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i have a prime number $p$ and an irreducible polynomial $R(x)$ of degree $n$ , $\alpha$ a root of $R$ then it's known that the field $\mathbb{F}_{p^n}$ is isomorphic to the field $\mathbb{F}_{p}[\alpha]$.

let $M$ be the companion matrix of $R$ in particular .

i wish to proof that $\mathbb{F}_{p}[M]$ is a field isomorphic to $\mathbb{F}_{p^n}$.

i've noticed that $R(M)=O$ (Carley-Hamilton theorem) but i don't know whether or not $M$ lies in the algebraic closure of $\mathbb{F}_{p}$ if it dose that will give the result by construction of finite fields so i got stuck there...

Next i tried to give an explicit isomorphism (ring isomorphism) \begin{array}{ccccc} \psi : & \mathbb{F}_{p^{n}} & \rightarrow & \mathbb{F}_{p}[M] & \\ & 0 & \rightarrow & O & \\ & \alpha ^{i} & \rightarrow & M^{i} & 0<i<q^{n}-1% \end{array} with the assumption that $\alpha$ is a primitive element of $\mathbb{F}_{p^{n}}$

it's easy to poof that $\psi(a*b)=\psi(a)*\psi(b)$ but i couldent proof that $\psi(a+b)=\psi(a)+\psi(b)$

UPDATE: can i say that $M$ is algebraic over $\mathbb{F}_{p}$ ?

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Another approach is to prove that $\mathbb{F}_{p}[M]$ is a field. Since it has $p^n$ elements, it must be isomorphic to $\mathbb{F}_{p^n}$, by uniqueness of finite fields.

A typical element of $\mathbb{F}_{p}[M]$ is given by $f(M)$ with $\deg f < \deg R$. If $f(M)\ne0$, then $\gcd(f,R)=1$, because $R$ is irreducible and $\deg f < \deg R$. Then, we can write $1=a(x)f(x)+b(x)R(x)$ in $\mathbb{F}_{p}[x]$. Therefore, $1=a(M)f(M)$ and $f(M)$ is invertible.

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  • $\begingroup$ That's the clearest proof! $\endgroup$ – Amin235 Apr 13 '18 at 21:34
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First, you should define $\psi\left(\sum_ia_i\alpha^i\right)=\sum_ia_iM^i$ so that $\psi(a+b)=\psi(a)+\psi(b)$ holds automatically. However, you will also need to prove that $\psi$ is well defined, e.g., if $\sum_ia_i\alpha^i=0$, then you should have $\sum_ia_iM^i=0$ as well.

You can get around this issue by using the 1st isomorphism theorem. Let $\psi:\mathbb{F}_p[x]\to\mathbb{F}_p(M)$ be the unique ring homomorphism given by $\psi(x)=M$ (i.e. $\psi\left(\sum_ia_ix^i\right)=\sum_ia_iM^i$. This map is clearly surjective. Since $R(M)=0$, we know that $\langle R(x)\rangle\subseteq\ker\psi$. But $R(x)$ is irreducible, so $\langle R(x)\rangle$ is maximal and equality holds.

Finally, by the 1st isomorphism theorem $\mathbb{F}_p/\langle R(x)\rangle\cong\mathbb{F}_p[M]$. Since $\mathbb{F}_{p^n}\cong\mathbb{F}_p/\langle R(x)\rangle$, we are done.

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  • $\begingroup$ $\alpha$ is a primitive element of $\mathbb{F}_{p^{n}}$ so $\sum_ia_i\alpha^i=\alpha^j $ for some $j$ $\endgroup$ – user496502 Apr 13 '18 at 19:44
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    $\begingroup$ Yet you can't prove that $\psi$ is additive... $\endgroup$ – David Hill Apr 13 '18 at 20:08
  • $\begingroup$ using this definition we can see that $\psi$ is additive , yet i couldn't proof that it is multiplicative, for multiplicativeness of $\psi$ i need to show that $\psi(\alpha^i)=M^i$ for $1<i<q^m-1$. but the definition give us only $i<i<m-1$ i managed to proof it for i=m since $M$ and $\alpha$ are root of $R(x)$ but i couldn't go any further $\endgroup$ – user496502 Apr 21 '18 at 23:13

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