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I am a little confused about actions right now.

Say I have a group $G$ that acts on the roots of a polynomial, say $(x - x_1)(x - x_2)(x - x_3)$. We know that this group permutes the roots of a polynomial, that is you can get to any of the roots by having an element of the group act on another root. Does that mean the group fixes the polynomial?

I feel like it should, but I'm not sure how this actually works. Supposedly we would have to write $g \cdot [(x - x_1)(x - x_2)(x - x_3)] = (x - x_1)(x - x_2)(x - x_3)$. But is this actually possible? Because $G$ acts on the set of the roots, not the polynomial. I'm not sure if we can say a polynomial is fixed by certain things acting on the roots. When they say the polynomial is fixed, does that mean the coefficients are fixed?

Also, can you say something like $g \cdot (x_1x_2 + x_3)$ = $(g \cdot x_1)(g \cdot x_2) + g \cdot x_3$?


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Re your second question, the group action presumably must specify the properties you mentioned, i.e. each element acts as a homomorphism on the field. – Yuval Filmus May 22 '11 at 20:54
Awesome, thanks – badatmath May 22 '11 at 21:02
Note that "the group permutes the roots" is not equivalent to "you can get to any of the roots by having an element of the group act on another root." Saying the group permutes the roots is equivalent to saying it acts on them. The property that given any two roots, $r_1$ and $r_2$, there is an element of $g$ that maps $r_1$ to $r_2$ is described by saying that the group acts transitively on the roots. Also, if the group only acts on the roots, then it does not make sense to talk about action on the polynomial: the polynomial is not its roots. You are confusing a lot of things here! – Arturo Magidin May 22 '11 at 21:58
@Arturo: I see. If the group permutes the roots, must there be a bijection between the set of the roots and the set of their images? – badatmath May 22 '11 at 23:26
@badatmath: If the group "permutes the roots", then the set of roots is the set of its images. I'm actually writing a long-ish answer discussing group actions. I think part of the problem is you are not very clear on what "group action" means in general, or how it is being applied when discussing Galois groups and field extensions. – Arturo Magidin May 22 '11 at 23:27
up vote 6 down vote accepted

Let's back up a little; you seem to be confused about more fundamental things right now.

Fix a field $F$. If $f(x) \in F[x]$ is a monic polynomial, we can talk about the splitting field of $f$, which is a field extension $K/F$ obtained by adjoining all of the roots $r_1, ... r_n$ of $f$. For various reasons it is natural to consider a certain group $G = \text{Aut}(K/F)$, the group of all invertible field homomorphisms $g : K \to K$ which fix $F$. This means that

  • $g(a + b) = g(a) g(b)$
  • $g(ab) = g(a) g(b)$
  • $\forall a \in F : g(a) = a$.

This should hopefully answer your last question. The properties above imply that $g(f(r)) = f(g(r))$ for any $r \in K$, from which it follows that $g(f(r)) = 0$ if and only if $f(g(r)) = 0$. Since $g$ is an automorphism, this implies that $r$ is a root of $f$ if and only if $g(r)$ is a root of $f$, hence that $G$ permutes the roots of $f$.

Now, the elements of $G$ act by automorphisms $K \to K$, but we can extend this action to an action by automorphisms $K[x] \to K[x]$ by declaring that $x$ is sent to $x$ by every element of $G$ and extending by linearity and multiplicativity. So one way to make sense of the statement that elements of $G$ fix the polynomial $f(x)$ is that they fix the element $f(x) \in F[x] \subset K[x]$. There are (at least) two ways to see this.

First, elements of $G$ fix elements of $F$ by definition, and since $f(x)$ is a sum of elements of $F$ and powers of $x$ (which $G$ also fixes by definition), the result is clear.

Second, note that in $K[x]$ we can write

$$f(x) = (x - r_1)(x - r_2)...(x - r_n)$$

so it follows by the homomorphism property that

$$g(f(x)) = (x - g(r_1))(x - g(r_2))...(x - g(r_n))$$

and since $g$ permutes the roots of $F$ it follows that the expression on the RHS is just a reordering of the factors of $f(x)$.

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Awesome answer, thanks. I didn't realize you were supposed to extend the action. – badatmath May 22 '11 at 23:19
@badatmath It's good to remember that a homomorphism $F \rightarrow F$ extends naturally to one $F[X] \rightarrow F[X].$ In fact, many people use the same symbol for both homomorphisms. – Kopper May 23 '11 at 1:37
@Jay: I guess next time I'll remember. That got me on a final exam last term too. – badatmath May 23 '11 at 3:01

Let's talk a bit about "group actions" as well.

Given a group $G$ and a nonempty set $X$, a (left) group action of $G$ on $X$ means an operation $\cdot\colon G\times X\to X$ such that for all $g,h\in G$ and $x\in X$, $$\begin{align*} h\cdot(g\cdot x) &= (hg)\cdot x;\\ 1\cdot x &= x, \end{align*}$$ An easy example of a group action is given as follows: Let $X$ be a nonempty set, and let $S_X$ be the group of all bijections from $X$ to itself (the "permutations on $X$"). Then we can define an action of $S_X$ on $X$ by $\sigma\cdot x = \sigma(x)$.

Similarly, if $H\lt S_X$, then $H$ acts on $X$ in a natural way, by letting $h\cdot x = h(x)$ for every $h\in H$.

Another example: suppose that $G$ acts on $X$ and $H$ acts on $Y$. Then we can let $G\times H$ act on $X\times Y$ in an obvious way (coordinate-wise): given $(g,h)\in G\times H$ and $(x,y)\in X\times Y$, let $(g,h)\cdot(x,y) = (g\cdot x,h\cdot y)$.

Yet another example: If $G$ acts on both $X$ and $Y$, then $G$ acts on $X\times Y$ "coordinatewise" as well: $g\cdot (x_1,x_2) = (g\cdot x_1,g\cdot x_2)$. More generally, if $G$ acts on each set of a family $\{X_i\}_{i\in I}$, then $G$ acts on the cartesian product $\prod\limits_{i\in I}X_i$ coordinate-wise.

Another example: suppose that $G$ acts on $X$, and $\iota\colon X\hookrightarrow Y$ is a one-to-one function from $X$ to another set $Y$. Then we can define an action of $G$ on $Y$ by "extension" as follows: given any $g\in G$ and any $y\in Y$, we define $g\cdot y$ by: $$g\cdot y = \left\{ \begin{array}{ll} \iota(g\cdot x)&\mbox{if $y=\iota(x)$;}\\ y &\mbox{if $y\notin\iota(X)$.} \end{array}\right.$$

Another example: Let $G$ act on $X$. We say that a subset $Y\subseteq X$ is $G$-invariant if and only if $G\cdot Y = \{g\cdot y\mid y\in Y\}\subseteq Y$. Then the restriction of the $G$-action to $Y$ gives us an action of $G$ on $Y$.

Another example: suppose that $G$ acts on $X$, and $f\colon K\to G$ is a homomorphism. Then $K$ acts on $X$ "via $f$", by letting $k\cdot x = f(k)\cdot x$.

In a sense, every action can be thought of as an action this way, in the following sense:

Theorem. Let $G$ be a group, and $X$ be a set. Then the following data are equivalent:

  1. A left action of $G$ on $X$, $\cdot\colon G\times X\to X$.
  2. A group homomorphism $\phi\colon G\to S_X$ from $G$ to the permutation group of $X$.

Proof. Assume $\cdot\colon G\times X\to X$ is a group action. For each $g\in G$, let $\phi_g\colon X\to X$ be the map $\cdot(g,-)\colon X\to X$ (currying $\cdot$) defined by $\phi_g\colon x\mapsto g\cdot x$. This is a bijection, because $\phi_{g^{-1}}$ is its inverse: $$\begin{align*} \phi_{g^{-1}}\circ\phi_g(x) &= g^{-1}\cdot(g\cdot x) = (g^{-1}g)\cdot x = 1\cdot x = x,\\ \phi_g\circ\phi_{g^{-1}}(x) &= g\cdot(g^{-1}\cdot x) = (gg^{-1})\cdot x = 1\cdot x= x.\end{align*}$$ That is, the map $g\longmapsto \phi_g$ is a function from $G$ to $S_X$. Morover, $\phi_g\circ\phi_h = \phi_{gh}$, since $g\cdot(h\cdot x) = (gh)\cdot x$, so the map is a homomorphism. Thus, an action of $G$ on $X$ yields a homomorphism $\phi\colon G\to S_X$ by $\phi(g) = \phi_g$.

Conversely, suppose that $\phi\colon G\to S_X$ is a group homomorphism. This induces an action of $G$ on $X$ by $g\cdot x = \phi(g)(x)$, as above.

Moreover, let $\psi\colon G\to S_X$ be a homomorphism, and let $\cdot_{\psi}$ be the induced action. Then $\phi_g(x) = g\cdot_{\psi}(x) = \psi(g)(x)$, so the homomorphism we get from the action induced by $\psi$ is $\psi$. Likewise, if you start with an action $\cdot$, and we let $\phi\colon G\to S_X$ be the induced homomorphism as above, then the action $\cdot_{\phi}$ induced by this homomorphism is $g\cdot_{\phi}x = \phi_g(x) = g\cdot x$, so we get back our original action. That is, the data are equivalent. QED

Another way of defining actions is by restriction: suppose $G$ acts on a set $X$; given $x\in X$, the orbit of $x$ is the set $$G\cdot x = \{ g\cdot x\mid g\in G\}.$$ Then $G$ acts on $G\cdot x$ in a natural way: we think of this as a *restriction * of the action of $G$, from considering what it does to all of $X$ to considering just what it does to $G\cdot x$.

An action of $G$ on $X$ induces an equivalence relation on $X$ by $x\sim y$ if and only if there exists $g\in G$ such that $g\cdot x = y$; equivalently, if and only if $y$ is in the orbit of $x$. This equivalence relation partitions $X$ into "$G$-orbits."

Two important properties of an action are faithfulness and transitivity:

Definition. Let $G$ act on $X$. The action of $G$ is faithful if $$g\cdot x = x\text{ for all }x\in X\Longleftrightarrow g=1.$$ Equivalently, if and only if the homomorphism $G\to S_X$ induced by the action is one-to-one.

Definition. Let $G$ act on $X$. The action of $G$ is transitive if for every $x,y\in G$ there exists $g\in G$ such that $g\cdot x = y$. Equivalently, if and only if $X$ consists of a single $G$-orbit.

To see an example of an action that is not transitive, let $G=S_3$ and let $G$ act on the set $\{1,2,3,4,5,6\}$ as follows: given $\sigma\in S_3$, let $\sigma\cdot i = \sigma(i)$ if $1\leq i\leq 3$, and let $\sigma\cdot j = 3+\sigma(j-3)$ if $4\leq j\leq 6$. For example, if $\sigma=(1,3)$, then $$\sigma(1)=3,\quad \sigma(2)=2,\quad \sigma(3)=1,\quad \sigma(4)=6,\quad \sigma(5)=5,\quad \sigma(6)=4.$$ The action is not transitive because $X$ has two $G$-orbits: $\{1,2,3\}$ and $\{4,5,6\}$.

For a more interesting example, let $X=\mathbb{Z}_2\oplus \mathbb{Z}_2$ be the Klein $4$-group, and let $G=\mathrm{Aut}(C_2\times C_2)$. Then $G$ acts on $X$ in the obvious way (apply the homomorphism). If $x=(0,0)$, then for every $\phi\in G$ we have $\phi(x)=x$ (since group automorphisms must send the identity to the identity). On the other hand, we have an automorphism that exchanges $(1,0)$ and $(0,1)$ and fixes $(1,1)$; an automorphism that exchanges $(1,0)$ and $(1,1)$ and fixes $(0,1)$; and an automorphism that exchanges $(0,1)$ and $(1,1)$ and fixes $(1,0)$. So that for every $x\in X$, $x\neq (0,0)$, and every $y\in X$, $y\neq (0,0)$, there exists $g\in G$ such that $g\cdot x = y$. That is, $X$ is partitioned into two $G$-orbits, $\{(0,0)\}$ and $\{(1,0), (0,1), (1,1)\}$.

So you see why, in your question, saying the group "permutes the roots" is not the same as saying that for any two roots there is an element of the group that maps the first root to the second. The first statement is saying that there is an action of $G$ on the set of roots; the second statement is saying that the action is transitive, and not every action is transitive.

Now, suppose that $G$ acts on a field $F$; the action may have absolutely nothing to do with the field structure of $F$ (it may just be an action on the underlying set of $F$), but these kinds of actions are not very interesting when you think of $F$ as a field. Instead, if you want to think of $F$ as a field, then you want to think of the action of $G$ as being an action "by automorphisms"; that is, we identify $G$ with a subgroup of the group $\mathrm{Aut}(F)$ of all field automorphisms of $F$, which is itself a subgroup of the group $S_F$ of all permutations on the underlying set of $F$.

Now suppose that $G$ acts on the field $F$. Then the action of $G$ on $F$ naturally induces an action of $G$ on $F[x]$, the polynomials with coefficients on $F$: remember that we can think of polynomials with coefficients in $F$ as the almost-null sequences of elements of $F$, that is, the infinite tuples $(a_0,a_1,a_2,\ldots,a_n,\ldots)$ with $a_i\in F$ and $a_i=0$ for almost all $i$; the action of $G$ on $F[x]$ is defined coordinatewise, as before.

But if the action of $G$ on $F$ is by field automorphisms, then we get more: the action of $G$ commutes with the evaluation homomorphism! Remember that for ever $a\in F$ we have an "evaluate at $a$" homomorphism $\varepsilon_a\colon F[x]\to F$ by $\varepsilon_a(p(x)) = p(a)$. If the action of $G$ is by field automorphisms, then for every $$p(x) = \alpha_nx^n + \cdots +\alpha_0\in F[x]$$ and for every $a\in F$, and we denote by $\phi_g$ the automorphism corresponding to the action of $g\in G$, we have: $$\begin{align*} g\cdot\Bigl(\varepsilon_a\bigl(p(x)\bigr)\Bigr) &= g\cdot\Bigl(p(a)\Bigr)\\ &= g\cdot\Bigl( \alpha_na^n + \cdots + \alpha_0\Bigr)\\ &= \phi_g\Bigl( \alpha_na^n + \cdots + \alpha_0\Bigr)\\ &= \phi_g(\alpha_n)\phi_g(a)^n + \cdots + \phi_g(\alpha_0)\\ &= (g\cdot p)(g\cdot a)\\ &= \varepsilon_{g\cdot a}\circ(g\cdot p(x)) \end{align*}$$ where $g\cdot p$ is the image of $p(x)$ under the action of $g$, namely, $$g\cdot p = (g\cdot \alpha_n)x^n + \cdots + (g\cdot \alpha_0).$$

Because any automorphism of a field must send $0$ to $0$, we have:

Theorem. Let $F$ be a field, and let $G$ act on $F$ by automorphisms. Then for every $p(x)\in F[x]$, every $a\in F$, and every $g\in G$, $a$ is a root of $p(x)$ if and only if $g\cdot a$ is a root of $g\cdot p(x)$.

Proof. $a$ is a root of $p(x)$ if and only if $p(x)$ lies in the kernel of the evaluation map $\varepsilon_a$, if and only if $\varepsilon_a(p(x))=0$, if and only if $g\cdot\varepsilon_a(p(x)) = g\cdot 0 = 0$, if and only if $\varepsilon_{g\cdot a}(g\cdot p(x)) = 0$, which holds if and only if $g\cdot a$ is a root of $g\cdot p(x)$. QED

In particular, if $g\cdot p(x) = p(x)$ (that is, if $g$ fixes the coefficients; note that even with a faithful action you can have $g\cdot a = a$ for some $g\in G$ and some $a\in X$), then $a$ is a root of $p(x)$ if and only if $g\cdot a$ is a root of $p(x)$. That is, the roots of $p(x)$ are a $G$-invariant subset of $F$, and therefore the action of $G$ on $F$ restricts to an action of $G$ on the roots of $p(x)$.

Conversely, that the action of $G$ on $F$ is by automorphisms, that $p(x)$ is a polynomial with coefficients on $F$ that splits over $F$, and that the set of roots of $p(x)$ is a $G$-invariant subset of $F$. Since the coefficients of $p(x)$ are symmetric polynomials on the roots of $p(x)$, and symmetric polynomials are invariant under any permutation of the arguments, then it follows that $G$ fixes every coefficient of $p(x)$.

However, even in this situation the action of $G$ on the set of roots need not be transitive. For example, consider the case of the complex numbers, $F=\mathbb{C}$, and let $G$ be the group consisting of the identity and complex conjugation (so $G$ is isomorphic to the cyclic group of order $2$). If $p(x)$ is a polynomial with real coefficients, then the set of roots of $p(x)$ is $G$-invaraiant (since a complex number is a root of a real polynomial if and only if its complex conjugate is a root as well), so that the action of $G$ induces an action on the roots of $p(x)$. If $p(x) = (x^2+1)(x^2+4)(x^2-1)$, for instance, then $G$ acts on the set of roots of $p(x)$, $X=\{1,-1,i,-i, 2i, -2i\}$: the identity fixes every element, and complex conjugation fixes $1$ and $-1$, exchanges $i$ and $-i$, and exchanges $2i$ and $-2i$; the $G$-orbits of $X$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, and $\{2i,-2i\}$. The action is not transitive.

However, the situation that usually arises in Galois Theory is even more restricted. There, we are usually considering not one field, but a pair of fields in a field extension, $F\subseteq K$. The group whose action we are considering is a group of automorphisms of $K$, but not any old automorphisms, but rather automorphisms that fix $F$ pointwise; that is, for every $g\in G$, $g$ acts on $K$ by automorphisms in such a way that $g\cdot f = f$ for every $f\in F$. In that situation, any polynomial $p(x)$ with coefficients in $F$ will be fixed by the (induced) action of $G$. In particular, the set of roots of $p(x)$ that are in $K$ is $G$-invariant, and so $G$ acts on the set of roots of $p(x)$ that are in $K$; the action is necessarily "by permutations", since every action works via permutations.

Conversely, if you have a finite $G$-invariant subset $S$ of $K$, then the polynomial $$p(x) = \prod_{s\in S}(x-s)$$ is fixed by the action of $G$, so that the coefficients lie in the "fixed field of $G$", which is the set of all $k\in K$ such that $g\cdot k = k$ for all $g\in G$. When the extension is a Galois extension, and $G$ is the collection of all automorphisms of $K$ that fix $F$ pointwise, then the fixed field of $G$ is precisely $F$ (in fact, that is one possible definition of "the extension is a Galois extension"), so that the polynomial $p(x)$ above must have coefficients in $F$.

Always, if the action of $G$ on $K$ is "by automorphisms", then you necessarily have $g\cdot(a+b) = g\cdot a + g\cdot b$ and $g\cdot(ab) = (g\cdot a)(g\cdot b)$. However, if the action is not "by automorphisms", then you generally cannot say that.

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Thank you, it was very nice of you to write that. You are amazing and should write textbooks. – badatmath May 23 '11 at 2:42

The coefficients of the polynomial are symmetric functions of the roots. In your particular case the polynomial is $$ x^3 - (x_1+x_2+x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3) x + x_1x_2x_3. $$ Here $x$ is a formal indeterminate - your group doesn't act on it. So all that matters is that the symmetric functions are invariant under permutations of their arguments.

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Thank you! In my textbook (when the group is $S_n$), they say that if you act on one of those symmetric functions, it just permutes the subscripts. Is that true for any group that happens to permute the arguments of the symmetric functions, or just $S_n$? And why is it that with $S_n$ you can "distribute through" the functions and make the permutation act on the arguments instead of the entire expression? Thanks (: – badatmath May 22 '11 at 21:01
@badmath It's up to you to define how $S_n$ acts on $F[x_1,\ldots,x_n]$. In order for it to be called an "action" you need that applying $\sigma$ and then $\tau$ is the same as applying $\sigma\tau$ (or $\tau\sigma$, depending on your convention). You'll probably also want the actions to be homomorphic. The rest is up to you, and up to your application. After all, these things are not arbitrary but serve some purpose. – Yuval Filmus May 22 '11 at 21:38
Cool, thanks. – badatmath May 22 '11 at 23:19

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