# If field $K/F$ is generated by the $\alpha_1,…,\alpha_n$, then an $\sigma\in$ Aut$(K/F)$ of $K$ uniquely determined

Does this proof seem correct? I'm having second doubts concerning the bolded material.

Show that if the field $K$ is generated over $F$ by the elements $\alpha_1,...,\alpha_n$, then an automorphism $\sigma$ of $K$ fixing $F$ is uniquely determined by $\sigma(\alpha_1),...,\sigma(\alpha_n)$. In particular show that an automorphism fixes $K$ iff it fixes a set of generators for $K$.

By hypothesis, $K=F(\alpha_1, ...,\alpha_n)=\{f_0+f_1\alpha_1+...+f_n\alpha_n:f_i\in F\}$. Now let $\sigma\in$ Aut$(K/F)$. When $n=0$, then $\sigma=id$ and so Aut$(K/F)$ is trivial. It follows that $\sigma$ is uniquely determined. Suppose now that $L=F(\alpha_1,...,\alpha_{n-1})$, where $\sigma \in$ Aut$(L/F)$ is uniquely determined by $\sigma(\alpha_1),...,\sigma(\alpha_{n-1})$. Then $L(\alpha_n)=\{l_0+l_1\alpha_n: l_i \in L\}$, and so for any $l\in L(\alpha_n)$ we can write $l=l_0+l_1\alpha_n$. It follows that $\sigma(l)=\sigma(l_0)+\sigma(l_1)\sigma(\alpha_n)$, and so

$\sigma(\alpha_n)=(\sigma(l)-\sigma(l_0))(\sigma(l_1))^{-1}$. ..................(1)

Using the induction hypothesis, $\sigma(l),\; \sigma(l_0),\; \sigma(l_1)$ uniquely determine $\sigma$. Therefore, (1) gives that $\sigma(\alpha_n)$ also uniquely determines $\sigma$.

• Your description of $\;F(a_1,.,.,a_n)\;$ is a little weird to me, not to say wrong. In fact, $\;\Bbb Q(\sqrt2,\,\sqrt3)\neq\{a+b\sqrt2+c\sqrt3\;:\;a,b,c\in\Bbb Q\}\;$ , for example ! – Timbuc Apr 5 '15 at 21:58
• Oh right, because you have to include $\sqrt{6}$. That makes $K$ slightly harder to describe using sets. Regardless, my representation for $K$ doesn't really affect the proof until I represent $L(\alpha_n)$ using the same notation. But I think $L(\alpha_n)=\{l_0+l_i\alpha_n:l_i\in L\}$ is correct, right? – pretzelman Apr 5 '15 at 22:03
• I don't really need to represent elements of $K$ as much as I do elements of $L(a_n)$. – pretzelman Apr 5 '15 at 22:03
• In accordance with an exchange of comments that we had, I should point out that it’s not true that an automorphism fixes $K$ iff it fixes a set of generators for $K$. The correct statement is that $\sigma$ fixes $K$ iff it fixes $F$ and a set of generators (over $F$) of $K$. – Lubin Apr 6 '15 at 1:26

## 3 Answers

You could write

$$F(a_1,...,a_n)=\left\{\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)}:f(X_1,...,X_n),\,g(X_1,...,X_n)\in F[X_1,...,X_n],\,g(a_1,...,a_n)\neq 0\right\}$$

and then, since the coefficients of each such rational function $\;\frac fg\;$ are in the fixed field of Aut$\,(K/F)\;$ , we get that

$$\sigma\in\,\text{Aut}\,(K/F)\implies \sigma(\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)})=\frac{f(\sigma(a_1),...,\sigma(a_n))}{g(\sigma(a_1),...,\sigma(a_n))}$$

and since any $\;k\in K\;$ is of the form $\;\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)}\;$ , we're done

If $\;a_1,...,a_n\;$ are algebraic over $\;F\;$ the above reduces to polynomials instead of rational functions.

• This only works if all $a_i$ are algebraic. But it is easy to fix. – lhf Apr 5 '15 at 22:11
• True. I assumed they're algebraic but it isn't given. As lhf says it is easy though to fix: instead of polynomials we must take rational functions $\;\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)}\;$ , with $\;f,g\in F[X_1,...,X_n]\;,\;\;g(a_1,...,a_n)\neq 0\;$ – Timbuc Apr 5 '15 at 22:15
• I'm sort of hazy as to why every element of $K$ can be expressed as a rational polynomial of $a_1,...,a_n$. – pretzelman Apr 5 '15 at 22:58
• is it because if $a_i$ were algebraic, then you get the field axioms from them and so you need to account for the inverses in the case that some of the $a_i$ are not algebraic? – pretzelman Apr 5 '15 at 23:00

To start with, it’s useful to think of everything happening inside a field $\Omega$ that contains $F$ and all the $\alpha_i$. For a proof of this kind, I think it’s most efficient to think of $F(\alpha_1,\cdots,\alpha_n)$ as the smallest subfield of $\Omega$ containing $F$ and the alphas, or, if you like, the intersection of all subfields of $\Omega$ containing $F$ and the alphas. This avoids the explicit description of elements of $F(\alpha_1,\cdots,\alpha_n)$ that you were hoping to use.

Now to prove the claim: certainly if $\sigma$ fixes $K=F(\alpha_1,\cdots,\alpha_n)$, then it fixes $F$ and the alphas. Conversely, suppose $\sigma$ is identity on $F$, and $\sigma(\alpha_i)=\alpha_i$ for all $i$. Now let $S$ be the set $\lbrace z\in\Omega\colon\forall i,\sigma(\alpha_i)=\alpha_i\rbrace$. You immediately see that this $S$ is a subfield of $\Omega$, contains $F$ and the $\alpha_i$’s, and therefore is one of those fields mentioned in the intersection-definition of $K$. Thus $K\subset S$, in other words, $\sigma$ fixes all elements of $K$.

Of course if you haven’t seen what I used as the definition of the field generated over $F$ by a set of elements, then you have to sit down and prove that my definition is equivalent to the one that @Timbuc used. As he says, you have to allow $F$-rational expressions in the generators, not just $F$-linear ones. In any event, it’s not at all hard to cook up a proof.

• I like this a lot. The only problem I have is why you get to assume $\sigma$ is the identity on $F$ in the converse? – pretzelman Apr 5 '15 at 23:18
• Oh my, if $\sigma$ mixes up the elements of $F$, that will be bad, and it will not leave all elements of $K$ fixed. – Lubin Apr 6 '15 at 1:23

$K$ is an $F$-algebra generated by $a_1, a_2, \ldots, a_n$. Any $F$-algebra homomorphism $\sigma$ of $K$ into an $F$-algebra $L$ is determined by $\sigma(a_1), \sigma(a_2), \ldots, \sigma(a_m)$ (because any $x \in K$ can be written as a rational function with coeefficients in $F$ of $a_1, a_2, \ldots, a_n$, and then $\sigma(x)$ will be the same rational function of $\sigma(a_1), \sigma(a_2), \ldots, \sigma(a_n)$). Apply this with $L = K$.