From Hartshorne's Algebraic Geometry, exercise 1.3:
Let $Y$ be the algebraic set in $\mathbb A^3$ defined by the two polynomials $x^2 - yz$ and $xz - x$. Show that $Y$ is a union of three irreducible components. Describe them and find their prime ideals.
I'm aware this problem has been looked at before, but my concern with it is a little different. Specifically, every solution I can see comes up with a different set of prime ideals from the ones I did, and I can't figure out exactly why the ideals I came up with don't work (assuming they don't, which is my intuition based on a vaguely-remembered idea of how the uniqueness of primary decomposition is supposed to work). I agree with the logic of the correct answers I've seen, but I can't figure out whether that means my logic was wrong, and if it does, where the issue is.
I'm inclined to start from the direction of algebra, so my reasoning went like this: $Y$ is the zero set of the two given polynomials by definition, so it's the zero set of the ideal generated by them, $(x^2 - yz, xz - x)$. The generating element $x^2-yz$ is irreducible and the other generating element $xz - x$ factors into the irreducibles $x(z - 1)$; since these three elements are irreducible in $k[x, y, z]$ and this ring is a UFD, the ideals generated by them are prime.
The product of coprime ideals is the same thing as their intersection, and clearly the product of any pair of these three ideals is coprime to the third one, which means that the product of these three prime ideals is equal to their intersection, and it seems like that should be exactly $(x^2 - yz, x, z - y) = (x^2 - yz, xz - x)$. (I say "seems like" because this is the best candidate I have for where my reasoning is incorrect). This ideal is also a radical ideal, since it's an intersection of prime ideals.
This means that $I(Y) = (x^2 - yz) \cap (x) \cap (z - 1) = I(Z(x^2 - yz) \cup Z(x) \cup Z(z - 1))$ where each of the three "Z(ideal)"'s is irreducible since the ideal is prime. Thus $\overline{Y} = Z(I(Y)) = Z(I(Z(x^2 - yz) \cup Z(x) \cup Z(z - 1))) = Z(x^2 - yz) \cup Z(x) \cup Z(z - 1)$, which should finish the problem since $Y$ is defined as an algebraic set, so it's already closed by definition.
I think the only confusion is that the ideal generated by two elements is not the product of the two ideals generated by each of the elements respectively. Thus the line where you write $I(Y)$ as the intersection of three ideals is wrong; intuitively, this corresponds to the fact, that $Y$ will be the intersection of the surfaces corresponding to the defining equations, thus $I(Y)$ should be bigger than the ideals corresponding to each of the defining equations, since more functions will vanish on $Y$ (in fact be generated by both of them, and not their product). Or in other words, you know that your solution in the end can not be right for dimension reasons: $Y$ will be the intersection of surfaces in $A^3$, whereas what you wrote is the union of three surfaces.