Consider $K[X,Y,Z]$ over a field $K$ of characteristic $0$. I want to show that
- EDIT (thanks to user26857): This equality does not hold, for example $X^2Y \notin (XY+XZ+YZ,XYZ)$. I want to show that $(XY+XZ+YZ,XYZ) = (X,Y)(Y,Z)(X,Z)$ in order to conclude that the vanishing sets of the left hand side ideal is equal to the union of the three ideals on the right hand side,
- the vanishing sets of $(X,Y),(Y,Z)$ and $(X,Z)$ are irreducible.
Concerning Part 1 EDIT (thanks to user26857): This equality does not hold, for example $X^2Y \notin (XY+XZ+YZ,XYZ)$. I have shown explicitely that $$V_{(XY+XZ+YZ,XYZ)}(K) = V_{(X,Z)} \cup V_{(X,Y)} \cup V_{(Y,Z)} $$ by solving the systems of equations (for example the vanishing set of $(X,Y)$ is the set of points $(0,0,z)$).
However, I wanted to try and show it using the property that the vanishing set of a product of ideals equals the union of the vanishing sets of the respective ideals.
An element in $(XY+XZ+YZ,XYZ)$ is of the form $f(XY+XZ+YZ)+gXYZ$ where $f,g$ are polynomials in $K[X,Y,Z]$. Now $gXYZ$ is in $(X,Y)(Y,Z)(X,Z)$, since, for instance, $gX\in (X,Y)$, $Y\in (Y,Z)$ and $Z \in (X,Z)$. I have problems for the term f(XY+XZ+YZ), because there are only two factors in each summand and seeing $XY$ as $XY1$ does not help because $1$ is not in any of the ideals. I also tried rewriting the expression as
$$(fY+fZ+gYZ)X + fYZ\text{ or }(fY+fZ)X + (fY+gXY)Z$$ but the main problem is that I do not see how any of the forms corresponds to an element in the product ideal $(X,Y)(Y,Z)(X,Z)$. I hope that my definition of the product of ideals $a,b,c$ is correct:
$$abc= \{\sum_{j=1}^m (\sum_{i=1}^n a_ib_i) c_j : a_i\in a, b_i\in b, c_i\in c, n,m\in \mathbb{N}\},$$ which I derived from the definition $$ab=\{\sum_{i=1}^n a_ib_i:a_i\in a, b_i\in b, n\in \mathbb{N}\}.$$
Concerning Part 2 Take for example $V_{(X,Z)} = \{(0,y,0)\}$.
- My first idea was to use the fact that $$V_{(X,Z)}(K) \subseteq V_{XYZ}(K).$$ The polynomial $XYZ$ is irreducible, which implies that the coordinate ring of $V_{XYZ}(K)$ is an integral domain. So $V_{(X,Z)}(K)$ is a subset of an irreducible set. This does not allow me to conclude, because as far as I know one can only say that open subsets of irreducible sets are irreducible. Now $V_{(X,Z)}(K)$ is closed and I would guess that it is not open (I have no argument for this though).
- As a second attempt, I tried to prove that the vanishing ideal of $V_{(X,Z)}(K)$ is prime, which is equivalent to the set being irreducible. First of all, what is the vanishing ideal of $V_{(X,Z)}(K)$? Is it $(X,Z)$? The inclusion $(X,Z) \subseteq I_{V_{(X,Z)}(K)}$ is obvious and for the other inclusion I argue that any polynomial in $K[X,Y,Z]$ that vanishes for any $(0,y,0)$ must have a zero constant term and in every term there must be at least an $X$ or $Z$, so that we can factor out $X$ and $Z$ to form a polynomial in $(X,Z)$. As for the primeness, I do not have a rigorous argument, however, I think that the ideal should be prime, since the product of two polynomial that are not in the ideal can never lie in the ideal, for there will be a constant term or term without an $X$ or $Y$ in it that will mess things up.