# Minimal polynomial of an element of the field extension $L/ \mathbb{F}_2$.

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!

• What is $K$ in this? I mean how is it related to $L$? – Anurag A Jun 24 '19 at 19:00
• @AnuragA I am very sorry! It was a typo – 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 $$p(T)=T^2+T+1$$