Proof that every polynomial has a root in some extension field

Proof

1. Let $K$ be a field and $f \in K[x]$ with $\deg f \ge 1$. The units of $K[x]$ are the nonzero constants, by the degree of a product.
2. Among the monic divisors of $f$ of degree at least $1$, choose one, $g$, of least degree. One exists: $f$ times the inverse of its leading coefficient.
3. The polynomial $g$ is irreducible. It is nonzero and not a unit, since $\deg g \ge 1$.
4. Suppose $g = a b$ with $a, b \in K[x]$ both of degree at least $1$. Then $\deg a < \deg g$, and $a$ times the inverse of its leading coefficient is a monic divisor of $f$. This contradicts the choice of $g$.
5. So in every factorisation $g = a b$, one factor has degree $0$, so it is a unit. Write $f = g k$ with $k \in K[x]$.
6. Let $L = K[x]/(g)$, the quotient ring. Since $g$ is irreducible, $L$ is a field.
7. The map $K \to L$, $c \mapsto c + (g)$, is a ring homomorphism from a field to a nonzero ring. So it is injective. As in adjoining a root, we identify $K$ with its image, so $L/K$ is a field extension.
8. Let $\alpha = x + (g) \in L$. For every $h \in K[x]$, the quotient map is a ring homomorphism that fixes $K$, so $h + (g) = h(\alpha)$.
9. In particular $g(\alpha) = g + (g) = 0$. Then $f(\alpha) = g(\alpha) k(\alpha) = 0$, so $\alpha$ is a root of $f$ in $L$.
10. Since $g$ is monic, irreducible and $g(\alpha) = 0$, the minimal polynomial $m_\alpha$ of $\alpha$ divides $g$. As $\deg m_\alpha \ge 1$, the cofactor is a unit, and since both are monic, $g = m_\alpha$.
11. By Step 8, $L = \set{ h(\alpha) : h \in K[x] }$. By the structure of simple extensions, this set is $K(\alpha)$, so $L = K(\alpha)$.
12. By the degree of a simple extension, $[L : K] = \deg m_\alpha = \deg g$.
13. Since $f = g k$ with $f \ne 0$, we have $k \ne 0$ and $\deg f = \deg g + \deg k \ge \deg g$. So $[L : K] \le \deg f$, and $f$ has the root $\alpha$ in $L$.