Discussion
Let \(\mathbb{R}[X]\) denote the ring of polynomials in X with real coefficients. Then, the quotient ring \(\mathbb{R}[X]/(X^{4} + 4)\) is
Answer: C
The quotient ring is not an integral domain, but has 0 as the only nilpotent element.
\(X^{4} + 4\) factors over \(\mathbb{R}\) (Sophie Germain's identity):
\[ X^{4} + 4 = (X^{2} + 2X + 2)(X^{2} - 2X + 2) \]
Both quadratics have discriminant 4 - 8 = -4 \(\lt\) 0, so they are irreducible, and they are distinct, hence coprime. By the Chinese remainder theorem the quotient splits into two factors, each a copy of \(\mathbb{C}\):
\[ \begin{aligned} \mathbb{R}[X]/(X^{4}+4) &\cong \mathbb{R}[X]/(X^{2}+2X+2) \\ &\times \mathbb{R}[X]/(X^{2}-2X+2) \\ &\cong \mathbb{C} \times \mathbb{C} \end{aligned} \]
A product of two fields has zero divisors, such as (1, 0)(0, 1) = 0, so it is neither a field nor an integral domain. It has no non-zero nilpotents, because a nilpotent pair would need each coordinate nilpotent in a field, that is zero. Non-zero nilpotents would appear only if \(X^{4} + 4\) had a repeated factor.
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