For a field $K$, a polynomial $f \in K[X]$ and its splitting field $L$, we define the Galois group of the polynomial as $$\text{Gal}(f) := \text{Aut}(L/K)$$

The elements of $\text{Aut}(L/K)$ are the $K$-automorhisms of $L$ which means that they act as the identity on $K$. Why do we require this condition? What goes wrong in analyzing polynomials if we used arbitrary automorphisms from $L \rightarrow L$ instead?

I edited the question slightly because I didn't know the definition of $K$-isomorhism when I initially asked the question.

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    $\begingroup$ By definition. $\operatorname{Aut}\left(L/K\right)$ is defined to be the set of all $K$-linear field automorphisms of $L$. A $K$-linear field automorphism acts as the identity on $K$. $\endgroup$ Mar 7 '15 at 16:46
  • $\begingroup$ Maybe I don't get the notation Aut$(L/K)$ yet. I can't find the place where it has been defined in the lecture notes I'm using. I thought it was simply the automorphisms from L to L. I also don't know what a $K$-linear field automorphism is. $\endgroup$
    – Marc
    Mar 7 '15 at 16:53
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    $\begingroup$ A field automorphism which is $K$-linear as a map! Recall that $L$ is a $K$-vector space. If your lecture defines $\operatorname{Aut}\left(L/K\right)$ as the set of all field automorphisms of $L$, then your lecture is wrong. $\endgroup$ Mar 7 '15 at 17:00
  • $\begingroup$ Ah, thanks. No, I said that I can't find the definition of Aut$(L/K)$ in my lecture notes at all. It just appears in the definition of the Galois group. This is really sloppy because the definition you gave isn't motivated naturally by the definitions of $L/K$ and the automorphism group alone. $\endgroup$
    – Marc
    Mar 7 '15 at 17:07

We want $\text{Gal}(f)$ to be "tied" to $f$, somehow. Note that any automorphism $\phi: L \to L$ is going to induce an automorphism:

$\tilde{\phi}: L[x] \to L[x]$ given by, for $g(x) = a_0 + a_1x +\cdots + a_nx^n$:

$\tilde{\phi}(g(x)) = \phi(a_0) + \phi(a_1)x + \cdots + \phi(a_n)x^n$

If, in particular, $g(x) \in F[x]$ (that is, all the coefficients are in $F$), and we want $\tilde{\phi}(g(x)) = g(x)$, for any $g(x) \in F[x]$, we need $\phi(a) = a$, for all $a \in F$.

At a more basic level, we want to examine the nature of roots of $f$, obtained by enlarging our field $F$. A field extension of $F$ is a vector space over $F$, so it is "natural" to look at field automorphisms that are $F$-vector space isomorphisms, as well (our starting field is the "scalars", and our larger field is the "vectors"). So we restrict our attention to the $L$-automorphisms $\phi$ that are also $F$-linear transformations. So, for $a \in F$:

$\phi(a) = \phi(a\cdot 1_K) = a\phi(1_K) = a\cdot 1_K = a$.

(Note as well, that since $F^{\ast}$ is a multiplicative subgroup of $K^{\ast}$, that $1_F = 1_K$).

If our $L$-automorphisms don't "fix" $F$, there is no reason to suppose these automorphisms have any connection to $f$, since it is quite likely they will send $f$ to a polynomial outside of $F[x]$.


This boils down to the definition of a $K$-homomorphism $L\rightarrow L$. This is by definition a $K$-algebra homomorphism, or in other words a ring morphism and an $K$-module morphism, or in other words a ring morphism which is the identity on $K$.

  • $\begingroup$ Sorry Heinz, I can't really follow this because "morphism" and "module" are terms with which I am not very familiar. $\endgroup$
    – Marc
    Mar 7 '15 at 17:08
  • $\begingroup$ as $K$ is a field, a $K$-module is the same as a $K$-vector space. morphism is short for homomorphism. $\endgroup$
    – user220467
    Mar 7 '15 at 17:43
  • $\begingroup$ Ok, thanks. But why do we need the automorphisms to be $K$-homomorphisms. What goes wrong in analyzing polynomials if we have simple automorphisms from $L \rightarrow L$? (I've also edited my question) $\endgroup$
    – Marc
    Mar 7 '15 at 17:47

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