Discussion
Consider the following statements :
(I) 3 is prime in the integral domain \(\mathbf{Z}[i\sqrt{5}]\)
(II) \(\overline{2}\) is a prime element in \(\mathbf{Z}/10\,\mathbf{Z}\) but not irreducible in \(\mathbf{Z}/10\,\mathbf{Z}\)
Which is correct ?
Answer: B
Only (II) is true.
(I) In \(\mathbf{Z}[i\sqrt{5}]\),
\[ 3 \cdot 3 = 9 = (2 + i\sqrt{5})(2 - i\sqrt{5}) \]
3 divides the product but neither factor, since \((2 \pm i\sqrt{5})/3\) is not in the ring. So 3 is not prime there.
(II) In \(\mathbf{Z}/10\mathbf{Z}\) the ideal generated by \(\overline{2}\) is {0, 2, 4, 6, 8}, and the quotient by it is \(\mathbf{Z}/2\mathbf{Z}\), a field, so \(\overline{2}\) is prime. But \(\overline{2} = \overline{2} \cdot \overline{6}\) (since 12 leaves remainder 2), and neither \(\overline{2}\) nor \(\overline{6}\) is a unit, so \(\overline{2}\) is not irreducible. Statement (II) is therefore true.
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