being defined componentwise and multiplication by distributing and using the rule $c t^n = t^n c^{\alpha^n}$ [where, of course, $c^{\alpha^n}$ denotes the image of $c$ under the $n$-th power of $\alpha$].

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    $\begingroup$ As an aged mathematician with failing vision, I add my oft-repeated plaint and request that you not use both $\alpha$ and $a$ in a formula, especially where one of them may appear in superscripts. $\endgroup$
    – Lubin
    Commented Jul 5, 2018 at 18:47
  • $\begingroup$ @Lubin Indeed, I normally try to avoid this, but in this particular case I wanted to copy the notation from Cohn's book. Also, it shouldn't be such a problem on a web page since browsers have a zoom feature; when writing PDF files (where formatting is fixed), I try to keep my fonts large and readable, as I also have a bad eyesight. $\endgroup$
    – Gro-Tsen
    Commented Jul 5, 2018 at 20:01
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    $\begingroup$ All you say is true, but it surely couldn’t have confused the reader for you to have called the coefficients $c_i$. $\endgroup$
    – Lubin
    Commented Jul 6, 2018 at 1:33

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