In each item, a commutative ring \(R\) and an ideal \(I \subseteq R\) are given. Determine whether \(I\) is prime, maximal, both, or neither.
\(R = \mathbb{C}[x]\), \(I = (x^2+1)\).
\(R = \mathbb{Z}[x]\), \(I = (6, x)\).
\(R = \mathbb{C}[x,y]\), \(I = (y^2 - x^3)\).
Solution
An ideal \(I\) is prime if and only if \(R/I\) is an integral domain, and maximal if and only if \(R/I\) is a field.
1. In part (a), \((x^2+1)\) is \(\boxed{\text{neither prime nor maximal}}\).
Over \(\CC\), \(x^2+1=(x-i)(x+i)\). Neither factor lies in \((x^2+1)\), since every nonzero element of that ideal has degree at least \(2\), but their product does. So \((x^2+1)\) is not prime, and hence not maximal.
2. In part (b), \((6,x)\) is \(\boxed{\text{neither prime nor maximal}}\).
The surjection \(\ZZ[x]\to\ZZ/6\ZZ\), \(f\mapsto f(0)\bmod 6\), has kernel \((6,x)\): a polynomial \(f\) lies in the kernel exactly when \(f=f(0)+xg\) with \(6\mid f(0)\). Hence \(\ZZ[x]/(6,x)\cong\ZZ/6\ZZ\), which has the zero divisors \(2\cdot3=0\) and is not a domain.
3. In part (c), \((y^2-x^3)\) is \(\boxed{\text{prime but not maximal}}\).
View \(y^2-x^3\) as a monic polynomial in \(y\) over \(\CC[x]\). A factorization would be \((y-g)(y+g)\) with \(g\in\CC[x]\) and \(g^2=x^3\), which is impossible since \(x^3\) has odd degree. Hence \(y^2-x^3\) is irreducible in the UFD \(\CC[x,y]\), so it is a prime element and \((y^2-x^3)\) is a prime ideal. It is properly contained in the proper ideal \((x,y)\), since \(y\notin(y^2-x^3)\) by degree, so it is not maximal.
4. Q.E.D.
Steps 1, 2, and 3 treat parts (a), (b), and (c).