Is $\mathbb R[x]/(x^2+a)$ isomorphic to $\mathbb R[x]/(x^2+b)$?

Question

Is $\mathbb R[x]/(x^2+a)$ isomorphic to $\mathbb R[x]/(x^2+b)$?

I think $\mathbb R[x]/(x^2+1)$ is isomorphic to $\mathbb C$. Is $\mathbb R[x]/(x^2+n)$ also isomorphic to $\mathbb C$ for $n \in \mathbb N$?

Thanks!

Answer 1

$\mathbb C$ is the algebraic closure of $\mathbb R$ (fundamental theorem of algebra) and is generated over $\mathbb R$ by any element not in $\mathbb R$ (easy to show — it's generated even additively).

So if quadratic polynomial $f$ is irreducible, then $\mathbb R[X]/(f) \cong \mathbb C$, if it's reducible with distinct roots, you get $\mathbb R \oplus \mathbb R$, and when $f = l^2$ for a linear $l$ the result will be two-dimensional algebra over $\mathbb R$ of the form, well, $\mathbb R[X]/X^2$. It maps to $\mathbb R$ via sending $X$ to $0$ with kernel $X \cdot \mathbb R$ but doesn't decompose as product of rings.

Anything of the form $x^2 + a$ with $a > 0$ is clearly irreducible, regardless of whether $a$ is natural or not.