# why we only need to consider $\lambda(\alpha)$ when we want to write the exact isomorphism?

I am referring to this question Constructing Isomorphism between finite field

Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.

I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.

I know I have to determine where $\alpha$ is mapped to under the isomorphism map but I can't figure it out.

======================================================= I know that two fields $F(\alpha)$ and $F(\beta)$ are isomorphic. My question is, why we only need to consider mapping of $\alpha$ when we write down the exact isomorphism?

• Are you aware of the fact that $\mathbb F_3\left(\alpha\right) \cong \mathbb F_3\left[X\right] / \left(X^3-X+1\right)$, and the equivalence class of $X$ on the right corresponds to the $\alpha$ on the left? Defining homomorphisms out of $\mathbb F_3\left[X\right] / \left(X^3-X+1\right)$ is a matter of specifying where $X$ goes and checking that $X^3-X+1$ goes to $0$. – darij grinberg Feb 16 '15 at 17:07
• To make it even more clear: Giving an explicit isomorphism $\mathbb F_3(\alpha) \to \mathbb F_3(\beta)$ is the same as finding an element in $\mathbb F_3(\beta)$, which satisfies $x^3-x+1=0$. You should test the elements $a\beta^2+b\beta$ (We don't a constant term, keyword: Artin-Schreier polynomial), there is a unique solution for $a,b$. – MooS Feb 16 '15 at 17:54

If $\alpha$ exists in an extension ring of $R=\Bbb Z/n\Bbb Z$ (with $n$ possibly $0$), then any homomorphism of unital rings from $R[\alpha]$ is determined by where it sends $\alpha$. This is because we know $1\mapsto1$ hence we know where any integer multiple of $1$ is mapped to, and if we know where $\alpha$ is mapped to then we know where any positive power of $\alpha$ is mapped to, and hence know where any polynomial expression in $\alpha$ is mapped to. But every element of $R[\alpha]$ is expressible as a polynomial in $\alpha$.
Edit: Scrapping the second paragraph of my original answer because I didn't read the problem closely enough; I thought $\alpha$ and $\beta$ satisfied the same polynomial, but they're different. I almost didn't catch this issue, in which case I wouldn't have been able to fix it and readers wouldn't have been alerted.
In order to specify an isomorphism $\Bbb F_3(\alpha)\to\Bbb F_3(\beta)$, we've seen that it's enough to say where $\alpha$ goes, but wherever it goes to it must still satisfy $X^3-X+1$, so we need an element of $\Bbb F_3(\beta)$ which satisfies this polynomial. As it happens, a clever trick is possible for this task: what happens if you rewrite the given condition $\beta^3-\beta^2+1$ as a polynomial in $\beta^{-1}$ (divide it by $\beta^3$, in other words)?