# Linear maps of polynomial roots over finite fields

Take an arbitrary degree $n$ monic irreducible polynomial $f(x)$ over $\mathbb{F}_p$ with roots $$\alpha, \alpha^p,\alpha^{p^2},\dots,\alpha^{p^{n-1}}$$ in $\mathbb{F}_p[x] / \langle f(x) \rangle$. Now, consider another degree $n$ monic irreducible $g(x)$ over $\mathbb{F}_p$. Is there a linear map which sends the roots of $f$ to the roots of $g$ in $\mathbb{F}_{p^n}\cong \mathbb{F}_p[x] / \langle f(x) \rangle$ (viewed as a vector space)?

Edit: More specifically, if $\alpha,\dots,\alpha^{p^{n-1}}$ are not a normal basis for $\mathbb{F}_{p^n}$, what (if anything) can be said?

## 1 Answer

I will use this example: Constructing an explicit isomorphism between finite extensions of finite fields

Let $p=2$, $f(x) = x^3+x+1$, $g(x)=x^3+x^2+1$ so that $L\simeq \mathbb{F}_2[x]/(x^3+x+1)$ and $L'\simeq \mathbb{F}_2[x]/(x^3+x^2+1)$ are isomorphic as fields. Let $\alpha$ be a root of $f(x)$ in $L$, and $\beta$ be a root of $g(x)$ in $L'$.

Since $f$ does not have $x^2$ term, we see that $\alpha^4+\alpha^2+\alpha =0$. For the roots of $g$, we have $\beta^4+\beta^2+\beta = 1 \in \mathbb{F}_2$.

We assume that $T:L\rightarrow L'$ is a $\mathbb{F}_2$-linear map that maps roots of $f$ to roots of $g$, i. e. $$T(\{\alpha,\alpha^2,\alpha^4\})\subseteq \{\beta,\beta^2,\beta^4\}.$$

Now, we consider $\alpha^4=\alpha^2+\alpha$. So, $T(\alpha^4)=T(\alpha^2)+T(\alpha)$.

If $T(\alpha)=T(\alpha^2)$, then we have $T(\alpha^4)=0$. This is not possible.

Then we must have $T(\alpha)\neq T(\alpha^2)$. We check the three possibilities:

Case 1: $T(\{\alpha,\alpha^2\})=\{\beta,\beta^2\}$.

Since $T(\alpha^4) = \beta^2+\beta$ and $\beta^4=\beta^2+\beta+1$, the element $\beta^2+\beta$ is not a root of $g$, this case is impossible.

Case 2: $T(\{\alpha,\alpha^2\})=\{\beta,\beta^4\}$.

Since $T(\alpha^4)= \beta^4+\beta = \beta^2+1$ is not a root of $g$, it is not possible.

Case 3: $T(\{\alpha,\alpha^2\})=\{\beta^2,\beta^4\}$.

Since $T(\alpha^4) = \beta^4+\beta^2 = \beta+1$ is not a root of $g$, it is not possible.

• @user1952009 Even when dimensions of spans of powers are the same, their linear dependence structure might differ. – i707107 Jun 30 '17 at 1:55
• @user1952009 Hmm...what if we know there exists at least one nontrivial linear relation of the form $\alpha+\alpha^{p^k}+\alpha^{p^m}=0$ and $\beta+\beta^{p^\ell}+\beta^{p^n}=0$ for $f$ and $g$ respectively? – user80108 Jun 30 '17 at 2:03
• Yes forget it. But it is true that if it was always true then every $\{\alpha^{p^i}\}$ would be (normal) basis. And when the coef of $x^{deg(f)-1}$ is $0$ then the roots of $f$ can't be a basis. – reuns Jun 30 '17 at 3:00