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The polynomial ring $F[X]$ is an integral domain, since the product of two nonzero polynomials has nonzero leading coefficient. Let $I$ be an ideal. If $I=0$, it is generated by zero. Otherwise choose a nonzero polynomial $g\in I$ of minimum degree. For any $f\in I$, the polynomial division algorithm over the field gives $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r=f-qg\in I$, minimality forces $r=0$. Hence every element of $I$ is a multiple of $g$, and $I=(g)$. Thus $F[X]$ is a principal ideal domain.

The nonprincipal coordinate ideal in two variables provides the contrast:If it were $(g)$, then $g$ would divide $X$ and $Y$. Total degrees add when nonzero polynomials over a field are multiplied. A nonconstant divisor of $X$ consequently has degree one, its cofactor is constant, and it is a nonzero scalar multiple of $X$. Such a polynomial cannot divide $Y$, as setting $X=0$ shows. Thus $g$ would be a nonzero constant and the ideal would contain one. But all elements of $(X,Y)$ vanish at $(0,0)$, so this is impossible.