3

The multiplicativity formula for degrees of a tower of fields is well-known. I wonder if the same formula still holds if we consider division rings instead of fields, namely:

Let $A \subseteq B \subseteq C$ be three division algebras. From the comments to this question follows that the notion of the rank (as modules) is well-defined.

Denote the rank of $B$ as an $A$-module by $r_A(B)$, etc.

Is it true that $r_A(C)=r_B(C) r_A(B)$? I think (but may be wrong) that the answer is yes, and the proof is similar to the commutative case (taking bases etc.).

Thank you very much for any help. I apologize if this question is trivial.

Community
  • 1
user237522
  • 6,467

1 Answers1

2

Yes, the proof from the commutative case still works. Explicitly, let $(b_1,\dots,b_n)$ be a basis for $B$ over $A$ and let $(c_1,\dots,c_m)$ be a basis for $C$ over $B$. The claim is then that the products $b_ic_j$ are a basis for $C$ over $A$.

To prove they span, let $x\in C$ and write $x=\sum x_jc_j$ for $x_j\in B$. We can then write $x_j=\sum x_{ij}b_i$ for $x_{ij}\in A$, so we have $x=\sum x_{ij}b_ic_j$.

To prove they are linearly independent, suppose $\sum x_{ij}b_jc_j=0$ for $x_{ij}\in A$. We can group the terms according to $j$ to get $\sum_j\left(\sum_i x_{ij}b_j\right)c_j=0$. Since $\sum x_{ij}b_j\in B$ and the $c_j$ are linearly independent over $B$, we have $\sum x_{ij}b_j=0$ for all $i$. Now linear independence of the $b_j$ over $A$ gives $x_{ij}=0$ for all $i,j$.

Eric Wofsey
  • 330,363