Here is some further context for the notion of an algebraically closed field, connecting it with irreducible polynomials and finite-degree field extensions.

The first of these is probably familiar at least in the case of
polynomials over **Q**.
A polynomial $P \in F[T]$ is said to be *irreducible*
(over the

As with primes, irreducible polynomials are characterized
by the property that for $M,N \in F[T]$ the product $MN$ is a multiple
__iff__ either $M$ or $N$ is a multiple
**Z** is a principal-ideal domain.
We take the latter route here. If $I \subseteq F[T]$ is an ideal
other than $\{0\}$ then it contains a nonzero polynomial $A$ of minimal degree.
Then $I$ contains all multiples of $A$, and we claim that this is
all

What has all this to do with algebraic closure?
Well, a polynomial of degree $1$ is automatically irreducible,
because in a factorization one or the other factor must have degree zero.

Claim: *$F$ is algebraically closed if and only if
every irreducible polynomial in $F[T]$ has degree $1.$*

Proof: $\Rightarrow$ If $F$ is algebraically closed then any $P \in F[T]$
has a

$\Leftarrow$ By induction over $\deg A$, every nonconstant polynomial
$A \in F[T]$ has a factorization $A = \prod_{i=1}^m P_i$ where
$m>0$ and each $P_i$ is irreducible. Writing each $P_i$ as $a_i (t-t_i)$
gives $A = a \prod_{i=1}^m (t-t_i)$ with $a = \prod_{i=1}^m a_i$.
So indeed $A$ factors completely.

A *field extension* of a field $F$ is some other $F'$
that contains $F$ (with the same field operations, so in particular
$0_F = 0_{F'}$ and $1_F = 1_{F'}).$ We observed already that
$F'$ is then a vector space *finite* if $F'$ is finite-dimensional as a
vector space *degree* of the extension and also denoted by $[F':F]$.
The familiar example is $F=\bf R$, $F'=\bf C$, with $[{\bf C} : {\bf R}] = 2$.
This is connected with irreducible polynomials as follows (which also
explains the “degree” terminology): if $P \in F[T]$ is
irreducible then $F' := F[T] \big/ P F[T]$ is a field extension of degree
equal to $\deg P$. (This generalizes the familiar construction of $\bf C$
as ${\bf R}[i]$ modulo the relation $i^2 + 1 = 0$; here we in effect
set $P(T) = 0$.) We already know that $\dim_F^{\phantom0} F' = \deg P$,
and the main thing to check is that when $P$ is irreducible $F'$ is
not just a commutative __iff__ it is injective. [This replaces
the familiar argument modulo a prime $p$, where we use the fact that
a map from the *finite* set ${\bf Z}/p{\bf Z}$ is surjective
__iff__ it is injective.] But injectivity of $X \mapsto MX$
comes down to what we showed above: if $N \bmod P$ is in the kernel
then $MN \equiv 0 \bmod P$, and since $M$ is not a multiple

In particular, if $F$ is *not* algebraically closed then
it has a finite extension of degree $\gt 1$. Remarkably the converse
is true: if $F$ *is* algebraically closed then it has
no finite extensions other than $F$ itself. (This is thus
our fourth equivalent characterization of algebraically closed fields.)
Indeed if $F'/F$ is a finite extension, say of **QED**.