3
$\begingroup$

I have another abstract algebra question. I stated it in the title, but here it is in more detail:

Let $F\subseteq R \subseteq E$ where $E$ is an algebraic extension of the field $F$. If $R$ is an $F$-subspace of $E$, $\text{char}R \ne 2$, and $u\in R$ implies that $u^k\in R$ for each $k\geq 2$, show that $F$ is a field.

I've already gotten that, if $R$ is a subring of $E$, then it is a field. So, all I need to do is show that $R$ is closed under multiplication. I think it has to do with minimal polynomials, especially because of the $u^k$ assumption, but I can't figure out how to get $r_1 r_2\in R$ if $r_1,r_2\in R$. Thanks in advance!

  • 0
    why the fact that$R$is a subring of E can imply that R is also a field?2017-03-28

1 Answers 1

5

If $x, y \in R$, $(x + y)^2 = x^2 + 2xy + y^2 \in R$. Hence $2xy \in R$. If $char(F) \neq 2$, $xy \in R$.

  • 0
    @Samuel: FYI, this is a standard trick used to convert between quadratic functions and bilinear functions, specifically to recover the bilinear function (in this case, $xy$) from its associated quadratic form (in this case, $x^2$).2012-11-22