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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!

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    $\begingroup$ 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
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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

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