On genus of division algebras

TL;DR

This paper investigates the properties of the genus of division algebras, showing that sharing maximal subfields does not imply similar properties for matrix algebras or after tensoring with certain algebras, revealing nuanced distinctions.

Contribution

It demonstrates that having the same maximal subfields does not guarantee similar maximal subfields in matrix algebras or after tensoring, providing new insights into the structure of division algebras.

Findings

Quaternion division algebras with the same maximal subfields can differ in their matrix algebra extensions.

Existence of fields where tensoring with certain division algebras changes the genus relationship.

The genus concept does not always extend predictably under matrix or tensor operations.

Abstract

The genus $gen(D)$ of a finite-dimensional central division algebra $D$ over a field $F$ is defined as the collection of classes $[D']\in Br(F)$, where $D'$ is a central division $F$-algebra having the same maximal subfields as $D$. We show that the fact that quaternion division algebras $D$ and $D'$ have the same maximal subfields does not imply that the matrix algebras $M_l(D)$ and $M_l(D')$ have the same maximal subfields for $l>1$. Moreover, for any odd $n>1$, we construct a field $L$ such that there are two quaternion division $L$-algebras $D$ and $D'$ and a central division $L$-algebra $C$ of degree and exponent $n$ such that $gen(D) = gen(D')$ but $gen(D \otimes C) \ne gen(D' \otimes C)$.

Equations

$$\mathcal{C}:={\mathcal{A}}_{F(x)}\otimes (c,x)$$

$$\mathcal{D}:={\mathcal{B}}_{F_1(y)}\otimes (c,y)$$

$$E^{<c}:=\bigcup _{c^{\prime }<c}{E}^{c^{\prime }}\text{mbox}and{E}^c:=E_c^{<c},$$

Full text

The genus ${\bf gen}({\cal D})$ of a finite-dimensional central division algebra ${\cal D}$ over a field $F$ is defined as the collection of classes $[{\cal D}^{\prime}]\in\text{\normalshape Br}(F)$, where ${\cal D}^{\prime}$ is a central division $F$-algebra having the same maximal subfields as ${\cal D}$. This means that ${\cal D}$ and ${\cal D}^{\prime}$ have the same degree $n$, and a field extension $K/F$ of degree $n$ admits an $F$-embedding $K\hookrightarrow{\cal D}$ if and only if it admits an $F$-embedding $K\hookrightarrow{\cal D}^{\prime}$. Different variations of the notion of the genus are mentioned in [2].

The following questions were formulated in [3, footnote 1 and Remark 2.2]:

*Does the fact that division algebras ${\cal D}$ and ${\cal D}^{\prime}$ have the same maximal subfields imply that the matrix algebras $M_{l}({\cal D})$ and $M_{l}({\cal D}^{\prime})$ have the same maximal subfields / étale subalgebras for any (or even some) $l>1$? *

*Let $n_{1}$ and $n_{2}$ be relatively prime positive integers. Let also ${\cal D}_{i}$ and ${\cal D}^{\prime}_{i}$ be central division algebras of degree $n_{i}$ over a field $F$ for $i=1,2$. Is it true that if ${\bf gen}({\cal D}_{i})={\bf gen}({\cal D}^{\prime}_{i})$ for $i=1,2$, then ${\bf gen}({\cal D}_{1}\otimes{\cal D}_{2})={\bf gen}({\cal D}^{\prime}_{1}\otimes{\cal D}^{\prime}_{2})$? *

Negative answers to these questions are given in Theorem 5 and Corollary 6 below.

We use the following notation. For a field $F$, $F^{*}$ denotes the multiplicative group of $F$. ${F^{*}}^{2}$ denotes the subgroup of squares in $F^{*}$. For a field extension $K/F$ and a central simple $F$-algebra ${\cal A}$, ${\cal A}_{K}$ denotes the tensor product ${\cal A}\otimes_{F}K$ and ${\normalshape{res}}_{K/F}:\text{\normalshape Br}(F)\longrightarrow\text{\normalshape Br}(K)$ denotes the restriction homomorphism.

Let $F$ be a field, $\text{\normalshape char}(F)\neq 2$. Assume that there are two non-isomorphic quaternion division $F$-algebras ${\cal A}$ and ${\cal B}$. We will be interested in finite separable field extensions $K/F$ that satisfy the following three conditions:

(A) There is $d\in K^{*}\backslash{K^{*}}^{2}$ such that there is no $a\in F^{*}$ such that $da\in{K^{*}}^{2}$;

(B) $K$ does not split ${\cal B}$;

(C) $K(\sqrt{d})$ splits ${\cal B}$ but does not split ${\cal A}$.

In the notation above, we have the following

(1) the homomorphism ${\normalshape{res}}_{F_{c}/F}:\text{\normalshape Br}(F)\longrightarrow\text{\normalshape Br}(F_{c})$ is injective;

(2) the field $F_{c}(\sqrt{c})$ splits the algebras ${\cal A}_{F_{c}}$ and ${\cal B}_{F_{c}}$.

Moreover, if a field extension $K/F$ satisfies conditions (A) - (C) and $c\not\in{K^{*}}^{2}$, then

(3) there is no $a\in F_{c}$ such that $da\in{F_{c}K^{*}}^{2}$;

(4) the composite $F_{c}K$ does not split ${\cal B}_{F_{c}}$;

(5) the composite $F_{c}K(\sqrt{d})$ splits ${\cal B}_{F_{c}}$ but does not split ${\cal A}_{F_{c}}$.

Proof. Let $F(x)$ be a purely transcendental extension of $F$ of transcendence degree 1. Let also

[TABLE]

be a biquaternion $F(x)$-algebra and $F_{1}$ the function field of the Severi-Brauer variety of the algebra ${\cal C}$.

Now let $F_{1}(y)$ be a purely transcendental extension of $F_{1}$ of transcendence degree 1 and

[TABLE]

a biquaternion $F_{1}(y)$-algebra. Let also $F_{c}$ be the function field of the Severi-Brauer variety of the algebra ${\cal D}$.

Since the kernel of the restriction homomorphism ${\normalshape{res}}_{F_{1}/F(x)}$ is generated by the class of the algebra ${\cal C}$, then the homomorphism ${\normalshape{res}}_{F_{1}/F}$ is injective. Indeed, ${\cal C}$ ramifies at the discrete valuation (trivial on $F$) of $F(x)$ defined by the polynomial $x$. Hence $[{\cal C}]\neq[{\cal Q}_{F(x)}]$ for any central simple $F$-algebra ${\cal Q}$.

Note that $F_{1}$ splits ${\cal C}$. Then ${\cal A}_{F_{1}}\cong(c,x)_{F_{1}}$ and $F_{1}(\sqrt{c})$ splits ${\cal A}_{F_{1}}$.

Let $K/F$ be a field extension satisfying conditions (A) - (C). Since $F_{1}/F$ is a regular field extension, then there is no $a\in F_{1}$ such that $da\in{F_{1}K^{*}}^{2}$. In particular, this means that $dc\not\in{F_{1}K^{*}}^{2}$. Moreover, $F_{1}K/K$ is a regular extension of $K$. Thus, if $c\not\in{K^{*}}^{2}$, then $c\not\in{F_{1}K^{*}}^{2}$.

The composite $F_{1}K(\sqrt{d})$ is the function field of the Severi-Brauer variety of the $K(\sqrt{d})(x)$-algebra ${\cal C}_{K(\sqrt{d})(x)}$. Hence the kernel of the restriction homomorphism ${\normalshape{res}}_{F_{1}K(\sqrt{d})/K(\sqrt{d})(x)}$ is generated by the class of ${\cal C}_{K(\sqrt{d})(x)}$. Since $dc\not\in{K^{*}}^{2}$, then $c\not\in K(\sqrt{d})^{*2}$ and ${\cal C}_{K(\sqrt{d})(x)}$ ramifies at the discrete valuation (trivial on $K(\sqrt{d})$) of $K(\sqrt{d})(x)$ defined by the polynomial $x$, but ${\cal A}_{K(\sqrt{d})(x)}$ is unramified at this valuation, hence $[{\cal A}_{K(\sqrt{d})(x)}]\neq[{\cal C}_{K(\sqrt{d})(x)}]$ and $F_{1}K(\sqrt{d})$ does not split ${\cal A}_{F_{1}}$. Analogously, the composite $F_{1}K$ does not split ${\cal B}_{F_{1}}$.

Thus the extension $F_{1}K/F_{1}$ satisfies conditions (A)-(C) with respect to the algebras ${\cal A}_{F_{1}}$ and ${\cal B}_{F_{1}}$.

The field $F_{c}$ satisfies conditions (1)-(5) of the proposition by the same arguments as for the field $F_{1}$. We just replace the ground field $F$ by $F_{1}$ and the extension $K/F$ by $F_{1}K/F_{1}$. ∎

In the notation above, we also have the following

(1) the homomorphism ${\normalshape{res}}_{E(F)/F}:\text{\normalshape Br}(F)\longrightarrow\text{\normalshape Br}(E(F))$ is injective;

(2) the field $E(F)(\sqrt{c})$ splits the algebras ${\cal A}_{E(F)}$ and ${\cal B}_{E(F)}$ for any $c\in U$.

Moreover, if a field extension $K/F$ satisfies conditions (A) - (C), then

(3) there is no $a\in E(F)$ such that $da\in{E(F)K^{*}}^{2}$;

(4) the composite $E(F)K$ does not split ${\cal B}_{E(F)}$;

(5) the composite $E(F)K(\sqrt{d})$ splits ${\cal B}_{E(F)}$ but does not split ${\cal A}_{E(F)}$.

Proof. Note that for any field extension $K/F$ satisfying conditions (A) - (C), $U\cap{K^{*}}^{2}={\varnothing}$ since $K$ does not split ${\cal A}$ and ${\cal B}$.

Let $<$ be a well-ordering on the set $U$ and let $c_{0}$ denote its least element. Set $E^{c_{0}}:=F_{c_{0}}$, where the field $F_{c_{0}}$ is constructed in Proposition 2.

For $c\in U$, $c\neq c_{0}$, set

[TABLE]

where the field $E^{c}$ is obtained by applying Proposition 2 to the field $E^{<c}$ and the element $c\in E^{<c}$ and the algebras ${\cal A}_{E^{<c}}$ and ${\cal B}_{E^{<c}}$. Define also $E(F):=\bigcup_{c\in U}E^{c}$.

By Proposition 2 and transfinite induction, the field $E(F)$ satisfies conditions (1)-(5) of the proposition. ∎

(1) ${\cal A}_{L}$ and ${\cal B}_{L}$ are division algebras and ${\bf gen}({\cal A}_{L})={\bf gen}({\cal B}_{L})$;

(2) If $K/F$ is a field extension of degree $n$ satisfying properties (A) - (C) with respect to the algebras ${\cal A}$ and ${\cal B}$, then the matrix algebras $M_{n}({\cal A}_{L})$ and $M_{n}({\cal B}_{L})$ do not have the same maximal subfields;

(3) If $K$ is a field from the previous item and $C$ is a central division $F$-algebra of exponent $n$ which is split by $K$, then $C_{L}$ is a division algebra of exponent $n$ and the algebras ${\cal A}_{L}\otimes{\cal C}_{L}$ and ${\cal B}_{L}\otimes{\cal C}_{L}$ do not have the same maximal subfields.

Proof. Let $K_{0}:=F$. We recursively define $K_{i}$, $i\in{\mathbb{Z}}_{>0}$, to be the field $E(K_{i-1})$ constructed by applying Proposition 4 to the field $K_{i-1}$ and the algebras ${\cal A}_{K_{i-1}}$ and ${\cal B}_{K_{i-1}}$. Let also $L:=\bigcup_{i\geq 0}K_{i}$.

By induction and Proposition 4, the homomorphism ${\normalshape{res}}_{L/F}:\text{\normalshape Br}(F)\longrightarrow\text{\normalshape Br}(L)$ is injective. Hence ${\cal A}_{L}$ and ${\cal B}_{L}$ are non-isomorphic division algebras.

Assume that $M$ is a maximal subfield of ${\cal A}_{L}$. Then there exists $i\geq 0$ such that $M=LM^{\prime}$, where $M^{\prime}$ is a quadratic extension of $K_{i}$ that splits ${\cal A}_{K_{i}}$. By the construction of $K_{i+1}$, the field $K_{i+1}M^{\prime}$ splits the algebra ${\cal B}_{K_{i+1}}$. Hence $M$ splits ${\cal B}_{L}$. Analogously, every maximal subfield of ${\cal B}_{L}$ splits ${\cal A}_{L}$. Thus the algebras ${\cal A}_{L}$ and ${\cal B}_{L}$ have the same family of maximal subfields, i.e., ${\bf gen}({\cal A}_{L})={\bf gen}({\cal B}_{L})$.

Assume that $K/F$ is a field extension of degree $n$ satisfying conditions (A) - (C) with respect to the algebras ${\cal A}$ and ${\cal B}$. By induction and Proposition 4, the composite $LK(\sqrt{d})$ splits ${\cal B}_{L}$ but does not split ${\cal A}_{L}$. Then $LK(\sqrt{d})$ embeds in $M_{n}({\cal B}_{L})$ but does not embed in $M_{n}({\cal A}_{L})$. Hence $M_{n}({\cal A}_{L})$ and $M_{n}({\cal B}_{L})$ do not have the same maximal subfields.

Finally, let ${\cal C}$ be a central division $F$-algebra of exponent $n$ which is split by $K$. Since the homomorphism ${\normalshape{res}}_{L/F}$ is injective, then the exponent of ${\cal C}_{L}$ is $n$. Since the exponent of ${\cal C}_{L}$ divides its index, then ${\cal C}_{L}$ is a division algebra. The composite $LK(\sqrt{d})$ splits ${\cal B}_{L}\otimes{\cal C}_{L}$ but does not split ${\cal A}_{L}\otimes{\cal C}_{L}$. This means that ${\cal A}_{L}\otimes{\cal C}_{L}$ and ${\cal B}_{L}\otimes{\cal C}_{L}$ do not have the same maximal subfields. ∎

Proof. Let $F$ be a field such that there are two non-isomorphic quaternion $F$-algebras ${\cal A}$ and ${\cal B}$ and for any $n>1$, the field $F$ has an extension of degree $n$ satisfying conditions (A) - (C) with respect to the algebras ${\cal A}$ and ${\cal B}$. Let $L$ be the field constructed in Theorem 5. By Theorem 5, the algebras ${\cal D}:={\cal A}_{L}$ and ${\cal D}^{\prime}:={\cal B}_{L}$ have the required properties.

∎