How to compute the intersection of ideals $(w,y,xz+x+z), (w,x,yz+y+z), (z,y,xw+x+w), (z,x,yw+y+w), (x+y,zw+z+w,y^2)$ over $\mathbb{F}_p$?

Question

Let $p$ be any prime number in $\mathbb{Z}$, and consider the ideals $(w,y,xz+x+z)$, $(w,x,yz+y+z)$, $(z,y,xw+x+w)$, $(z,x,yw+y+w)$, $(x+y,zw+z+w,y^2)$ in $\mathbb{F}_p[x,y,z,w]$.

My question is, how to compute the intersection of these ideals?

I may intoduce the backgroud for these ideals. As ideals in $\mathbb{Q}[x,z,y,w]$, they are the primary decomposition of ideal $(xy,xz+yz+xw+yw+zw+x+y+z+w,xzw+yzw)$. And by using the answer of @Michael Burr

(see Is the inverse of $(x+y, zw+z+w, y^2)\subset Q[x,y,z,w]$ under inclusion $g: Z[x,y,z,w]\to Q[x,y,zw]$ in $Z[x,y,z,w]$ still be $(x+y, zw+z+w, y^2)$?)

this intersection equality still holds in $\mathbb{Z}[x,y,z,w]$. Now my question is turn to whether this equality holds in $\mathbb{F}_p[x,y,z,w]$?

Answer 1

With $R := \mathbb{Z}[x,y,z,w]$, let $M$ be the submodule of $R^5$ spanned by the rows of: $$\begin{bmatrix} w&w&w&w&w\\ y&y&y&y&y\\ xz+x+z&xz+x+z&xz+x+z&xz+x+z&xz+x+z\\ w&0&0&0&0\\ x&0&0&0&0\\ yz+y+z&0&0&0&0\\ 0&z&0&0&0\\ 0&y&0&0&0\\ 0&xw+x+w&0&0&0\\ 0&0&z&0&0\\ 0&0&x&0&0\\ 0&0&yw+y+w&0&0\\ 0&0&0&x+y&0\\ 0&0&0&zw+z+w&0\\ 0&0&0&y^2&0 \end{bmatrix}. $$ We now consider the intersection with the submodule $0\oplus 0\oplus 0\oplus 0\oplus R$. Pretty much by construction, the last component must be in the ideal $\langle w, y, xz+x+z \rangle$; namely, if the scalars used are $c_1,\ldots,c_{15}$ then the last component is $c_1 w + c_2 y + c_3(xz+x+z)$. Now, to get the first component to be zero, we must have $c_1 w + c_2 y + c_3(xz+x+z) = -(c_4 w + c_5 x + c_6(yz+y+z)) \in \langle w, x, yz+y+z \rangle$ also. Similarly, the second component being zero gives $c_1 w + c_2 y + c_3(xz+x+z) \in \langle z, y, xw+x+w \rangle$, and so on. Therefore, any last component of such a vector is in the desired intersection of ideals. It is straightforward to see that the converse is also true.

Thus, if we calculate a Groebner basis of $M$, with an order eliminating towards the last component, then the elements of this Groebner basis whose first four elements are zero will have last elements give a Groebner basis of the desired intersection.

However, in the calculation I did using Singular, what I see is that in fact, all the leading coefficients of the overall Groebner basis of $M$ are equal to 1. Also, all the calculations done over $\mathbb{Z}[x,y,z,w]$ will translate straightforwardly to valid calculations over $\mathbb{F}_p[x,y,z,w]$ that show the given elements span the corresponding module. Furthermore, the fact of all the leading coefficients being 1 means that the Buchberger criterion being satisfied over $\mathbb{Z}[x,y,z,w]$ will translate to the Buchberger criterion being satisfied over $\mathbb{F}_p[x,y,z,w]$.

(Note that in a more general situation, the same argument should work for any prime $p$ which does not divide any of the leading coefficients of the Groebner basis of the corresponding version of $M$.)

So, the conclusion is: yes, the given intersection is equal to the desired ideal also over $\mathbb{F}_p[x,y,z,w]$.

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Below, see the Singular input I used to verify the calculations.

ring R=ZZ,(x,y,z,w),(c,lp);
module M=[w,w,w,w,w],[y,y,y,y,y],[xz+x+z,xz+x+z,xz+x+z,xz+x+z,xz+x+z],
[w,0,0,0,0],[x,0,0,0,0],[yz+y+z,0,0,0,0],[0,z,0,0,0],[0,y,0,0,0],
[0,xw+x+w,0,0,0],[0,0,z,0,0],[0,0,x,0,0],[0,0,yw+y+w,0,0],
[0,0,0,x+y,0],[0,0,0,zw+z+w,0],[0,0,0,y2,0];
groebner(M);
ideal I=xy,xz+yz+xw+yw+zw+x+y+z+w,xzw+yzw;
groebner(I);

(Then compare the first six entries of the output of groebner(M) to the output of groebner(I).)