Groups acting on polynomials

I am a little overwhelmed concerning activities now.

Claim I have a team $G$ that acts upon the origins of a polynomial, claim $(x - x_1)(x - x_2)(x - x_3)$. We understand that this team permutes the origins of a polynomial, that is you can reach any one of the origins by having a component of the team act upon an additional origin. Does that suggest the team solutions the polynomial?

I seem like it should, yet I'm not exactly sure just how this in fact functions. Apparently we would certainly need to write $g \cdot [(x - x_1)(x - x_2)(x - x_3)] = (x - x_1)(x - x_2)(x - x_3)$. Yet is this in fact feasible? Due to the fact that $G$ acts upon the set of the origins, not the polynomial. I'm not exactly sure if we can claim a polynomial is dealt with by particular points acting upon the origins. When they claim the polynomial is dealt with, does that suggest the coefficients are dealt with?

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

Many thanks!

5
2022-06-07 14:33:48
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Answers: 3

The coefficients of the polynomial are symmetrical features of the origins. In your certain instance 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 an official indeterminate - your team does not act upon it. So all that issues is that the symmetrical features are stable under permutations of their debates.

3
2022-06-07 15:05:58
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Let is chat a little bit concerning "group actions" too.

Offered a team $G$ and also a nonempty set $X$, a (left) team activity of $G$ on $X$ suggests a procedure $\cdot\colon G\times X\to X$ such that for all $g,h\in G$ and also $x\in X$, $$\begin{align*} h\cdot(g\cdot x) &= (hg)\cdot x;\\ 1\cdot x &= x, \end{align*}$$ A very easy instance of a team activity is offered as adheres to: Let $X$ be a nonempty set, and also allow $S_X$ be the team of all bijections from $X$ to itself (the "permutations on $X$"). After that we can specify an activity of $S_X$ on $X$ by $\sigma\cdot x = \sigma(x)$.

In a similar way, if $H\lt S_X$, after that $H$ act upon $X$ in an all-natural means, by allowing $h\cdot x = h(x)$ for every single $h\in H$.

An additional instance: intend that $G$ acts upon $X$ and also $H$ acts upon $Y$. After that we can allow $G\times H$ act upon $X\times Y$ in a noticeable means (coordinate - sensible): offered $(g,h)\in G\times H$ and also $(x,y)\in X\times Y$, allow $(g,h)\cdot(x,y) = (g\cdot x,h\cdot y)$.

Yet an additional instance: If $G$ act upon both $X$ and also $Y$, after that $G$ acts upon $X\times Y$ "coordinatewise" too: $g\cdot (x_1,x_2) = (g\cdot x_1,g\cdot x_2)$. Extra usually, if $G$ acts upon each set of a family members $\{X_i\}_{i\in I}$, after that $G$ acts upon the cartesian item $\prod\limits_{i\in I}X_i$ coordinate - sensible.

An additional instance: intend that $G$ acts upon $X$, and also $\iota\colon X\hookrightarrow Y$ is a one - to - one function from $X$ to an additional set $Y$. After that we can specify an activity of $G$ on $Y$ by "extension" as adheres to: offered any kind of $g\in G$ and also any kind of $y\in Y$, we specify $g\cdot y$ 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 instance: Let $G$ act upon $X$. We claim that a part $Y\subseteq X$ is $G$ - stable if and also just if $G\cdot Y = \{g\cdot y\mid y\in Y\}\subseteq Y$. After that the constraint of the $G$ - activity to $Y$ offers us an activity of $G$ on $Y$.

An additional instance: intend that $G$ acts upon $X$, and also $f\colon K\to G$ is a homomorphism. After that $K$ act upon $X$ "via $f$", by allowing $k\cdot x = f(k)\cdot x$.

In a feeling, every activity can be taken an activity in this manner, in the adhering to feeling:

Theorem. Allow $G$ be a team, and also $X$ be a set. After that the adhering to information are equal:

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

Proof. Think $\cdot\colon G\times X\to X$ is a team activity. For each and every $g\in G$, allow $\phi_g\colon X\to X$ be the map $\cdot(g,-)\colon X\to X$ (currying $\cdot$) specified by $\phi_g\colon x\mapsto g\cdot x$. This is a bijection, due to the fact that $\phi_{g^{-1}}$ 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, $\phi_g\circ\phi_h = \phi_{gh}$, given that $g\cdot(h\cdot x) = (gh)\cdot x$, so the map is a homomorphism. Hence, an activity of $G$ on $X$ generates a homomorphism $\phi\colon G\to S_X$ by $\phi(g) = \phi_g$.

Alternatively, intend that $\phi\colon G\to S_X$ is a team homomorphism. This generates an activity of $G$ on $X$ by $g\cdot x = \phi(g)(x)$, as above.

In addition, allow $\psi\colon G\to S_X$ be a homomorphism, and also allow $\cdot_{\psi}$ be the stimulated activity. After that $\phi_g(x) = g\cdot_{\psi}(x) = \psi(g)(x)$, so the homomorphism we obtain from the activity generated by $\psi$ is $\psi$. Furthermore, if you start with an activity $\cdot$, and also we allowed $\phi\colon G\to S_X$ be the generated homomorphism as above, after that the activity $\cdot_{\phi}$ generated by this homomorphism is $g\cdot_{\phi}x = \phi_g(x) = g\cdot x$, so we come back our initial activity. That is, the information are equal. QED

Another means of specifying activities is by constraint: intend $G$ acts upon a set $X$ ; offered $x\in X$, the orbit of $x$ is the set $$G\cdot x = \{ g\cdot x\mid g\in G\}.$$ Then $G$ act upon $G\cdot x$ in an all-natural means: we consider this as a *constraint * of the activity of $G$, from considering what it does to every one of $X$ to taking into consideration simply what it does to $G\cdot x$.

An activity of $G$ on $X$ generates an equivalence relationship on $X$ by $x\sim y$ if and also just if there exists $g\in G$ such that $g\cdot x = y$ ; equivalently, if and also just if $y$ remains in the orbit of $x$. This equivalence relationship dividings $X$ right into "$G$ - orbits."

Two vital buildings of an activity are loyalty and also transitivity :

Definition. Allow $G$ act upon $X$. The activity of $G$ is loyal if $$g\cdot x = x\text{ for all }x\in X\Longleftrightarrow g=1.$$ Equivalently, if and also just if the homomorphism $G\to S_X$ generated by the activity is one - to - one.

Definition. Allow $G$ act upon $X$. The activity of $G$ is transitive if for every single $x,y\in G$ there exists $g\in G$ such that $g\cdot x = y$. Equivalently, if and also just if $X$ contains a solitary $G$ - orbit.

To see an instance of an activity that is not transitive, allow $G=S_3$ and also allow $G$ act upon the set $\{1,2,3,4,5,6\}$ as adheres to: offered $\sigma\in S_3$, allow $\sigma\cdot i = \sigma(i)$ if $1\leq i\leq 3$, and also allow $\sigma\cdot j = 3+\sigma(j-3)$ if $4\leq j\leq 6$. As an example, if $\sigma=(1,3)$, after that $$\sigma(1)=3,\quad \sigma(2)=2,\quad \sigma(3)=1,\quad \sigma(4)=6,\quad \sigma(5)=5,\quad \sigma(6)=4.$$ The activity is not transitive due to the fact that $X$ has 2 $G$ - orbits: $\{1,2,3\}$ and also $\{4,5,6\}$.

For an extra intriguing instance, allow $X=\mathbb{Z}_2\oplus \mathbb{Z}_2$ be the Klein $4$ - team, and also allow $G=\mathrm{Aut}(C_2\times C_2)$. After that $G$ act upon $X$ in the noticeable means (use the homomorphism). If $x=(0,0)$, after that for every single $\phi\in G$ we have $\phi(x)=x$ (given that team automorphisms have to send the identification to the identification). On the various other hand, we have an automorphism that exchanges $(1,0)$ and also $(0,1)$ and also solutions $(1,1)$ ; an automorphism that exchanges $(1,0)$ and also $(1,1)$ and also solutions $(0,1)$ ; and also an automorphism that exchanges $(0,1)$ and also $(1,1)$ and also solutions $(1,0)$. To make sure that for every single $x\in X$, $x\neq (0,0)$, and also every $y\in X$, $y\neq (0,0)$, there exists $g\in G$ such that $g\cdot x = y$. That is, $X$ is separated right into 2 $G$ - orbits, $\{(0,0)\}$ and also $\{(1,0), (0,1), (1,1)\}$.

So you see why, in your inquiry, claiming the team "permutes the roots" is not the like claiming that for any kind of 2 origins there is a component of the team that maps the first origin to the 2nd. The first declaration is claiming that there is an activity of $G$ on the set of origins ; the 2nd declaration is claiming that the activity is transitive , and also not every activity is transitive.


Currently, intend that $G$ acts upon an area $F$ ; the activity might have definitely nothing to do with the area framework of $F$ (it might simply be an activity on the underlying set of $F$), yet these sort of activities are not really intriguing when you consider $F$ as an area. Rather, if you intend to consider $F$ as an area, after that you intend to consider the activity of $G$ as being an activity "by automorphisms" ; that is, we recognize $G$ with a subgroup of the team $\mathrm{Aut}(F)$ of all area automorphisms of $F$, which is itself a subgroup of the team $S_F$ of all permutations on the underlying set of $F$.

Currently intend that $G$ acts upon the area $F$. After that the activity of $G$ on $F$ normally generates an activity of $G$ on $F[x]$, the polynomials with coefficients on $F$: bear in mind that we can consider polynomials with coefficients in $F$ as the virtually - null series of components of $F$, that is, the boundless tuples $(a_0,a_1,a_2,\ldots,a_n,\ldots)$ with $a_i\in F$ and also $a_i=0$ for mostly all $i$ ; the activity of $G$ on $F[x]$ is specified coordinatewise, as in the past.

Yet if the activity of $G$ on $F$ is by area automorphisms, after that we get extra: the activity of $G$ commutes with the analysis homomorphism! Bear in mind that for ever before $a\in F$ we have an "evaluate at $a$" homomorphism $\varepsilon_a\colon F[x]\to F$ by $\varepsilon_a(p(x)) = p(a)$. If the activity of $G$ is by area automorphisms, after that for every single $$p(x) = \alpha_nx^n + \cdots +\alpha_0\in F[x]$$ and also for every single $a\in F$, and also we represent by $\phi_g$ the automorphism representing the activity 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 $g\cdot p$ favors $p(x)$ under the activity of $g$, particularly, $$g\cdot p = (g\cdot \alpha_n)x^n + \cdots + (g\cdot \alpha_0).$$

Because any kind of automorphism of an area have to send $0$ to $0$, we have:

Theorem. Allow $F$ be an area, and also allow $G$ act upon $F$ by automorphisms. After that for every single $p(x)\in F[x]$, every $a\in F$, and also every $g\in G$, $a$ is an origin of $p(x)$ if and also just if $g\cdot a$ is an origin of $g\cdot p(x)$.

Evidence. $a$ is an origin of $p(x)$ if and also just if $p(x)$ hinges on the bit of the analysis map $\varepsilon_a$, if and also just if $\varepsilon_a(p(x))=0$, if and also just if $g\cdot\varepsilon_a(p(x)) = g\cdot 0 = 0$, if and also just if $\varepsilon_{g\cdot a}(g\cdot p(x)) = 0$, which holds if and also just if $g\cdot a$ is an origin of $g\cdot p(x)$. QED

In certain, if $g\cdot p(x) = p(x)$ (that is, if $g$ solutions the coefficients ; keep in mind that despite having a loyal activity you can have $g\cdot a = a$ for some $g\in G$ and also some $a\in X$), after that $a$ is an origin of $p(x)$ if and also just if $g\cdot a$ is an origin of $p(x)$. That is, the origins of $p(x)$ are a $G$ - stable part of $F$, and also consequently the activity of $G$ on $F$ limits to an activity of $G$ on the origins of $p(x)$.

Alternatively, that the activity of $G$ on $F$ is by automorphisms, that $p(x)$ is a polynomial with coefficients on $F$ that divides over $F$, which the set of origins of $p(x)$ is a $G$ - stable part of $F$. Given that the coefficients of $p(x)$ are symmetrical polynomials on the origins of $p(x)$, and also symmetrical polynomials are stable under any kind of permutation of the debates, after that it adheres to that $G$ solutions every coefficient of $p(x)$.

Nonetheless, also in this scenario the activity of $G$ on the set of origins require not be transitive. As an example, take into consideration the instance of the intricate numbers, $F=\mathbb{C}$, and also allow $G$ be the team containing the identification and also facility conjugation (so $G$ is isomorphic to the cyclic team of order $2$). If $p(x)$ is a polynomial with actual coefficients, after that the set of origins of $p(x)$ is $G$ - invaraiant (given that an intricate number is an origin of an actual polynomial if and also just if its intricate conjugate is an origin too), to make sure that the activity of $G$ generates an activity on the origins of $p(x)$. If $p(x) = (x^2+1)(x^2+4)(x^2-1)$, as an example, after that $G$ acts upon the set of origins of $p(x)$, $X=\{1,-1,i,-i, 2i, -2i\}$: the identification solutions every component, and also intricate conjugation solutions $1$ and also $-1$, exchanges $i$ and also $-i$, and also exchanges $2i$ and also $-2i$ ; the $G$ - orbits of $X$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, and also $\{2i,-2i\}$. The activity is not transitive.


Nonetheless, the scenario that generally emerges in Galois Theory is a lot more limited. There, we are generally taking into consideration not one area, yet a set of areas in an area expansion, $F\subseteq K$. The team whose activity we are taking into consideration is a team of automorphisms of $K$, yet none old automorphisms, yet instead automorphisms that deal with $F$ pointwise ; that is, for every single $g\in G$, $g$ acts upon $K$ by automorphisms as if $g\cdot f = f$ for every single $f\in F$. Because scenario, any kind of polynomial $p(x)$ with coefficients in $F$ will certainly be dealt with by the (generated) activity of $G$. Specifically, the set of origins of $p(x)$ that remain in $K$ is $G$ - stable, therefore $G$ act upon the set of origins of $p(x)$ that remain in $K$ ; the activity is always "by permutations", given that every activity functions using permutations.

Alternatively, if you have a limited $G$ - stable part $S$ of $K$, after that the polynomial $$p(x) = \prod_{s\in S}(x-s)$$ is dealt with by the activity of $G$, to make sure that the coefficients hinge on the "fixed area of $G$", which is the set of all $k\in K$ such that $g\cdot k = k$ for all $g\in G$. When the expansion is a Galois expansion, and also $G$ is the collection of all automorphisms of $K$ that deal with $F$ pointwise, after that the dealt with area of $G$ is specifically $F$ (actually, that is one feasible definition of "the expansion is a Galois extension"), to make sure that the polynomial $p(x)$ over has to have coefficients in $F$.


Constantly, if the activity of $G$ on $K$ is "by automorphisms", after that you always have $g\cdot(a+b) = g\cdot a + g\cdot b$ and also $g\cdot(ab) = (g\cdot a)(g\cdot b)$. Nonetheless, if the activity is not "by automorphisms", after that you usually can not claim that.

8
2022-06-07 14:53:36
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Let is back up a little ; you appear to be perplexed around extra basic points now.

Deal with an area $F$ . If $f(x) \in F[x]$ is a monic polynomial, we can speak about the splitting area of $f$ , which is an area expansion $K/F$ gotten by adjacent every one of the origins $r_1, ... r_n$ of $f$ . For numerous factors it is all-natural to take into consideration a particular team $G = \text{Aut}(K/F)$ , the team of all invertible area homomorphisms $g : K \to K$ which deal with $F$ . This suggests that

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

This need to with any luck address your last inquiry. The buildings over indicate that $g(f(r)) = f(g(r))$ for any kind of $r \in K$ , where it adheres to that $g(f(r)) = 0$ if and also just if $f(g(r)) = 0$ . Given that $g$ is an automorphism, this indicates that $r$ is an origin of $f$ if and also just if $g(r)$ is an origin of $f$ , therefore that $G$ permutes the origins of $f$ .

Currently, the components of $G$ act by automorphisms $K \to K$ , yet we can expand this activity to an activity by automorphisms $K[x] \to K[x]$ by proclaiming that $x$ is sent out to $x$ by every component of $G$ and also expanding by linearity and also multiplicativity. So one means to understand the declaration that components of $G$ deal with the polynomial $f(x)$ is that they deal with the component $f(x) \in F[x] \subset K[x]$ . There are (at the very least) 2 means to see this.

First, components of $G$ solution components of $F$ necessarily, and also given that $f(x)$ is an amount of components of $F$ and also powers of $x$ (which $G$ additionally solutions necessarily), the outcome is clear.

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

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

so it adheres to by the homomorphism building that

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

and also given that $g$ permutes the origins of $F$ it adheres to that the expression on the RHS is simply a reordering of the variables of $f(x)$ .

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2022-06-07 14:52:53
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