2
$\begingroup$

Does there exist a field $K$ and a finite-dimensional $K$-division algebra $D$ possessing two maximal separable subfields of different dimensions?

Remark: If $D$ is separable ($Z(D)$ a separable field extension of $K$), then all such subfields have the same dimension.

$\endgroup$
  • 3
    $\begingroup$ No. Since $D$ is a division algebra, separable subfields correspond to $K$-subtori of the associated algebraic group of units (a Galois-twisted form of the Weil restriction of ${\rm{GL}}_d$ from $Z(D)$ down to $K$, where $d^2 = [D:Z(D)]$), so maximal separable subfields correspond to maximal $K$-tori. But in any smooth affine group over any field $k$ whatsoever, it is a theorem of Grothendieck that all maximal $k$-tori remain maximal after any field extension and so have the same dimension (though the context of separable division algebras is more classical; see Jacobson "Basic Algebra II"). $\endgroup$ – nfdc23 Oct 12 '16 at 15:28
  • $\begingroup$ Thanks for this comment. Is the argument also true for $D^{n\times n}$? In addition, do you have a reference for your argumentation because I am not familiar with this theory. $\endgroup$ – Sven Wirsing Oct 12 '16 at 15:56
  • $\begingroup$ Rename $Z(D)$ as $K$ so $D$ has center $K$, as is assumed in most literature. Since $K$ may not have degree-$n$ extensions, the link between subfields and maximal tori can break if $n>1$. For $d^2=[D:K]$, all maximal subfields of $D$ have degree $d$: the proof of Prop. 1.2.3.1 in the recent book "Complex Multiplication and Lifting Problems" is a tour through Jacobson's BAII to prove this classic fact. You want more: any separable subfield $F\subset D$ is inside a separable maximal one. The only proof I know is via Grothendieck's result, so hard-core algebraic groups; I'll make an answer. $\endgroup$ – nfdc23 Oct 13 '16 at 20:09
  • $\begingroup$ I do not think we can assume that $D$ is central because the maximal separable subfields do not need to contain the center of $D$. $\endgroup$ – Sven Wirsing Oct 14 '16 at 7:01
  • $\begingroup$ If $F \subset D$ is a commutative subfield over $K$ then since it commutes with $Z(D)$ we get a $K$-algebra map $F \otimes_K Z(D) \rightarrow D$ whose image is a $Z(D)$-subalgebra of $D$ that is a quotient of $F \otimes_K Z(D)$, and that quotient cannot have nonzero zero-divisors (being a subring of $D$). The only such quotients are the factor fields since $Z(D)$ is $K$-separable, and these are separable over $K$ (since $F$ and $Z(D)$ are). Thus, every $K$-separable subfield is contained in one that contains $Z(D)$, so the maximal ones all contain $Z(D)$. Thus, one reduces to the central case. $\endgroup$ – nfdc23 Oct 14 '16 at 19:19
3
$\begingroup$

No, there are no such examples, but I don't know any way to attack this by methods of ring theory. The theory of linear algebraic groups gives a very illuminating insight into this matter, by recasting the problem in a wider context.

To explain this, let's work more generally for the moment with finite-dimensional associative $K$-algebras $C$ with identity such that the center $Z$ is a finite reduced $K$-algebra (i.e., a product of finitely many finite extension fields of $K$). Note that $Z$ is a field if $C$ is simple, and we specifically avoid limiting ourselves to division algebras because that way our setup is preserved under separable extension of the ground field (which is not the case for division algebras).

Note that separable extension on $K$ will typically make the center decompose into several more factor fields, which is why we do not assume $Z$ is a field either (but just a finite product of such). In particular, $$C \otimes_K K_s = C \otimes_Z (Z \otimes_K K_s) = \prod (C \otimes_Z Z_i)$$ where $\prod Z_i$ is the decomposition of the nonzero finite reduced $K_s$-algebra $Z \otimes_K K_s$ into a product of fields. Each $Z_i$ is separably closed, being finite over $K_s$, so each $C \otimes_Z Z_i$ is a matrix algebra over $Z_i$ (though not over $K_s$!) since Brauer groups of separably closed fields are trivial.

The crucial thing is to relate separable (commutative) subfields of $C$ over $K$, or more generally finite etale (commutative) $K$-subalgebras of $C$, to $K$-tori in the associated "algebraic $K$-group of units" of $C$. By definition, this linear algebraic group represents the functor on $K$-algebras defined by $R \rightsquigarrow (C \otimes_K R)^{\times}$. More concretely, if we consider the affine $d^2$-space $\underline{C}$ over $Z$ associated to $C$ then $G$ is the Weil restriction ${\rm{R}}_{Z/K}(\underline{C}^{\times})$ of the affine open subspace $\underline{C}^{\times} \subset \underline{C}$ of "units over $Z$": the complement of the zero-locus of the reduced norm ${\rm{Nrd}}:\underline{C} \rightarrow \mathbf{A}^1_Z$ that is a polynomial map over $Z$.

If $A \subset C$ is a finite etale $K$-subalgebra then the algebraic group of units $T_A := \underline{A}^{\times} \subset G$ (defined functorially on points valued in a $K$-algebra $R$ by $(A \otimes_K R)^{\times} \hookrightarrow (C \otimes_K R)^{\times}$) is a $K$-torus, and conversely if $T \subset G$ is a $K$-torus then the $K_s$-subalgebra $K_s[T(K_s)] \subset C_{K_s}$ is ${\rm{Gal}}(K_s/K)$-stable and a direct product of copies of $K_s$ (here one uses that $T_{K_s}$ is a split torus) so it descends to a $K$-subalgebra $A_T \subset C$ that is $K$-etale. These two procedures are inclusion-preserving inverse bijections between the set of such $A$ and the set of $K$-tori in $C$, inclusion-preserving in both directions. By considerations over $K_s$ one sees that $\dim_K A = \dim T_A$, so equality of the $K$-dimensions of maximal $K$-etale commutative subalgebras of $C$ is the same as the equality of the dimensions of all maximal $K$-tori in $G$.

By construction, the formation of $G$ commutes with any separable extension on $K$. In particular, $G_{K_s} = \prod {\rm{R}}_{Z_i/K_s}({\rm{GL}}_{n_i}))$ where $n_i^2 = \dim_{Z_i} (C \otimes_Z Z_i)$, so $G$ is a smooth connected affine $K$-group. If $Z$ is $K$-etale then $G$ is even reductive (as finite etale Weil restriction preserves reductivity), but this always fails when $Z$ has a factor field not separable over $K$ (though $G$ is always pseudo-reductive, but we don't need this). In what follows the link to reductive groups won't be used; nonetheless, it does clarify the significance of $Z$ being $K$-etale or not.

In case $C$ is a division algebra, a $K$-etale subalgebra cannot have nontrivial idempotents (as $C$ has no non-zero zero-divisors) and hence must be a field, moreover separable over $K$ by the etale condition. Thus, we see that our task is a special case of one that has nothing at all to do with division algebras: show that every maximal $K$-torus in $C$ has the same dimension.

More generally, if $G$ is any smooth affine group over a field $K$ then we claim that all maximal $K$-tori in $G$ have the same dimension. Over algebraically closed fields it is a basic fact in the theory of algebraic groups that all maximal tori are conjugate and so have the same dimension. Thus, it is enough to prove that if $T \subset G$ is a maximal $K$-torus then $T_{\overline{K}}$ is a maximal $\overline{K}$-torus in $G_{\overline{K}}$. The centralizer $Z_G(T)$ is a smooth closed $K$-subgroup of $G$ by the general theory of linear algebraic groups (this is part of the Corollary to Theorem 9.2 in Borel's book "Linear Algebraic Groups", though that might not be apparent from reading the formulation there), and $H := Z_G(T)/T$ certainly contains no nontrivial $K$-tori, so it suffices to show that $H_{\overline{K}}$ contains no nontrivial $\overline{K}$-tori.

Finally, it suffices to show that if $H$ is a smooth affine $K$-group then there exists some $K$-torus $S \subset H$ that remains maximal over $\overline{K}$ (as then if $H$ has no nontrivial $K$-tori the same must hold for $H_{\overline{K}}$, as desired). This final assertion is an important theorem of Grothendieck proved as 18.2(i) in Borel's book "Linear Algebraic Groups" (a simplification on the original proof in SGA3).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.