Can I take the vector space generated by {$1 , x, x^2$} over the field $F_3$?

Let $$V$$ be the vector space over the field $$F_3$$ (field with $$3$$ elements) of dimension $$3$$. Define an action of the symmetric group $$S_3$$ on $$V$$ by $$\sigma.e_i$$= $$e_{\sigma (i)}$$ where {$$e_1$$, $$e_2$$, $$e_3$$} is a basis for $$V$$.

Can anyone please make me understand what this problem is saying?

I do not understand what is a vector space over the field $$F_3$$ (field with $$3$$ elements) ? Can I take the vector space generated by {$$1 , x, x^2$$} over the field $$F_3$$?

If the action is only on he basis elements then how would we find the operations like $$\sigma (e_1 + e_2)$$?

• Look up the definition of a vector space over an arbitrary field $K$ and then just take $K=\Bbb{F}_3$. – Dietrich Burde Dec 6 at 16:31
• I did not get you.. Can you please go through my whole question?@DietrichBurde – cmi Dec 6 at 16:54
• I meant your statement "I do not understand what is a vector space over the field $\Bbb{F}_3$". This is just the definition. – Dietrich Burde Dec 6 at 17:59

If the action is only on he basis elements then how would we find the operations like $$σ(e_1+e_2)$$?

This seems to be the key to your question.

You are right: at the outset, the action is explicitly only defined on the set $$\{e_1, e_2, e_3\}$$. Given a group element $$\sigma$$, and one of these basis elements $$b$$, we know what $$\sigma\cdot b$$ is, and that is all.

But every mapping defined on the basis of a vector space gives rise to a unique linear transformation extending that mapping to the entire vector space spanned by the elements. Furthermore, if the mapping is a bijection and the vector space is finite dimensional, the extension is an isomorphism.

So what is happening is that by definition the action on an arbitrary element of the vector space is defined to be the linear extension: $$\sigma\cdot (\sum\lambda_ie_i):=\sum\lambda_i(\sigma\cdot e_i)$$.

You can check all the axioms to confirm that indeed, the originally defined group action on the basis has now been promoted to a group action of $$S_3$$ on the entire vector space spanned by the basis.

This "lifting" of the group action is what you were missing, I think.

There's another way to see how this happens. You can rephrase a group action of $$G$$ on a set $$X$$ to be a homomorphism from $$G\to Sym(X)$$.

If $$X$$ is the finite basis of an $$n$$ dimensional $$F$$ vector space, we have another group homomorphism from $$Sym(X)\to GL(n,F)$$. Composing these two homomorphisms we get a group homomorphism from $$G\to GL(n,F)$$, and that defines an action of $$G$$ on $$span(X)$$.

It says, essentially, that every element of $$G$$ acts like a linear automorphism on $$span(X)$$ (very nice elements of $$Sym(span(X))$$!)

A vector space of dimension $$3$$ with the underlying field $$F_3$$ is isomorphic to $$F_3^3$$. A standard basis for this vector space is $$\{e_1=(1,0,0)^{T},e_2=(0,1,0)^{T},e_3=(0,0,1)^{T}\}$$. As we know, A linear map $$f$$ from the vector space to itself can be represented by a matrix of this form $$\begin{bmatrix} . & . & . \\ f(e_1) &f(e_2) & f(e_3) \\ \ . & .&. \end{bmatrix}$$

Since a permutation just permute the basis vectors, The resulting linear maps are exactly the 3by3 permutation matrices. Foe example, the permutation $$(1,2) \in S_3$$ can be represented by the matrix

$$\begin{bmatrix} 0 & 1 & 0 \\ 1 &0 & 0 \\ \ 0 & 0&1 \end{bmatrix}$$. So you can really think of your group as the set of permutation matrices and a group element acting on the vectors by left multiply the matrix with the vector. For example, $$(1,2)$$ act on the vector $$(1,2,2)^{T}$$ gives the vector $$(2,1,2)^{T}$$.

The elements of $$V$$ simply consist of triples of elements from $$F_3$$. So a typical element of $$V$$ would be $$(1,0,2)$$ or $$(2,1,2)$$ etc. Addition of vectors in $$V$$ follows the rules of addition in $$F_3$$ so $$(1,0,2)+(2,1,2) = (0,1,1)$$. Scalar multiplication of vectors follows the rules of multiplication in $$F_3$$, so $$2(1,0,2) = (1,0,2) + (1,0,2) = (2,0,1)$$.

I think there is an unstated assumption that each action $$\sigma$$ will be linear on $$V$$, so

$$\sigma(e_1 + e_2) = \sigma(e_1) + \sigma(e_2) = e_{\sigma(1)} + e_{\sigma(2)}$$

• So I think this question math.stackexchange.com/questions/2592909/… is not complete.@gandalf61 – cmi Dec 6 at 16:53
• @cmi This is an answer to your question. Why do yo refer to some other question? Furthermore, the linked question seems complete to me. Why not? – Christoph Dec 6 at 17:18
• Is it illegal here to refer other question?@Christoph – cmi Dec 6 at 17:26
• Can you tell me how to define $\sigma (c_1 e_1 + c_2 e_2)$?@Christoph – cmi Dec 6 at 17:28
• @cmi as multiple responders have stated, the group action should be linear on the vector space, as it can be represented that way. Therefore it splits as you expect. A Google search shows that Wolfram mathworld confirms this with a general statement about group actions I don't confess to understand, but I believe it is mathnoob's answer here. – theREALyumdub Dec 6 at 17:49

$$\mathbb F_3[x]/(p(x))$$ will be just that vector space, if $$p(x)\in\mathbb F_3[x]$$ is irreducible and $$\operatorname{deg}p=3$$.

So for instance, take $$p$$ that doesn't have a root in $$\mathbb F_3$$. Say, $$p(x)=x^3+2x^2+1$$.

To answer your last question, use the property of group actions: $$\sigma (e_1+e_2)=\sigma (e_1)+\sigma (e_2)=e_{\sigma (1)}+e_{\sigma (2)}$$, for $$e_1,e_2\in V$$.

• Nothing is comprehensible .. Where did you get $p(x)$? where did you get this property of group action? – cmi Dec 6 at 17:11
• A group action preserves the structure of the vector space. As for $p$, I guess I could have waited on that. – Chris Custer Dec 6 at 17:17
• "Group action preserves the structure of vector space". Can you please share a link or something so that I can read the proof of this statement?@Chris Custer – cmi Dec 6 at 17:27
• You have referred me to ocean to find a fish.@Chris Custer – cmi Dec 6 at 17:34