The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not principal [closed]

The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.

I don't know how to consider it. Any suggestions?

A generator $g$ of $\langle x,y \rangle$ would divide $x$ and $y$, which are irreducible and non-associated. Thus $g$ would be a unit, and $\langle x,y \rangle$ would be the unit ideal, contradiction.

Suppose it would be principal. Then there would exist $p(x,y)$ such that for all polynomials $g(x,y)$ and $h(x,y)$ in $k[x,y]$, we would have some $i(x,y)$ such that $$ xg + yh = ip. $$ In particular, $x = i_1 p$ and $y = i_2 p$, so that the degree of $ p$ is at most $1$. It can't be zero, because that would mean p is a unit or zero and we would lose the fact that $\langle p \rangle = \langle x,y \rangle$. So $\deg p = 1$, hence $p=ax+by+c$ for $a,b,c \in k$. But then $x = i_1(ax+by+c)$ means $b= 0$ and $ y =i_2 (ax+c)$ means $a=0$, hence $p$ has not degree $1$, contradiction.

Hope that helps,