$(x^2-5)$ is not a prime ideal of $\mathbb{R}[x]$

Question

This question might be very naive but I still want to varify if my idea is right, since I am only beginning ring theory.

I have to show $(x^2-5)$ is not a prime ideal of $\mathbb{R}[x]$. My thought is that, $x^2-5$ can be reduced as $x^2-5=(x-\sqrt{5})(x+\sqrt{5})$ for $x-\sqrt{5},x+\sqrt{5}\in\mathbb{R}[x]$ and since neither of them is necessarily in $(x^2-5)$, we have $(x^2-5)$ not a prime ideal.

Is my idea ok?

Answer 1

Yes, and you can see that neither $x - \sqrt5$ or $x + \sqrt 5$ is contained in $\mathfrak p =(x^2 -5)$ since they are degree $1$ and $\mathfrak p$ only contains polynomials of degree $\geq 2$ and the zero polynomial