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Let $A$ be a noncommutative division ring, and let $B$ be a sub division ring (here, $B$ is allowed to be commutative) of degree $2$. Are there easy examples known for which $B$ is not globally fixed by any element of $A \setminus B$ (action by conjugation) ?

Famous non-example: consider $\mathbb{C}$ as a subfield of $\mathbb{H}$ (classical quaternions, classical embedding). The degree is $2$, but obviously conjugating $\mathbb{C}$ by elements of $\mathbb{H} \setminus \mathbb{C}$ leaves it invariant some of the time.

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    $\begingroup$ At least going by your non-example, any commutative and separable $B$ will have a nontrivial automorphism which is stable under conjugation by an element of $A\setminus B$ by Skolem-Noether (stacks.math.columbia.edu/tag/074P).\\ On the other hand, if $B$ is a purely inseparable extension of degree 2 inside, say $A$ a quaternion division algebra over characteristic 2 field, then every automorphism of $B$ is trivial. If some element $x\in A$ conjugated $B$ to $B$ then it would be the identity on $B$, hence in the centralizer of $B$, which is $B$ in this case. $\endgroup$
    – Eoin
    Commented Sep 14, 2023 at 18:04
  • $\begingroup$ What are you working over? If you are working over a field, isn't B forced to be a quadratic field extension? $\endgroup$
    – Kimball
    Commented Sep 15, 2023 at 3:03
  • $\begingroup$ @Eoin: can you make your remark into an answer (with more details) ? $\endgroup$
    – THC
    Commented Sep 15, 2023 at 11:46
  • $\begingroup$ Just a remark: you're considering the normalizer $N_{A^*}(B^*)$ of $B^*$ in $A^*$, and asking whether there are examples with $N_{A^*}(B^*)/B^*=\{1\}$. In the case of $\mathbf{C}\subseteq\mathbf{H}$ you are mentioning, this quotient is a 2-element group. $\endgroup$
    – YCor
    Commented Sep 18, 2023 at 5:16
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    $\begingroup$ A small remark about terminology: the word "degree" for central simple algebras is used for the square root of the dimension (because the dimension over the center is always a square), so it is better to use "dimension" instead. For instance, quaternion algebras have degree $2$. $\endgroup$
    – Tom De Medts
    Commented Sep 18, 2023 at 9:10

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(This is basically a more detailed version of Eoin's comment.)

I assume that you are considering division algebras over a field $k$, i.e., $Z(A) = k$. If $B$ is a subalgebra of dimension $2$ of $A$, then $B$ is always commutative, and since it lives inside the division algebra $A$, it must be a quadratic field extension of $k$.

If $B/k$ is a separable extension, then it admits a non-trivial automorphism $\sigma$. The Skolem-Noether theorem tells us that every automorphism of a simple subalgebra extends to an inner automorphism of the whole algebra. In particular, $\sigma$ extends to an inner automorphism of $A$, say conjugation by $a \in A$, and of course we must have $a \not\in B$ since $\sigma$ is non-trivial.

If $B/k$ is an inseparable extension, then we have $\operatorname{char}(k)=2$ and $\operatorname{Gal}(B/k)=1$. So any inner automorphism of $A$ normalizing $B$ must, in fact, centralize $B$. Now let $C = C_A(B)$ be this centralizer. Then, by the centralizer formula, we have $\dim_K(A) = \dim_K(B) \dim_K(C)$. In particular, if $\dim_K(A) = 4$ (the smallest possible since the dimension is always a square), we have $C = B$, and this provides the kind of examples you are looking for. On the other hand, if $\dim_K(A) > 4$, then $\dim_K(C) > 2$ so $C$ strictly contains $B$. In other words, there are elements of $C \setminus B$ that conjugate $B$ to itself, so there are no new examples in this situation.

(For a reference, both the Skolem-Noether theorem and the centralizer formula can be found in section 8.4 of W. Scharlau's book "Quadratic and hermitian forms" from 1985.)

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  • $\begingroup$ Why is B commutative ? $\endgroup$
    – THC
    Commented May 25 at 10:17
  • $\begingroup$ Simply because it is $2$-dimensional over $k = Z(A)$... (Write $B = k.1 + k.b$.) $\endgroup$
    – Tom De Medts
    Commented May 27 at 7:53

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