Question:
Which one of the following options is correct for the ideal \(I=\langle x^2+5\rangle\) in the ring \(\mathbb{Q}[x]\)?
Which one of the following options is correct for the ideal \(I=\langle x^2+5\rangle\) in the ring \(\mathbb{Q}[x]\)?
Show Hint
Check whether \(x^2+5\) is irreducible over \(\mathbb{Q}\); irreducibles generate maximal (hence prime) ideals in a PID.
Updated On: Jul 3, 2026
- I is a prime ideal, but not a maximal ideal.
- I is a maximal ideal, but not a prime ideal.
- I is both a prime ideal and a maximal ideal.
- I is neither a prime ideal nor a maximal ideal.
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The Correct Option is C
Solution and Explanation
Step 1: \(\mathbb{Q}[x]\) is a Euclidean domain and hence a PID. In a PID, the ideal generated by a nonzero, non-unit polynomial \(f(x)\) is maximal if and only if \(f(x)\) is irreducible, and every maximal ideal is automatically prime.
Step 2: Check irreducibility of \(x^2+5\) over \(\mathbb{Q}\). Being degree 2, it is irreducible over \(\mathbb{Q}\) iff it has no root in \(\mathbb{Q}\).
Step 3: Solving \(x^2+5=0\) gives \(x^2=-5\), i.e. \(x=\pm\sqrt{-5}\), which is not real, hence not rational. So \(x^2+5\) has no root in \(\mathbb{Q}\), and it is irreducible over \(\mathbb{Q}\).
Step 4: Since \(x^2+5\) is irreducible in the PID \(\mathbb{Q}[x]\), \(I=\langle x^2+5\rangle\) is a maximal ideal, and therefore also a prime ideal.
\[\boxed{I \text{ is both a prime ideal and a maximal ideal}}\]
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