Kronecker's theorem hands you one root at a time. Keep going and a polynomial eventually breaks into linear factors; the smallest field where that happens is its splitting field. Push it to the limit, with every polynomial at once, and you get an algebraic closure: a field where nothing is left to factor. Along the way we'll need a test for repeated roots, which is where a derivative shows up in a world with no limits.
If $f \in F[x]$ is a non-constant polynomial, the splitting field of $f$ over $F$ is the smallest field extension of $F$ containing all roots of $f$; that is, the smallest extension over which $f(x)$ factors into linear factors.
"Smallest" means: it's generated over $F$ by the roots. If $f$ factors as $c(x - \alpha_1)\cdots(x - \alpha_n)$ over some big field, the splitting field is $F(\alpha_1, \dots, \alpha_n)$.
Existence is Kronecker's theorem on repeat (Field Extensions): adjoin a root of one irreducible factor, factor out $(x - \alpha_1)$, and do it again to what's left.
Find the splitting fields over $\mathbb{Q}$ of $x^2 - 2$, $(x^2 - 2)(x^2 - 3)$, and $x^3 - 2$, with their degrees.
A splitting field always depends on the base field. $x^2 + 1$ splits over $\mathbb{R}$ in $\mathbb{C}$ (degree $2$ over $\mathbb{R}$), over $\mathbb{Q}$ in $\mathbb{Q}(i)$ (degree $2$ over $\mathbb{Q}$, and much smaller than $\mathbb{C}$), and over $\mathbb{F}_5$ it already splits: $x^2 + 1 = (x - 2)(x - 3)$.
How big can a splitting field get? Each new root costs at most the degree of what's left to factor.
If $f \in F[x]$ has degree $n$, and $K$ is a splitting field of $f$ over $F$, then $[K:F] \le n!$.
Proof. Induct on $n$; $n = 1$ is clear ($K = F$). Let $\alpha_1 \in K$ be a root of $f$. Its minimal polynomial divides $f$, so $[F(\alpha_1):F] \le n$. Over $F(\alpha_1)$, $f = (x - \alpha_1)g$ with $\deg g = n - 1$, and $K$ is a splitting field of $g$ over $F(\alpha_1)$ (it's generated by $\alpha_1$ and the roots of $g$). By induction $[K:F(\alpha_1)] \le (n-1)!$, and the Tower Law gives $[K:F] \le n \cdot (n-1)! = n!$. $\blacksquare$
The bound is reached by $x^3 - 2$ ($3! = 6$). Next: is "the" splitting field really unique? Different constructions could, in principle, produce different fields. They don't, and the proof gives something stronger that Galois theory will lean on.
Let $\tau\colon F_1 \to F_2$ be a field isomorphism, and let $f_1 \in F_1[x]$ and $f_2 = \tau(f_1) \in F_2[x]$ (apply $\tau$ to each coefficient). If $K_1$ is a splitting field for $f_1$ over $F_1$, and $K_2$ is a splitting field for $f_2$ over $F_2$, then $\tau$ extends to an isomorphism $\sigma\colon K_1 \to K_2$.
Proof sketch. Induct on $[K_1:F_1]$. If it's $1$, $f_1$ splits over $F_1$, so $f_2 = \tau(f_1)$ splits over $F_2$, $K_2 = F_2$, and $\sigma = \tau$.
Otherwise, pick an irreducible factor $p_1$ of $f_1$ of degree $\ge 2$, and a root $\alpha \in K_1$. Then $p_2 = \tau(p_1)$ is an irreducible factor of $f_2$ (an isomorphism $F_1[x] \cong F_2[x]$ preserves factorizations); let $\beta \in K_2$ be a root of it. Then
$$F_1(\alpha) \cong F_1[x]/(p_1) \cong F_2[x]/(p_2) \cong F_2(\beta),$$
an isomorphism extending $\tau$ that sends $\alpha \mapsto \beta$. Now $K_1$ is a splitting field of $f_1$ over $F_1(\alpha)$, $K_2$ is one of $f_2$ over $F_2(\beta)$, and $[K_1:F_1(\alpha)] < [K_1:F_1]$, so induction extends this isomorphism to all of $K_1$. $\blacksquare$
Taking $F_1 = F_2 = F$ and $\tau = \operatorname{id}$: any two splitting fields of $f$ over $F$ are isomorphic by an isomorphism that fixes $F$. So we can say the splitting field.
Look at the freedom in the proof: $\beta$ can be any root of $p_2$. Different choices give different isomorphisms. When $K_1 = K_2$, those different choices are automorphisms of the splitting field that shuffle the roots of $p$. Counting them is exactly what Galois Theory does.
Now go all the way: a field where every polynomial has a root.
A field $K$ is algebraically closed if every non-constant polynomial $f \in K[x]$ has a root in $K$.
Peel off one root at a time (Bezout's theorem) and you see that in an algebraically closed field, every non-constant polynomial factors completely into linear factors. The Fundamental Theorem of Algebra says $\mathbb{C}$ is algebraically closed (proved with analysis or topology; we'll use it).
A field $\overline{F}$ is an algebraic closure of $F$ if $\overline{F}/F$ is algebraic and every polynomial $f \in F[x]$ splits completely into linear factors over $\overline{F}$.
Is $\mathbb{C}$ an algebraic closure of $\mathbb{R}$? Of $\mathbb{Q}$?
Of $\mathbb{R}$: yes. $[\mathbb{C}:\mathbb{R}] = 2$ is finite, so $\mathbb{C}/\mathbb{R}$ is algebraic, and every real polynomial splits over $\mathbb{C}$.
Of $\mathbb{Q}$: no. Every rational polynomial splits over $\mathbb{C}$, but $\mathbb{C}/\mathbb{Q}$ isn't algebraic ($\pi$ is transcendental). An algebraic closure of $\mathbb{Q}$ is the field $\mathbb{A}$ of algebraic numbers from the Spicy Bonus on Algebraic Extensions: it's algebraic over $\mathbb{Q}$, and every root of a rational polynomial is, by definition, algebraic.
Don't mix up the two definitions. "Algebraically closed" is about polynomials over $K$ itself. "Algebraic closure of $F$" is about polynomials over the smaller field $F$, plus a size limit (algebraic over $F$). $\mathbb{C}$ is algebraically closed but is not an algebraic closure of $\mathbb{Q}$. (It's a theorem that an algebraic closure is always algebraically closed, but that needs proof.)
Does every field have one? For $\mathbb{Q}$ we got lucky: $\mathbb{C}$ was already there. For $\mathbb{F}_p$ or $\mathbb{F}_p(t)$, there's no analysis to lean on. Artin's proof builds an algebraically closed field out of pure algebra, using every polynomial at once.
Every field $F$ is contained in an algebraically closed field.
Proof (Artin). Let $\mathcal{F}$ be the set of all non-constant monic polynomials in $F[x]$. For each $f \in \mathcal{F}$, associate an indeterminate $x_f$ (one variable per polynomial!), and form the polynomial ring $R = F[\{x_f : f \in \mathcal{F}\}]$. Each element of $R$ involves only finitely many variables. Let $I$ be the ideal in $R$ generated by $\{f(x_f) : f \in \mathcal{F}\}$.
Claim: $I$ is a proper ideal. If not, $1 \in I$, so there exist $a_1, \dots, a_k \in R$ and $f_1, \dots, f_k \in \mathcal{F}$ such that
$$a_1 f_1(x_{f_1}) + \dots + a_k f_k(x_{f_k}) = 1.$$
By Kronecker's Theorem (applied $k$ times), there exists a finite extension $E/F$ containing roots $\alpha_1, \dots, \alpha_k$ of $f_1, \dots, f_k$. Substitute $x_{f_i} = \alpha_i$ (and $0$ for the finitely many other variables appearing in the $a_i$). Every $f_i(x_{f_i})$ becomes $0$, so the left side becomes $0$ while the right side stays $1$: $0 = 1$ in $E$, a contradiction.
Since $I \ne R$, by Zorn's Lemma there exists a maximal ideal $\mathfrak{m} \subseteq R$ containing $I$. The quotient $K_1 = R/\mathfrak{m}$ is a field. The map $F \to K_1$, $a \mapsto a + \mathfrak{m}$, is injective (its kernel is an ideal of a field, and $1 \notin \mathfrak{m}$), so $K_1$ contains a copy of $F$. By construction every $f \in \mathcal{F}$ has a root in $K_1$, namely $x_f + \mathfrak{m}$, because $f(x_f) \in I \subseteq \mathfrak{m}$.
Repeat with $K_1$ in place of $F$ to get $K_2 \supseteq K_1$ in which every non-constant polynomial over $K_1$ has a root, and so on: $F \subseteq K_1 \subseteq K_2 \subseteq \dots$. Let $K = \bigcup_{n=1}^\infty K_n$ (a field, since any two elements lie in a common $K_n$). A non-constant $g \in K[x]$ has finitely many coefficients, all in some $K_n$, so $g$ has a root in $K_{n+1} \subseteq K$. So $K$ is algebraically closed and contains $F$. $\blacksquare$
One variable for every polynomial. Not every polynomial of degree $5$. Every polynomial. The polynomial ring has more variables than the alphabet has letters, more than there are integers if $F$ is uncountable, and Artin just shrugs and quotients it by a maximal ideal that Zorn promises exists but nobody will ever write down.
From an algebraically closed $K \supseteq F$, an algebraic closure of $F$ is easy to cut out: take $\overline{F} = \{\alpha \in K : \alpha \text{ is algebraic over } F\}$. It's a field (Algebraic Extensions), it's algebraic over $F$ by construction, and every $f \in F[x]$ splits in $K$ with roots that are algebraic over $F$, so they lie in $\overline{F}$. And like splitting fields, algebraic closures are unique.
Any two algebraic closures $L_1$ and $L_2$ of a field $F$ are isomorphic over $F$ (by an isomorphism fixing $F$).
Proof. Consider the collection $\mathcal{S}$ of all $F$-homomorphisms $\tau\colon K \to L_2$ where $F \subseteq K \subseteq L_1$ (ring homomorphisms fixing $F$; they're injective because $K$ is a field). $\mathcal{S}$ is nonempty: it contains the inclusion $F \to L_2$. Partially order $\mathcal{S}$ by extension: $\tau_1 \le \tau_2$ if the domain of $\tau_2$ contains that of $\tau_1$ and $\tau_2$ restricts to $\tau_1$. Every chain has an upper bound, obtained by taking the union of the domains, so by Zorn's Lemma there is a maximal element $\tilde{\tau}\colon \tilde{K} \to L_2$.
A note on the notes: they say "$\tilde{\tau}(g)$ splits in $L_2$" in step 1, but $g$ has coefficients in $\tilde{K}$, not in $F$, and $L_2$ is only promised to split polynomials over $F$. The fix above routes through $h = \operatorname{min}_F(\alpha)$, which is over $F$. The notes' "identical argument applied to the image" for surjectivity is also spelled out.
Over $\mathbb{R}$, a double root is where the graph touches the axis and turns around: $f$ and $f'$ both vanish there. Over $\mathbb{F}_p$ there are no graphs and no limits, but the derivative can be defined by the power rule alone, and the double-root test survives.
Let $f(x) = \sum_{i=0}^n a_i x^i \in F[x]$. The formal derivative is
$$(D_x f)(x) = \sum_{i=1}^n i a_i x^{i-1}.$$
The standard rules hold: $D_x(f + g) = D_x f + D_x g$ and $D_x(fg) = (D_x f)g + f(D_x g)$. (Both sides are bilinear in $f$ and $g$, so it's enough to check the product rule on monomials, where it's $(i + j)x^{i+j-1} = ix^{i-1}x^j + x^i jx^{j-1}$.)
In characteristic $p$, the derivative can vanish on a non-constant polynomial: $D_x(x^p) = p x^{p-1} = 0$. "Derivative zero means constant" is false here. That one fact causes all of the strange behavior below.
A polynomial $f \in F[x]$ has a repeated root in a splitting field if and only if $\gcd(f, D_x f) \ne 1$.
Proof. First, a gcd doesn't change when you enlarge the field: $\gcd(f, D_x f)$ computed in $F[x]$ satisfies a Bézout identity $uf + vD_x f = d$ that still holds in $K[x]$, and the Euclidean algorithm runs the same in both. So we may compute it in $K[x]$, $K$ a splitting field of $f$.
($\implies$) Suppose $\alpha$ is a root of multiplicity $m \ge 2$, so $f(x) = (x - \alpha)^m g(x)$. Then
$$D_x f(x) = m(x - \alpha)^{m-1}g(x) + (x - \alpha)^m D_x g(x),$$
which is divisible by $(x - \alpha)$ since $m - 1 \ge 1$. So $(x - \alpha)$ divides both $f$ and $D_x f$, and $\gcd \neq 1$.
($\impliedby$) If $d = \gcd(f, D_x f) \neq 1$, then $d \mid f$ splits in $K$, so $d$ has a root $\alpha \in K$ with $f(\alpha) = 0$ and $D_x f(\alpha) = 0$. Write $f(x) = (x - \alpha)h(x)$. Then $D_x f(x) = h(x) + (x - \alpha)D_x h(x)$, and evaluating at $\alpha$ gives $h(\alpha) = 0$. So $(x - \alpha) \mid h(x)$ and $(x - \alpha)^2 \mid f(x)$. $\blacksquare$
Build a polynomial from its roots and watch the gcd. Drag the dots along the axis; stacking two dots on one spot makes a double root.
Polynomials with no repeated roots get a name, because they're the ones Galois theory can handle.
A polynomial $f \in F[x]$ is separable if all of its roots in a splitting field are distinct.
In characteristic $0$, irreducible polynomials are automatically separable. This isn't in the notes, but it's why the Galois theory examples over $\mathbb{Q}$ never have to worry.
If $\operatorname{char}(F) = 0$ and $f \in F[x]$ is irreducible, then $f$ is separable.
Proof. If $\deg f = n \ge 1$ with leading coefficient $a_n$, then $D_x f$ has leading term $n a_n x^{n-1} \neq 0$ (since $n \neq 0$ in characteristic $0$). So $D_x f$ is a nonzero polynomial of smaller degree. $\gcd(f, D_x f)$ divides the irreducible $f$, so it's $1$ or (a unit times) $f$; it can't be $f$, because $f \nmid D_x f$ by degrees. So the gcd is $1$, and the Lemma says $f$ is separable. $\blacksquare$
Let $F = \mathbb{F}_p(t)$ and $f(x) = x^p - t \in F[x]$. Show that $f$ is irreducible but not separable.
Irreducible: $f$ is Eisenstein at the prime $t$ of $\mathbb{F}_p[t]$, so it's irreducible over $\mathbb{F}_p(t)$ by Gauss's Lemma (same argument as $x^2 - t$ on Algebraic Extensions).
Not separable: $D_x f = p x^{p-1} = 0$, so $\gcd(f, D_x f) = \gcd(f, 0) = f \neq 1$. Concretely, if $\theta$ is a root in a splitting field, then $\theta^p = t$ and
$$x^p - t = x^p - \theta^p = (x - \theta)^p,$$
since $(a - b)^p = a^p - b^p$ in characteristic $p$. One root, repeated $p$ times.
The same derivative test, on $x^{p^n} - x$, is the key to building the field with $p^n$ elements in the Existence and Uniqueness section of Finite Fields.
Find the field, find the degree.
Find the splitting field $K$ of $x^4 - 2$ over $\mathbb{Q}$ and compute $[K:\mathbb{Q}]$.
The roots are $\pm\sqrt[4]2, \pm i\sqrt[4]2$, so $K = \mathbb{Q}(\sqrt[4]2, i\sqrt[4]2) = \mathbb{Q}(\sqrt[4]2, i)$ (divide the two roots to get $i$).
$x^4 - 2$ is Eisenstein at $2$, so $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}] = 4$. Since $\mathbb{Q}(\sqrt[4]2) \subseteq \mathbb{R}$ and $i \notin \mathbb{R}$, $i$ has degree $2$ over $\mathbb{Q}(\sqrt[4]2)$ (root of $x^2 + 1$). Tower Law: $[K:\mathbb{Q}] = 4 \cdot 2 = 8$.
Find the splitting field of $x^4 + 1$ over $\mathbb{Q}$ and its degree.
The roots are the primitive $8$th roots of unity $\zeta, \zeta^3, \zeta^5, \zeta^7$ with $\zeta = e^{2\pi i/8} = \frac{\sqrt2}{2}(1 + i)$. They're all powers of $\zeta$, so the splitting field is $\mathbb{Q}(\zeta)$.
$\zeta + \zeta^7 = \zeta + \bar\zeta = \sqrt2$ and $\zeta^2 = i$, so $\mathbb{Q}(\sqrt2, i) \subseteq \mathbb{Q}(\zeta)$; conversely $\zeta = \frac{\sqrt2}{2}(1 + i) \in \mathbb{Q}(\sqrt2, i)$. So the splitting field is $\mathbb{Q}(\sqrt2, i)$, of degree $4$ ($\sqrt2 \in \mathbb{R}$, $i \notin \mathbb{R}$). This agrees with $x^4 + 1$ being irreducible (shown on Polynomial Rings): one root already has degree $4$.
Use $\gcd(f, D_x f)$ to show that $f(x) = x^3 - 3x + 2$ has a repeated root, and find it.
$D_x f = 3x^2 - 3 = 3(x - 1)(x + 1)$. Check its roots in $f$: $f(1) = 1 - 3 + 2 = 0$, $f(-1) = -1 + 3 + 2 = 4$. So $\gcd(f, D_x f) = x - 1 \neq 1$, and $1$ is a repeated root. Indeed $f = (x - 1)^2(x + 2)$.
Algebraic closures of finite fields exist (Artin), but they can't be finite.
Prove that no finite field is algebraically closed. Then show that an algebraic closure of $\mathbb{F}_p$ is infinite but countable.
Proof. Let $F = \{a_1, \dots, a_q\}$ be finite and set $f(x) = (x - a_1)(x - a_2)\cdots(x - a_q) + 1 \in F[x]$. It's non-constant (degree $q$), and $f(a_i) = 0 + 1 = 1 \neq 0$ for every $i$. So $f$ has no root in $F$, and $F$ is not algebraically closed.
Let $\overline{\mathbb{F}_p}$ be an algebraic closure. It's infinite: for each $n$, the polynomial $g = x^{p^n} - x \in \mathbb{F}_p[x]$ splits in $\overline{\mathbb{F}_p}$, and $D_x g = p^n x^{p^n - 1} - 1 = -1$, so $\gcd(g, D_x g) = 1$ and $g$ has $p^n$ distinct roots there. So $|\overline{\mathbb{F}_p}| \ge p^n$ for every $n$. It's countable: $\mathbb{F}_p[x]$ is countable, every element of $\overline{\mathbb{F}_p}$ is a root of one of these polynomials, and each has finitely many roots. $\blacksquare$