= Solution
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 $\deg r<\deg g$. Since $r=f-qg\in I$, minimality forces $r=0$. Hence every element of $I$ is a multiple of $g$, and $I=(g)$. Thus \b[$F[X]$ is a <principal ideal domain>].
The <nonprincipal coordinate ideal in two variables> provides the contrast:
$$
\boxed{(X,Y)\subset F[X,Y]\text{ is not principal}.}
$$
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.
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