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 $\left\langle{f}\right\rangle$ generated by $f$ is maximal.

So by Maximal Ideal iff Quotient Ring is Field:
 * $L = K \left[{X}\right] / \left\langle{f}\right\rangle$ is a field

where $L$ is a field extension over $K \left[{X}\right]$.

Now:
 * $L = \left\{{g + \left\langle{f}\right\rangle: g \in K \left[{X}\right]}\right\}$

From Division Theorem for Polynomial Forms over Field:
 * $\forall g \in K \left[{X}\right]: \exists q, r \in K \left[{X}\right]: g = q f + r, \operatorname{deg} r < \operatorname{deg} f =: n$

Therefore:
 * $L = \left\{{r + \left\langle{f}\right\rangle: r \in K \left[{X}\right],\ \operatorname{deg} r < n}\right\}$

By Basis for Quotient of Polynomial Ring, this has basis:
 * $1 + \left\langle{f}\right\rangle, \ldots, X^{n-1} + \left\langle{f}\right\rangle$ 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:
 * $\left[{L : K}\right] = 1$

where $\left[{L : K}\right]$ is the degree of $L$ over $K$.

Hence $n = \operatorname{deg}f = 1$.

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 $K \left[{X}\right]$ of strictly positive degree.

From Polynomial Forms over Field form Principal Ideal Domain, $K \left[{X}\right]$ is a principal ideal domain.

From Principal Ideal Domain is Unique Factorization Domain, $K \left[{X}\right]$ 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 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:
 * $\mu_\alpha = \left({X - \beta}\right) g$ for some $g \in K \left[{X}\right]$.

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

Also:
 * $\mu_\alpha \left({\alpha}\right) = \alpha - \beta = 0$

so $\alpha = \beta$.

Therefore $\alpha \in K$.

Therefore $L = K$ as required.

Thus $K$ is algebraically closed by definition 1.