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The following fact is used in the Unitary space.

If $F$ is a field having an automorphism $\alpha$ of order 2. Let $F_0=\{a\in F: \alpha(a)=a\}$. Then $|F:F_0|=2$.

Is there any easy proof (or reference) for this fact?

Thank WimC give the perfect answer. I add my own idea about this problem:

Let $Char F \ne 2$, and write $\bar{a}$ for $\alpha(a)$. Note $x=1/2(x+\bar{x})+1/2(x-\bar{x})\ \ \ (*)$ and $\overline{x-\bar{x}}=-(x-\bar{x})$ and $x+\bar{x} \in F_0$. So there exists $a$ such that $\bar{a}=-a$ and $a \not \in F_0$. By (*), we just need to proof the elements $b$ such that $\bar{b}=-b$ is in the space $K$ generated by 1 and $a$ over $F_0$ (Clearly, $|K:F_0|=2$).

Since $\bar{a}=-a$, we get $k=a^2\in F_0$, and $a^{-1}=k^{-1}a \in K$. Now $\overline{ab}=\bar{a}\bar{b}=(-a)(-b)=ab$, we get $l=ab\in F_0$, and $b=la^{-1} \in K$. Hence $K=F$ and then $|F:F_0|=2$.

But I can not prove the case that $Char K=2$. Any idea about this case?

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2 Answers 2

up vote 10 down vote accepted

This is a special case of Artin's theorem from Galois theory. The field $F_0$ is strictly smaller than $F$ (not all elements are fixed by $\alpha$). If $a, b \in F \setminus F_0$ then

$$a - \frac{a-\alpha(a)}{b-\alpha(b)} b + \frac{a \, \alpha(b)-b\,\alpha(a)}{b-\alpha(b)} =0$$

which shows that $1, a, b$ are linearly dependent over $F_0$. Therefore $|F:F_0|=2$.

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Sweet! Artin's book is a gem, isn't it! +1! –  Robert Lewis Jul 17 at 6:58
    
Thanks. A beautiful proof. I have think about this for a long time. –  Wei Zhou Jul 17 at 8:03
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Here's a different demonstration which, like Wei Zhou's argument, works in the case $\text{char} F \ne 2$, although I admit part of it was inspired by WimC's most excellent answer. But, based upon that inspiration (and exactly what that is will shortly become manifest), we have a self-contained proof as follows:

First of all, we note that, unless $\alpha$ is the trivial automorphism, i.e., unless $\alpha$ fixes all of $F$, that there must exist $\aleph \in F$ with $\alpha(\aleph) \ne \aleph$; by definition, such an $\aleph \notin F_0$, and $\alpha(\aleph) \notin F_0$ either, lest we have $\aleph = \alpha(\alpha(\aleph)) = \alpha(\aleph) \in F_0$. Choosing such an $\aleph$, consider the field elements $\aleph + \alpha(\aleph), \aleph \alpha(\aleph) \in F$; we see they are both fixed by $\alpha$, for

$\alpha(\aleph + \alpha(\aleph)) = \alpha(\aleph) + \alpha^2(\aleph) = \aleph + \alpha(\aleph) \tag{1}$

since $\alpha^2 = 1$, and likewise

$\alpha(\aleph \alpha(\aleph)) = \alpha(\aleph) \alpha^2(\aleph) = \aleph \alpha(\aleph), \tag{2}$

again since $\alpha^2 = 1$. Thus we see that both $\aleph + \alpha(\aleph), \aleph \alpha(\aleph) \in F_0$, since $F_0$ is the fixed field of $\alpha$. Next consider the polynomial

$p_\aleph(x) = x^2 - (\aleph +\alpha(\aleph))x + \aleph \alpha(\aleph) \in F_0[x]; \tag{3}$

we have $p_\aleph(x) \in F_0[x]$ since its coefficients, as has been seen, are all fixed by $\alpha$. The roots of $p_\aleph(x)$ are easily seen to be $\aleph$ and $\alpha(\aleph)$; indeed we have

$p_\aleph(\aleph) = \aleph^2 - (\aleph + \alpha(\aleph)) \aleph + \aleph \alpha(\aleph) = \aleph^2 -\aleph^2 - \alpha(\aleph) \aleph + \aleph \alpha(\aleph) = 0, \tag{4}$

with a similar calculation showing that

$p_\aleph(\alpha(\aleph)) = 0 \tag{5}$

as well; alternatively, it may be observed that $p_\aleph(x)$ splits in $F$ as

$p_\aleph(x) = x^2 - (\aleph +\alpha(\aleph))x + \aleph \alpha(\aleph) = (x - \aleph)(x - \alpha(\aleph)), \tag{6}$

which also shows the roots are $\aleph$, $\alpha(\aleph)$. Based on these considerations, we may conclude that (i) $p_\aleph(x)$ is irreducible over $F_0$, since $\aleph, \alpha(\aleph) \notin F_0$; (ii.) $F_0(\aleph) \subset F$ is the splitting field of $p_\aleph(x)$ over $F_0$, since $\alpha(\aleph) = \aleph^{-1}(\aleph \alpha(\aleph)) \in F_0(\aleph)$ by virtue of $\aleph \in F_0(\aleph)$, $\aleph \alpha(\aleph) \in F_0 \subset F(\aleph)$; (iii.) $[F_0(\aleph):F_0] = 2$, since $\deg p_\aleph(x) = 2$.

Having $[F_0(\aleph):F_0] = 2$, we conclude by showing that $F = F_0(\aleph)$. Clearly $F_0(\aleph) \subset F$, so let $\beth \in F$ and consider the product $\gimel = (\beth - \alpha(\beth))(\aleph - \alpha(\aleph))$; we have

$\alpha((\beth - \alpha(\beth))(\aleph - \alpha(\aleph))) = (\alpha(\beth) - \beth)(\alpha(\aleph) - \aleph) = (\beth - \alpha(\beth))(\aleph - \alpha(\aleph)), \tag{7}$

that is, $\gimel = (\beth - \alpha(\beth))(\aleph - \alpha(\aleph))$ is fixed by $\alpha$, hence

$(\beth - \alpha(\beth))(\aleph - \alpha(\aleph)) = \gimel \in F_0. \tag{8}$

We now have

$\beth - \alpha(\beth) = (\aleph - \alpha(\aleph))^{-1} \gimel \in F_0(\aleph), \tag{9}$

and since

$\beth + \alpha(\beth) \in F_0, \tag{10}$

being fixed by $\alpha$ just as is $\aleph + \alpha(\aleph)$, we conclude that (and this is where we need the assumption $\text{char}F \ne 2$):

$2\beth = (\beth - \alpha(\beth)) + (\beth + \alpha(\beth)) \in F_0(\aleph), \tag{11}$

whence

$\beth \in F_0(\aleph) \tag{12}$

and hence $F \subset F_0(\aleph)$; thus in fact $F = F_0(\aleph)$ and finally

$[F:F_0] = [F_0(\aleph): F_0] = 2, \tag{13}$

the desired conclusion. QED.

Note: The inspiration I took from WimC's answer was to examine the quantity $\gimel = (\beth - \alpha(\beth))(\aleph - \alpha(\aleph))$; this originated in a careful scrutiny of the coefficient of $b$ in his equation

$a - \dfrac{a-\alpha(a)}{b-\alpha(b)} b + \dfrac{a \, \alpha(b)-b\,\alpha(a)}{b-\alpha(b)} =0, \tag{14}$

which is also invariant under $\alpha$. I too would like to see if and how the assumption $\text{char} F \ne 2$ could be circumvented in the context of the above argument. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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Interesting answer. –  Wei Zhou Jul 18 at 0:49
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