I am trying to prove the isomorphism $\mathbb R[X]/\langle X^4-1\rangle \cong \mathbb R \times \mathbb R \times \mathbb C$. I will write what I did so you can help me from there.
First notice that $x^4-1=(x-1)(x+1)(x^2+1)$. If I could justify $\mathbb R[X]/\langle X^4-1 \rangle \cong \mathbb R[X]/\langle X-1 \rangle \times \mathbb R[X]/\langle X+1 \rangle \times \mathbb R[X]/\langle X^2+1\rangle$, then I define the following morphisms: $$\phi_1: \mathbb R[X] \to \mathbb R$$$$f \to f(1),$$$$\phi_2: \mathbb R[X] \to \mathbb R$$$$f \to f(-1),$$$$\phi_1: \mathbb R[X] \to \mathbb C$$$$f \to f(i)$$ By the first isomorphism theorem we have $\mathbb R[X]/\langle X-1 \rangle \cong \mathbb R$,$\mathbb R[X]/\langle X+1 \rangle \cong \mathbb R$ and $\mathbb R[X]/\langle X^2+1 \rangle \cong \mathbb C$ so the original isomorphism follows from here.
I would appreciate if someone could tell me how do I justify $\mathbb R[X]/\langle X^4-1\rangle \cong \mathbb R \times \mathbb R \times \mathbb C$. I know of the existence of the chinese remainder theorem,$I,J \lhd R, I+J=R \implies R/IJ \cong R/I \times R/J$. If I could show $\langle x-1 \rangle + \langle x+1 \rangle=\mathbb R[X]$, and $\langle x^2-1 \rangle + \langle x^2+1 \rangle=\mathbb R[X]$, then I would be done.
Take any $f \in \mathbb R [x]$. Observe that $f = (x+1)(\frac{1}{2}f) + (x-1)(-\frac{1}{2}f)$.