Let $L$ be a field extension of $\mathbb{F}_2$ with $[L : \mathbb{F}_2]=2$ and let $\beta\in L-\mathbb{F}_2$.

Now I have to determine the minimal polynomial of $\beta$.

I know that for $\alpha \in L$ with $\alpha^2\in\mathbb{F}_2 $ it follows that $\alpha\in \mathbb{F}_2$, but I don't think that could help me here.

Basically I don't even know where to start things off, so I am grateful for any kind of help or advice!

  • 2
    $\begingroup$ What is $K$ in this? I mean how is it related to $L$? $\endgroup$
    – Anurag A
    Jun 24 '19 at 19:00
  • $\begingroup$ @AnuragA I am very sorry! It was a typo $\endgroup$
    – TwoStones
    Jun 24 '19 at 19:50

The minimal polynomial of $\beta$ is of degree 2 and irreducible. There is only one such polynomial in $\mathbb{F}_2$ : $T^2+T+1$

To show this, note that $p(T)=T^2+T+1$ has no root in $\mathbb{F}_2$ so it must be irreducible (if $p(T)=g(T)\cdot f(T)$, then $g(T)$ would have degree 1 and a root, therefore $p(T)$ should also have a root)

To show that this is the only one possible choice of $p(T)$, there are $4$ polynomial of degree $2$ in $\mathbb{F}[x]_2$ and $2$ of degree $1$. So the number of reducible polynomials of degree $2$ is those of the form $(T-a)\cdot(T-b)$ (of which there is 1) and of the form $(T-a)^2$(of which there are 2)

To sum up, there are 3 reducible polynomial of degree 2 in $\mathbb{F}[T]_2$, and a total of 4 polynomials of degree 2. So there is EXACTLY one polynomial of degree 2 in $\mathbb{F}[T]_2$ which is irreducuble, namely $p(T)=T^2+T+1$

PS: If you find this answer messy, please feel free to edit it for clarification. Thank you very much


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.