Let $\left<x^2,y^2\right>$ be the ideal generated over the ring $k[x,y]$, where $k$ is a field. Intuitively, I know that $x$ cannot be in this ideal, because if it were, then there would be some $p(x,y), q(x,y)$ with
$$x=x^2p(x,y)+y^2q(x,y).$$
Neither of the summands on the right hand side have any terms of degree $<2$, so it seems obviously to me that there are no $p,q$ that are going to work. Can someone help me do this in a more rigorous way?
If you want to add more to this argument, talk in detail about why that can't happen. When in doubt, this is often a good move.
The smallest non-zero term of $x^2p(x)$ has degree at least $2$. The same is true for $y^2q(x)$. But when you add polynomials you do it component wise, and so the coefficient of $x$ in the sum (known to be $1$) is equal to the sum of the corresponding coefficients in the summand (both known to be $0$). But $0+0=0\neq 1$