Polynomial X^2 + 1 is Irreducible in Ring of Real Polynomials

Theorem
Let $\R \sqbrk X$ be the ring of polynomials in $X$ over the real numbers $\R$.

Then the polynomial $X^2 + 1$ is an irreducible element of $\R \sqbrk X$.

Proof
$x^2 + 1$ has a non-trivial factorization in $\R \sqbrk X$.

Then:
 * $\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$

and from the Polynomial Factor Theorem:
 * $\alpha^2 + 1 = 0$

But that means:
 * $\alpha^2 = -1$

and such an $\alpha$ does not exist in $\R$.

Hence the result by Proof by Contradiction.