Question regarding algebraic closure of$ \mathbb{F_2}$

Question

Let $\mathbb{F_2} $ be the finite field of order $2$. Then which of the following statements are true?

$1.$ $\mathbb{F_2} [x]$ has only finitely many irreducible elements.

$2.$ $\mathbb{F_2} [x]$ has exactly one irreducible polynomial of degree $2.$

$3.$$\mathbb{F_2} [x]/<x^2+1>$ is a finite dimensional vector space over .

$4.$ Any irreducible polynomial in $\mathbb{F_2} [x]$ of degree $5$ has distinct roots in any algebraic closure of $\mathbb{F_2}$

My attempt:

• option 1) will True take $f(x) = x^2+x+1$

• option $2)$ will false because number irreducible polynomail of degree $2 = \frac{p^2-p}{p-1} = 2$

• option $3)$ will true dimension will be $4$

• option $4$ i don't have any hints to tackle this option

any hints/solution will be appreciated

thanks u

Answer 1

$\mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.

But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.

$\mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $\mathbb{F}_2$.

For 4. yes, see this question and answer

Answer 2

Since every finite field is a perfect field and every finite field extension over a perfect field is also perfect so $F_2[X]$ is perfect as it is a finite field extension over $F_2$ which is also finite. Now we know that every irreducible polynomial over a perfect field has distinct roots, therefore option 4 is correct.