Question

Which of the statements given below is/are true?
(I) \(\mathbb{Z}[i]\) is a Principal ideal domain and Euclidean domain.
(II) \(\mathbb{Z}[\sqrt{-5}]\) is neither a Principal ideal domain nor a Unique factorization domain.

• A. Only (I)
• B. Only (II)
• C. Neither (I) nor (II)
• D. Both (I) and (II)

Hint:

Check if Z[i] has a division algorithm, and test unique factorization of 6 in Z[sqrt(-5)].

Answer 1

The Correct Option is D

Step 1: Consider statement (I): \(\mathbb{Z}[i] = \{a+bi : a,b \in \mathbb{Z}\}\), the Gaussian integers. Define the norm \(N(a+bi) = a^2+b^2\). For any \(\alpha,\beta \in \mathbb{Z}[i]\) with \(\beta \neq 0\), rounding the real and imaginary parts of \(\alpha/\beta\) to the nearest integers gives \(q \in \mathbb{Z}[i]\) with \(N(\alpha - q\beta) < N(\beta)\), so a division algorithm exists.
Step 2: This makes \(\mathbb{Z}[i]\) a Euclidean domain, and every Euclidean domain is a Principal ideal domain, since any ideal is generated by a nonzero element of minimal norm inside it, using the division algorithm. So \(\mathbb{Z}[i]\) is both a Euclidean domain and a PID, and statement (I) is true.
Step 3: Consider statement (II): \(\mathbb{Z}[\sqrt{-5}] = \{a+b\sqrt{-5} : a,b \in \mathbb{Z}\}\). Look at two factorizations of 6: \[6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5})\] Using the norm \(N(a+b\sqrt{-5}) = a^2+5b^2\), we get \(N(2)=4\), \(N(3)=9\), and \(N(1\pm\sqrt{-5}) = 6\). Since no element of the ring has norm 2 or 3, as \(a^2+5b^2=2\) or \(3\) has no integer solutions, each of \(2, 3, 1+\sqrt{-5}, 1-\sqrt{-5}\) is irreducible, and none of them are associates of one another.
Step 4: This gives two genuinely different factorizations of 6 into irreducibles, so \(\mathbb{Z}[\sqrt{-5}]\) is not a UFD. Since every PID is a UFD, it cannot be a PID either. So statement (II) is also true.
Step 5: Both (I) and (II) are true. \[\boxed{\text{Both (I) and (II)}}\]