I am working on the following problem:
Describe the factorization of $x^4 + 1$ into irreducible polynomials in $\mathbb{R}[x]$. Prove that $\mathbb{R}[x]/(x^4+1) \cong \mathbb{C} \oplus \mathbb{C}$.
I first found the factorization of $x^4 + 1$ in $\mathbb{R}[x]$ : $x^4 + 1 = (x^2+1)^2 - 2x^2 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x -1)$, where these two factors are both irreducible polynomials in $\mathbb{R}[x]$.
I'm struggling with the second part of the question. I learned that $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$ : one can think of an element in the former ring as a polynomial $ax + b$, with $a,b \in \mathbb{R}$ and $x^2 = -1$, so that the map $\phi: \mathbb{R}[x]/(x^2+1) \longrightarrow \mathbb{C}: x \mapsto i$ defines an isomorphism. (This way, the polynomial $ax + b$ gets identified with the complex number $ai + b$.)
I tried to use this same logic in showing $\mathbb{R}[x]/(x^4 + 1) \cong \mathbb{C} \oplus \mathbb{C}$. One can think of an element in the former ring as a polynomial $ax^3 + bx^2 + cx + d$, where $a,b,c,d \in \mathbb{R}$ and $x^4 = -1$. I'm having trouble defining a map $\phi: \mathbb{R}[x]/(x^4 + 1) \longrightarrow \mathbb{C} \oplus \mathbb{C}$ that defines an isomorphism. It seems that $x^2$ should get identified with $i$, but what about $x^3$? How can I identify such a polynomial with a pair of appropriate complex numbers with a clever map?
Thanks!
You have $$\mathbb{R}[x]/(x^4+1)\cong \mathbb{R}[x]/(x^2-\sqrt{2}x+1)\oplus\mathbb{R}[x]/(x^2+\sqrt{2}x-1)\cong\mathbb{C}\oplus\mathbb{R}\oplus\mathbb{R}.$$ Note that in $\mathbb{R}[x]/(x^2-\sqrt{2}x+1)\cong \mathbb{C}$, $x^2\equiv \sqrt{2}x+1$, so $x$ is not mapped to $i$ under the isomorphism. Instead, $x$ is mapped to a root of $x^2-\sqrt{2}x+1$, which are $\frac{\sqrt{2}}{2}(1\pm i)$. On the other hand, $x^2+\sqrt{2}x-1$ is not irreducible. It has roots $\frac{-\sqrt{2}\pm\sqrt{6}}{2}$, so $\mathbb{R}[x]/(x^2+\sqrt{2}x-1)\cong\mathbb{R}\oplus\mathbb{R}$.
EDIT: As pointed out in the comments below, the factorization of $x^4+1$ in the post is not correct (I didn't check, yikes!). Instead, $$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$ Both quadratic factors are irreducible and the roots of $x^2-\sqrt{2}x+1$ are $-\frac{\sqrt{2}}{2}(1\pm i)$.