What is a principal ideal domain?
Consider the four rings \(\mathbb{Q}[x]\), \(\mathbb{Q}[x]/(x^2-2)\), \(\mathbb{Q}[x]/(x^2-4)\), and \(\mathbb{Q}[x,y]\). Which of these rings are principal ideal domains, and which are not? Explain.
Solution
1. A principal ideal domain is a commutative ring \(R\) with\(1\ne0\), no zero divisors, and every ideal generated by one element:for each ideal \(I\) there is \(a\in R\) such that \(I=(a)=aR\).The zero ideal is allowed and is generated by zero [@DF04].
2. The ring \(\mathbb Q[x]\) is a principal ideal domain.
The product of two nonzero polynomials is nonzero, because its leading coefficient is the product of two nonzero rational numbers. Thus \(\mathbb Q[x]\) is an integral domain.
Let \(I\) be a nonzero ideal, and choose a nonzero \(f\in I\) of smallest degree. For any \(g\in I\), polynomial division over the field \(\mathbb Q\) gives \(g=qf+r\), with \(r=0\) or \(\deg r<\deg f\). Since \(r=g-qf\in I\), the latter alternative contradicts the choice of \(f\). Therefore every \(g\in I\) belongs to \((f)\), while \((f)\subseteq I\) because \(f\in I\). Hence \(I=(f)\). Together with \((0)\), this accounts for every ideal.
3. The ring \(\mathbb Q[x]/(x^2-2)\) is a field and hence aprincipal ideal domain.
The quadratic \(f=x^2-2\) has no rational root. Indeed, if \((a/b)^2=2\) for relatively prime integers \(a,b\) with \(b\ne0\), then \(a^2=2b^2\) makes \(a\) even, and substitution then makes \(b\) even, a contradiction. A reducible quadratic over a field has a linear factor, hence a root, so \(f\) is irreducible.
For any nonzero residue class \(\bar g\) modulo \((f)\), the polynomial \(f\) does not divide \(g\). Irreducibility gives \(\gcd(f,g)=1\). The Euclidean algorithm in \(\mathbb Q[x]\) supplies polynomials \(a,b\) with \(ag+bf=1\), so \(\bar a\bar g=1\). Thus every nonzero element of the quotient is invertible. The quotient is nonzero because a polynomial of positive degree cannot divide \(1\). It is therefore a field. A field has only the ideals \((0)\) and \((1)\): any ideal containing a nonzero element contains its product with its inverse, namely \(1\).
4. The ring \(\mathbb Q[x]/(x^2-4)\) is not a principal ideal domain.
The residue classes of \(x-2\) and \(x+2\) are nonzero: neither degree-one polynomial is a multiple of \(x^2-4\). Their product is zero, since \((x-2)(x+2)=x^2-4\). The quotient has zero divisors and is not an integral domain, so it cannot be a principal ideal domain.
5. The ring \(\mathbb Q[x,y]\) is not a principal ideal domain.
Consider \(I=(x,y)\). Evaluation at \((0,0)\) sends every element of \(I\) to zero but sends \(1\) to \(1\), so \(I\) is proper.
Suppose \(I=(f)\). Since \(x\in I\), there is a nonzero polynomial \(g\) with \(fg=x\). Regard this identity as one in \(\mathbb Q[x][y]\). The coefficient ring \(\mathbb Q[x]\) is a domain by step 2, so degrees in \(y\) add for nonzero products. The right side has \(y\)-degree zero, hence \(f,g\in\mathbb Q[x]\). Their \(x\)-degrees add to one. If \(f\) has degree zero, it is a nonzero rational unit, contradicting that \(I\) is proper. Otherwise \(g\) is a nonzero rational constant, and \(f=cx\) for some \(c\in\mathbb Q^\times\).
But \(y\in(f)\) would then give \(y=cxh\) for a polynomial \(h\). Setting \(x=0\) yields the false identity \(y=0\) in \(\mathbb Q[y]\). Thus \(I\) is not principal. This rules out the principal ideal domain property.