[proofplan]
We prove by strong induction on $n = \deg f$ that every nonconstant polynomial $f \in K[t]$ admits a splitting field, i.e., an extension $L/K$ over which $f$ factors into linear factors and which is generated over $K$ by the roots of $f$.
**Base case.** $n = 1$: the polynomial is already linear, so $L = K$ works.
**Inductive step.** Assume the result holds for all nonconstant polynomials of degree less than $n$ over any field. Let $\deg f = n \ge 2$. Choose an irreducible factor $p \mid f$ in $K[t]$ with $\deg p \ge 2$. Form the quotient $K_1 = K[t]/(p)$, which is a field extension of $K$ containing a root $\alpha_1$ of $p$. In $K_1[t]$, divide out $(t - \alpha_1)$ to get a polynomial of degree $n - 1$. Apply the inductive hypothesis over $K_1$ to that quotient, then verify the resulting extension is a splitting field for $f$ over $K$.
[/proofplan]
[step:Base case: degree 1]
If $\deg f = 1$, then $f = a(t - b)$ for some $a, b \in K$ with $a \neq 0$. The polynomial already splits into linear factors over $K$, and $K$ is generated over itself by the (single) root $b$. So $L = K$ is a splitting field for $f$.
[/step]
[step:Choose an irreducible factor and build an extension containing a root]
Let $n \ge 2$ and assume the theorem holds for all nonconstant polynomials of degree less than $n$ over any field. Let $f \in K[t]$ with $\deg f = n$.
If $f$ already splits into linear factors over $K$, take $L = K$ and we are done. Otherwise $f$ has an irreducible factor $p \in K[t]$ with $d = \deg p \ge 2$.
[/step]
[guided]
Why must such a $p$ exist? Factor $f$ into irreducibles in $K[t]$ (this is possible because $K[t]$ is a UFD). If every irreducible factor were linear, $f$ would already split over $K$, contradicting our assumption.
[/guided]
[step:The quotient $K[t]/(p)$ is a field containing a root of $p$]
Since $p$ is irreducible over $K$, the ideal $(p)$ is maximal in the PID $K[t]$, so
\begin{align*}
K_1 = K[t]/(p)
\end{align*}
is a field. Let $\alpha_1 = t + (p) \in K_1$. Then $p(\alpha_1) = p(t) + (p) = 0$ in $K_1$, so $\alpha_1$ is a root of $p$ in $K_1$. Because $p \mid f$ in $K[t]$, we also have $f(\alpha_1) = 0$, so $\alpha_1$ is a root of $f$.
[/step]
[guided]
The key algebraic point: evaluation of $p$ at the coset $\alpha_1 = t + (p)$ sends each coefficient $c_k \in K$ to itself (since $K$ embeds into $K_1$ via $c \mapsto c + (p)$) and sends $t$ to $\alpha_1$. The result is $p(t) + (p) = 0 + (p)$, the zero element of $K_1$.
[/guided]
[step:Factor out the root and apply the inductive hypothesis]
Since $f(\alpha_1) = 0$ in $K_1$, the factor theorem gives
\begin{align*}
f(t) = (t - \alpha_1)\, g(t), \qquad g \in K_1[t], \quad \deg g = n - 1.
\end{align*}
Now $g$ is a nonconstant polynomial of degree $n - 1 \ge 1$ over the field $K_1$. By the inductive hypothesis applied to $g$ over $K_1$, there exists a splitting field $L / K_1$ for $g$. That is, there exist $\alpha_2, \dots, \alpha_n \in L$ and a leading coefficient $a \in K^\times$ such that
\begin{align*}
g(t) = a \prod_{i=2}^{n} (t - \alpha_i) \quad \text{in } L[t].
\end{align*}
Moreover $L = K_1(\alpha_2, \dots, \alpha_n)$.
[/step]
[step:Verify $L$ is a splitting field for $f$ over $K$]
Combining the two factorizations gives
\begin{align*}
f(t) = a\prod_{i=1}^{n}(t - \alpha_i) \quad \text{in } L[t],
\end{align*}
so $f$ splits completely over $L$. It remains to check minimality: $L$ is generated over $K$ by the roots of $f$. Indeed,
\begin{align*}
L = K_1(\alpha_2, \dots, \alpha_n) = K(\alpha_1)(\alpha_2, \dots, \alpha_n) = K(\alpha_1, \alpha_2, \dots, \alpha_n).
\end{align*}
Therefore $L$ is a splitting field for $f$ over $K$, completing the induction. $\blacksquare$
[/step]
