A question based on ring $\mathbb{F}_{2}$ [X]

Question

This question was asked in my abstract algebra quiz and I was unable to solve it.

Let $R = \mathbb{F}_{2}$ [X] . Then choose the correct option(s):
$1$. $R$ has uncountably many maximal ideals.
$2$. Every maximal ideal of $R$ has infinitely many elements.
$3$. For all maximal ideals, $m$ of $R$, $~R/m$ is a finite field.
$4$. For every integer $n$, every ideal of $R$ has only finitely many elements of degree $\leqslant n$.

Although I am not able to prove/ disprove any of the options kindly just tell me the reasoning behind only 1st options. Rest I would like to work by myself. If unable to do ask, then I will ask here later.

Answer 1

In order.

1. Since $\Bbb F_2$ is a field, $\Bbb F_2[x]$ is a PID, so there are at most as many ideals of $R$ as elements. Since $R$ itself is countable, the set of maximal ideals is countable.

2. If $I$ is maximal, then $I\neq(0)$, so $I$ contains $f(x)\in\Bbb F_2[x]$. Now $x^nf(x)$ gives a countable family of elements in $I$.

3. This is true: if $R/m$ is not a field, then it has a nonzero proper ideal $I$. This corresponds to a nonzero proper ideal $J$ of $R$ containing $m$, contradicting the fact that $m$ is maximal. Edit: a better method is to note that $R/m$ is a field if and only if $m$ is generated by an irreducible polynomial, in which case $R/m$ has order $2^d$, where $d$ is the degree of the generator of $m$.

4. This is true: a polynomial of degree $n$ consists of $n+1$ choices of coefficient. Since there are only two possible coefficients, and a degree-$n$ polynomial must have leading term $x^n$, there are $2^n$ polynomials of degree $n$.

Answer 2

$\Bbb F_2[x]$ is PID, so every ideal can be generated by one element. Thus since $\Bbb F_2[x]$ is a countable set, so too is the set of the ideals of $\Bbb F_2[x]$ and since the set of maximal ideals is a subset of the one before, it can’t be uncountable.

Answer 3

The ring of polynomials over a field is Euclidean, and thus a PID. So every ideal (including all maximal ideals) are of the form $(p)$, where $p$ is a polynomial in $\mathbb F_2[X]$. In particular, there are at most as many maximal ideals as there are polynomials in the first place. Can you show that their number is countable?