Discussion
Consider the polynomial \(p(X) = X^{4} + 4\) in the ring \(\mathbb{Q}[X]\) of polynomials in the variable \(X\) with coefficients in the field \(\mathbb{Q}\) of rational numbers. Then
Answer: B
The true statement is \(p(X)\) is reducible in the ring \(\mathbb{Q}[X]\).
Sophie Germain's identity factors it over the integers:
\[ \begin{aligned} X^{4} + 4 &= (X^{2} + 2)^{2} - 4X^{2} \\ &= (X^{2} - 2X + 2)(X^{2} + 2X + 2) \end{aligned} \]
The zeros are \(1 \pm i\) and \(-1 \pm i\). They all lie in \(\mathbb{Q}(i)\), so the splitting field is \(\mathbb{Q}(i)\), of degree 2 over \(\mathbb{Q}\), not 3 or 4. The zeros do not form a group under multiplication: 1 is not a zero, and \((1+i)(1-i) = 2\) is not a zero either.
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