UTM Ideals Varieties and Algorithm - Chapter 1 Section 5 Exercise 3

Problem:

Solution:

It is obvious that $x \in I$ and $y \in I$, if $I$ were a principal ideal then there exists $f \in I$ such that every element in $p \in I$ can be written as $p = f g_p$.

Now $f g_x = x$ and $f g_y = y$

Since $\deg(x)  = 1$, so either $f$ has degree 1 and $g_x$ has degree 0 or $f$ has degree 0 and $g$ has degree 1.

Suppose $f$ has degree 1 so $f = ax + by + c$ and therefore $g_x = d$ has degree 0. We found $fg_x = (adx + bdy + cd) = 0$, so $ad = 1$ and hence $d \ne 0$, we also know $b = c = 0$, so $f = ax$ and $g_x = d$. However, $f g_y = ax g_y = y$, there couldn't be a polynomial $g_y$ that satisfy this, so $f$ has to have degree 0.

Now $f = e$ has degree 0, but then it implies $e \in I$, but there is not way $e \in I$ because any constant cannot be written as $p(x, y)x + q(x, y)y$, so we conclude $\langle x, y. \rangle$ cannot be a principal ideal.