Discussion
For an odd prime \(p\), consider the ring \(\mathbb{Z}[\sqrt{-p}] = \{a + b\sqrt{-p} : a, b \in \mathbb{Z}\}\) \(\subseteq \mathbb{C}\). Then the element 2 in \(\mathbb{Z}[\sqrt{-p}]\) is
Answer: D
The element 2 is irreducible.
Use the norm \(N(a + b\sqrt{-p}) = a^{2} + pb^{2}\), which is multiplicative. \(N(2) = 4\). A factorization \(2 = \alpha\beta\) into non-units would need \(N(\alpha) = N(\beta) = 2\), that is \(a^{2} + pb^{2} = 2\), which has no integer solution when \(p \ge 3\). So 2 is irreducible.
It is not a unit, since units have norm 1. It is not a square: \((a + b\sqrt{-p})^{2} = 2\) forces \(ab = 0\) and then \(a^{2} = 2\) or \(-pb^{2} = 2\). It is not prime: \(p + 1\) is even, so 2 divides \((1 + \sqrt{-p})(1 - \sqrt{-p}) = 1 + p\), yet 2 divides neither factor, because \(\frac{1}{2} \pm \frac{1}{2}\sqrt{-p}\) is not in the ring.
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