I am reading Algebra written by Dummit and Foote. Let $F$ be a field and $(x,y)$ be an ideal of $F[x,y].$ On page 682 Examples (3) the book says that $$(x,y)^2 = (x^2, xy, y^2) \subsetneq (x^2, y) \subsetneq (x,y).$$
Is $(x,y)^2\subsetneq (x^2,y)$ because $y\not\in (x,y)^2$? But how can we express $xy$ by elements of $(x^2,y)?$
Is $(x^2,y)\subsetneq (x,y)$ because $x\not\in (x^2,y)$?
$y\notin (x,y)^{2}$: suppose $y\in (x,y)^{2}$. Then $y=f(x,y)x^{2}+g(x,y)xy+h(x,y)y^{2}$ for some $f,g,h\in F[x,y]$. So $y=h(0,y)y^{2}$, and $1=h(0,y)y$. $h(0,y)\neq 0$. The degree of LHS is $0$, but the degree of RHS is $\geq 1$, contradiction.
$xy\in (x^{2},y)$: because $xy=0\cdot x^{2}+x\cdot y$.
$x\notin (x^{2},y)$: suppose $x\in (x^{2},y)$. Then $x=f(x,y)x^{2}+g(x,y)y$ for some $f,g\in F[x,y]$. So $x=f(x,0)x^{2}$, and $1=f(x,0)x$. Similarly as before, this is a contradiction.