Let $\mathbb{R}[x]$ denote the ring of polynomials in $X$ with real coefficients. Then, the quotient ring $R=\mathbb{R}[x]/(x^4+4)$ is
(a) field
(b) an integral domain, but not a field
(c) not an integral domain, but has $0$ as the only nilpotent element
(d) a ring which contains non-zero nilpotent elements
My approach Since $x^4+4$ can be reducible into two irreduciable polynoimials over $\mathbb{R}$ as $(x^2+2x+2)$ and $(x^2-2x+2)$ and using Chinese Remainder theroem for ideals we can
$$R=\frac{\mathbb{R}[x]}{(x^4+4)}=\frac{\mathbb{R}[x]}{(x^2+2x+2)} \times \frac{\mathbb{R}[x]}{(x^2-2x+2)}$$
$$=\mathbb{F_1}\times\mathbb{F_2} \tag{1}$$
From $(1)$, we can say that it is not an integral domain as it has zero divisors so it can't be a field So first two options are Wrong.
Doubt 1 How to find that some nonzero element is nilpotent or not?
Maybe this is bit silly question but I'm curious to know,
Doubt 2 Is there any difference between two rings that are not integral domains like say $(\mathbb Z_4, +_4, ._4)$ and $(\mathbb Z_6, +_6, ._6)$ having no. of nilpotent elements are $2$ and $1$ respectively.
Thanks in advanced!
Edit Since neither $(x^2+2x+2)$ nor $(x^2-2x+2)$ can't be writen as square or any polynomial in $\mathbb{R[x]}$ (Both polynomials are degree $2$ and have complex root so irreduciable.) Nilpotent elements in the quotient ring of a polynomial ring
So $R$ has only $0$ as the nilpotent element. and correct option is (c).
Given factor ring $$R=\frac{\mathbb{R}[x]}{(x^4+4)}$$
$(x^4+4)$ is reducible into irreducible polynomials $(x^2+2x+2)$ and $(x^2-2x+2)$ in $\mathbb{R[x]}$ and using Chinese Remainder theroem,
$$R=\frac{\mathbb{R}[x]}{(x^4+4)}=\frac{\mathbb{R}[x]}{(x^2+2x+2)} \times \frac{\mathbb{R}[x]}{(x^2-2x+2)}$$
$$ R \approx \mathbb{R[\iota]}\times \mathbb{R[\iota]}$$ $$ R \approx \mathbb{C}\times \mathbb{C}$$ Clearly, $(0,0)$ is the only nilpotent element in $R$.