Spliting Field over $\mathbb{F}_3$ How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?
 A: I'll try to help you with one, you do the other one and try to build an isomorphism between the corresponding fields (knowing it exists must help...).
Since $\,f(x)=x^3-x+1\in\Bbb F_3[x]\,$ is irreducible, we get that $\,\Bbb F_{3^3}:=\Bbb F_3[x]/\langle f(x)\rangle\,$ is a field with $\,3^{\deg f}=27\,$ elements. Let $\,w\in \overline{\Bbb F_3}\,$ be a root of $\,f\,$ , so :
$$(**)\;\;\;\;w^3=w-1=w+2\pmod 3\implies$$
$$\implies \Bbb F_{27}=\{0,1,2,w, 2w,...\}=\{q(w)\;;\;q(x)\in\Bbb F_3[x]\;,\;\deg q\le 2\} $$
with the sum and product operations done modulo three and with the relation (**) taken into account.
Thus, for example, we have that
$$(2w+1)(w^2+w+1)=2w^3+\color{red}{2w^2}+\color{blue}{2w}+\color{red}{w^2}+\color{blue}w+1=2(w-1)+1=2w-1=2w+2$$
or also
$$w(aw^2+bw+c)=1\;,\;\;a,b,c\in\Bbb F_3\;\implies aw-a+bw^2+cw=1\implies$$
$$\implies a = 2\,,\,c=1\,,\,b=0\implies\;\text{in the field}\;\;\Bbb F_{27}\;,\;\;w^{-1}=2w^2+1$$
Well, try to do the same with the other polynomial and then try to come up with some isomorphism between the respective fields.
Further hint: each one of the above fields has an element (a primitive element) s.t. every non-zero element in the field is the power of that elements, which is just a lengthy way to say that the multiplicative groups of the non-zero elements of each field is a cyclic group of order $\;3^3-1=26\;$ .
If you understand how the product is done from the examples above then that'll help you to find those primitive elements in each field...
A: let $a$ is a root of $f(x)$ in the splitting field, 
then $f(a)=0=a^3-a+1=(a+1)^3-(a+1)+1=(a-1)^3-(a-1)+1$
so, $(x-a)(x-(a+1))(x-(a-1))=f(x)$
splitting field of $f(x)$ is $F_3[x]/<f(x)>=F(a)$ where $a$ is a root of $f$
similarly you can do for $g(x)$
A: First show that $K=\mathbb{F}_3[x]/\langle f(x)\rangle$ is a splitting field of $f(x)$. Answers by others will help you there.
Hint (for the isomorphism): You can easily verify that 
$$
f(-x)=-g(x).
$$
It follows that the coset of $-x$ in the field $\mathbb{F}_3[x]/\langle f(x)\rangle$ is a zero of $g(x)$.
The rest is routine, I hope.
A: You have check that $f,g$ are irreducible over $\mathbb{F}_{3}$.
You know that in this case $\mathbb{F}_{3}[x]/\langle f\rangle$ is a
field extension of degree $\deg(f)=3$ over $\mathbb{F}_{3}$. The
same applies for $g$.
Moreover, you know that if $x$ is a root of $f$ in the above extension
then $\{1,x,x^{\deg(f)-1}\}$ is a basis for this extension.
The same applies for $g$: if $x'$ is a root of $f$ in the above
extension then $\{1,x',(x')^{\deg(f)-1}\}$is a basis for $\mathbb{F}_{3}[x]/\langle g\rangle$.
Hint: Try to find an isomorphism that maps $1\mapsto 1$ and $x \mapsto x'$
A: The polynomials $x^3-x-1$ and $x^3-x+1$ are irreducible over $\mathbb F_3$ as pointed out
in the comment by @JyrkiLahtonen. 
Now suppose that $\alpha$ is a root of $x^3-x-1$,
that is, $\alpha^3 = \alpha+1$.
Then the set of $27$ elements of the form $a_0 + a_1\alpha + a_2\alpha^2$, 
$a_i \in \mathbb F_3$ is a three-dimensional vector space over $\mathbb F_3$.
Also, the set is closed under multiplication since the product of two elements,
being a polynomial of degree at most $4$ can be expressed as a polynomial
of degree at most $2$ by setting $\alpha^3 = \alpha+1$ and $\alpha^4 = \alpha^2+\alpha$
and collecting coefficients.  The proof of the
existence of unique multiplicative inverses takes
a little longer, but can be done, and the end result is that these $27$ elements
constitute a field of $27$ elements. 
Now, $\alpha^3 -\alpha = 1 \Rightarrow -(\alpha^3 -\alpha)=-1
\Rightarrow (-\alpha)^3 -(-\alpha) = -1$, that is, $-\alpha$ is a
a root of the other polynomial $x^x-x+1$. Writing $\beta$ for $-\alpha$,
we have, as before, that the elements $b_0 + b_1\beta+b_2\beta^2$, $b_i \in \mathbb F_3$ 
constitute a field of $27$ elements. But these two different-seeming
fields are in fact the same field since for all choices of $3$
elements $a_0,a_1,a_2 \in\mathbb F_3$, the element
$a_0 + a_1\alpha + a_2\alpha^2$ in the first field is the same as
the element $a_0 -a_1 \beta + a_2\beta^2$ in the second field:
$$a_0 + a_1\alpha + a_2\alpha^2 = a_0 -a_1(-\alpha) + a_2(-\alpha)^2 
= a_0 -a_1 \beta + a_2\beta^2$$
A: first observe that $f(x) = x^3 - x + 1$ is irreducible over $F_3$ since the derivative of $f(x)$ is equal to 2, which is nonzero in $F_3$. Therefore, the extension of $F_3$ given by $L = F_3[x]/(f(x))$ must have order equal to $3^{\deg(f)} = 27$ and hence is isomorphic to $F_{3^3}$. So for your isomorphism question, remark that any two finite fields of the same size are isomorphic.
To actually "see" the splitting field, the theorem of the primitive element says that $$L = F_3[y],$$ for some $y\in F_3^{sep}$. One way to try and find such a $y$ is to ask, first of all, if $x\in L$ with $x^3 - x + 1 = 0$ is a generator. To do so, see if you can write down all powers of $x$ as $$x^n = \alpha_n.x+\beta_n,$$ for some $\alpha_n, \beta_n\in F_3$ and $\alpha_n\neq 0$ for $n<\deg(f)$. We would need to check for $1< n < \deg(f)= 27$...It's instructive to work out what $x^2$ is...(this check for such $\alpha,\beta$ is from linear algebra, i.e. we are trying to see if $\langle x\rangle_{F_3}$ spans $L$ as an $F_3$-vector space; we know that if $y^n = \gamma$ for some $0<n<\deg(f)$ and $\gamma\in F_3$, then $y\in L$ is not a generator of $L/F_3$).
