Equivalence of Definitions of Algebraically Closed Field

Theorem
Let $K$ be a field. The following are equivalent:


 * $(1):\quad$ The only algebraic extension of $K$ is $K$ itself


 * $(2):\quad$ Every irreducible polynomial $f$ over $K$ has degree $1$


 * $(3):\quad$ Every polynomial $f$ over $K$ of strictly positive degree has a root in $K$

Proof
$(1) \implies (2)$

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

By Principal Ideal of Irreducible Element the ideal $\langle f \rangle$ generated by $f$ is maximal.

So by Maximal Ideal iff Quotient Ring is Field, $L = K[X]/\langle f \rangle$ is a field.

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

By the division theorem for polynomials for each $g \in K[X]$ there exist $q,r \in K[X]$ such that $g = qf + r$ and $\operatorname{deg}r < \operatorname{deg}f =: n$.

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

By Basis for Quotient of Polynomial Ring, this has basis $1 + \langle f \rangle,\ldots, X^{n-1} + \langle f \rangle$ span $L$, so $L$ is finite, and hence algebraic.

Also $K \subseteq L$, so by hypothesis $K = L$, which implies $[L:K] = 1$, hence $n = \operatorname{deg}f = 1$.

$(2) \implies (3)$

Suppose 2. and let $f$ be a polynomial in $K[X]$ of strictly positive degree.

Since the ring of polynomial forms is a PID, also it is a UFD.

So $f$ can be decomposed $f = u g_1\cdots g_r$ with $u$ a unit and $g_i$ 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 $K$.

$(3) \implies (1)$

Let $L/K$ be an algebraic extension of $K$, and $\alpha \in L$.

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

Therefore by the Polynomial Factor Theorem $\mu_\alpha = (X - \beta)g$ for some $g \in K[X]$.

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

Also $\mu_\alpha(\alpha) = \alpha - \beta = 0$, so $\alpha = \beta$, so $\alpha \in K$.

Therefore $L = K$ as required.