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

Aiming for a contradiction, suppose $x^2 + 1$ 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.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $58$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous): Chapter $9$: Rings: Exercise $24$