I have the following polynomial over $\mathbb F_3$: $$ f(T) = T^3+2T+1 $$ I would like to find out a field extension in order to add the roots of this polynomial.
Edit: by defining $\alpha\not\in \mathbb F_3$ such that $\alpha^3 + 2 \alpha + 1$ = 0, if I have another polynomial $$ g(T) = T^3 + T^2+ T + 2 $$ I would like to find out if $$\mathbb F_3(\alpha)\cong \mathbb F_3(\beta)$$ being $\beta\not\in \mathbb F_3$ such that $\beta^3+\beta^2+\beta + 2$.
Edit II: I think they are not isomorphic, since in that case, if $\phi : \mathbb F_3(\alpha)\to \mathbb F_3(\beta)$ is an isomorphism, $\phi(\alpha) = \beta$, and $$ 0 = \phi(0) = \phi(\alpha^3+2\alpha +1) = \beta^3 + 2\beta + 1 \quad\&\quad \beta^3 + \beta^2 + \beta + 2 = 0 $$ so $$ \beta^2 -\beta +1 =0= \beta^3+2\beta + 1 \implies \beta^3 = \beta^2$$ which is absurd.
The polynomial $f(T)=T^3+2T+1$ is irreducible over $\mathbb{F}_3$ because it has no roots (and has degree $3$). We can always build an extension field where $f$ has a root, namely $$ K=\mathbb{F}_3[T]/(f(T))=\mathbb{F}_3[\alpha] $$ where $\alpha^3+2\alpha+2=0$. Such a field $\mathbb{F}_3[\alpha]$ has dimension $3$ as a vector space over $\mathbb{F}_3$, so it has $27$ elements.
Note that $\mathbb{F}_3[\alpha]$ can also be described as a splitting field of the polynomial $T^{27}-T$ over $\mathbb{F}_3$, by general theory of finite fields.
The polynomial $g(T)=T^3+T^2+T+2$ is irreducible as well. Hence the field $\mathbb{F}_3[\beta]$, where $\beta^3+\beta^2+\beta+2=0$ has $27$ elements as well. Hence it is a splitting field of $T^{27}-T$ and therefore it is isomorphic to $\mathbb{F}_3[\alpha]$.
Of course the isomorphism doesn't map $\alpha$ to $\beta$. In $\mathbb{F}_3[\alpha]$ there is certainly a root for $g$. The elements of $\mathbb{F}_3[\alpha]$ are of the form $x=a+b\alpha+c\alpha^2$. Expand $$ x^3+x^2+x+2 $$ and determine $a,b,c\in\mathbb{F}_3$ such that the expression is $0$. Then you get an isomorphism by sending $x$ to $\beta$, $x^2$ to $\beta^2$ and extending by linearity.