Equivalence of Definitions of Algebraically Closed Field

Theorem
Let $K$ be a field.

Definition $(1)$ implies Definition $(2)$
Let $K$ be algebraically closed by definition 1.

Let $f$ be an irreducible polynomial over $K$.

By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, the ideal $\gen f$ generated by $f$ is maximal.

So by Maximal Ideal iff Quotient Ring is Field:
 * $L = K \sqbrk X / \gen f$ is a field

where $L$ is a field extension over $K \sqbrk X$.

Now:
 * $L = \set {g + \gen f: g \in K \sqbrk X}$

From Division Theorem for Polynomial Forms over Field:
 * $\forall g \in K \sqbrk X: \exists q, r \in K \sqbrk X: g = q f + r, \deg r < \deg f =: n$

Therefore:
 * $L = \set {r + \gen f: r \in K \sqbrk X, \deg r < n}$

By Basis for Quotient of Polynomial Ring, this has basis:
 * $1 + \gen f, \ldots, X^{n - 1} + \gen f$ span $L$.

Thus $L$ is finite.

By Finite Field Extension is Algebraic, $L$ is algebraic.

Also $K \subseteq L$.

So by hypothesis $K = L$.

This implies:
 * $\index L K = 1$

where $\index L K$ is the degree of $L$ over $K$.

Hence:
 * $n = \deg f = 1$

Thus $K$ is algebraically closed by definition 2.

Definition $(2)$ implies Definition $(3)$
Let $K$ be algebraically closed by definition 2.

Let $f$ be a polynomial in $K \sqbrk X$ of strictly positive degree.

From Polynomial Forms over Field form Principal Ideal Domain, $K \sqbrk X$ is a principal ideal domain.

From Principal Ideal Domain is Unique Factorization Domain, $K \sqbrk X$ is a unique factorization domain.

So $f$ can be factorized $f = u g_1 \cdots g_r$ such that:
 * $u$ is a unit

and:
 * $g_i$ are irreducible for $i = 1, \ldots, r$.

By hypothesis, $g_1$ has degree $1$.

Therefore by the Polynomial Factor Theorem $g_1$, and hence $f$, has a root in $K$.

Thus $K$ is algebraically closed by definition 3.

Definition $(3)$ implies Definition $(1)$
Let $K$ be algebraically closed by definition 3.

Let $L / K$ be an algebraic field extension of $K$.

Let $\alpha \in L$.

By hypothesis, the minimal polynomial $\mu_\alpha$ of $\alpha$ over $K$ has a root $\beta$ in $K$.

Therefore by the Polynomial Factor Theorem:
 * $\mu_\alpha = \paren {X - \beta} g$

for some $g \in K \sqbrk X$.

Since $\mu_\alpha$ is irreducible and monic (see Minimal Polynomial) it follows that:
 * $\mu_\alpha = X - \beta$

Also:
 * $\map {\mu_\alpha} \alpha = \alpha - \beta = 0$

so $\alpha = \beta$.

Therefore $\alpha \in K$.

Therefore $L = K$ as required.

Thus $K$ is algebraically closed by definition 1.