Question

Prove that polynomial ring of 2 independent variables over field is not a PID.

Answer 1

Question 1. Prove that polynomial ring of 2 independent variables over field is not a PID.

Solution. Let $R[x,y]$ be the mentioned polynomial ring. Consider the ideal $I = \langle x,y\rangle = \{xP_1 + yP_2\mid P_1,P_2\in R[x,y]\}$. Suppose there is a polynomial $P\in R[x,y]$ such that $I = P\cdot R[x,y]$. Then any element of $I$ should be divisible by $P$, in particular, $P$ should divide $x$ and $y$. But $x$ and $y$ have degree 1. Therefore, the degree of $P$ should not exceed 1. So, $P = ax + by + c$ for some scalars $a,b,c\in R$. Since $P$ divides $x$, we conclude that $b = c = 0$. But the fact that $P$ divides $y$ implies $a = c = 0$. Thus, $a = b = c = 0$ and so $P = 0$. Therefore, $I = \{0\}$, which is a contradiction (I contains, for example, $x$ and $y$).