For which \(n\) is the polynomial ring \(k[x_1,\dots,x_n]\) over a field \(k\) a PID?
Solution
For \(n=0\), the ring is the field \(k\), hence a PID. For \(n=1\), the ring \(k[x_1]\) is Euclidean, hence a PID.
Assume \(n\ge2\) and consider \[ I=(x_1,x_2). \] This is proper: evaluation at \(x_1=x_2=0\) sends every element of \(I\) to \(0\), while it sends \(1\) to \(1\).
Suppose \(I=(d)\) were principal. Since \(x_1,x_2\in(d)\), the polynomial \(d\) divides both \(x_1\) and \(x_2\). In the UFD \(k[x_1,\dots,x_n]\), the irreducibles \(x_1\) and \(x_2\) are nonassociate, so their only common divisors are units. Thus \(d\) is a unit, which would give \[ I=(1), \] contradicting properness.
Therefore \(k[x_1,\dots,x_n]\) is a PID exactly for \(n\le1\).