Using the following definition of irreducible:
An element $q$ in the ring of polynomials over the field of real polynomials $\mathbb R[x]$ is irreducible if, for all $a, b\in \mathbb R[x]$ s.t. $q = a * b$, $a$ is invertible OR $b$ is invertible, i.e. $a$ or $b$ is a scalar.
For our purposes assume all polynomials are monic.
So, I want to show that $f = x^2 + bx + c$ with $b^2 – 4c < 0$ is irreducible per this definition. We immediately know that there exists a root $\alpha$ that is complex. We also know that the complex conjugate of $\alpha$, $\bar\alpha$, is a root.
Case 1: If $\alpha = \bar\alpha$, then $0 = (x-\alpha) = (x-\alpha)^2 = f$
Case 2: If $\alpha \neq \bar\alpha$, then $0 = (x-\alpha)(x – \bar\alpha) = f$
This is where I am confused. Per this definition, for case 2, if $q = x^2 + bx + c$, and $a = (x – \alpha)$, and $b = (x – \bar\alpha)$, then $q = a * b$ and neither $a$ nor $b$ are invertible! Therefore, $f$ is not irreducible!
And for case 1, while $a = (x – \alpha)$ and $b = 1$ is clearly irreducible, I do not understand how $(x – \alpha)^2$ is irreducible, where $a = b = (x – \alpha)$! This definition of irreducibility does not specify whether $a$ may $= b$, so I assume that it may.
Can someone help me see what I am not seeing here?
Best Answer
If $b^2-4c < 0$, the roots $\alpha$ and $\overline{\alpha}$ are not real, so $z - \alpha$ and $z - \overline{\alpha}$ are not in $\mathbb R[z]$. You are factoring over $\mathbb R[z]$, not $\mathbb C[z]$.