Is $(\overline{x}t_1 - \overline{z}t_0, \overline{w}t_1 - \overline{y}t_0)$ a prime ideal in $\dfrac{k[x,y,z,w]}{(xy-zw)} [t_0, t_1]$?

Question

Consider the affine coordinate ring $A:= \dfrac{k[x,y,z,w]}{(xy-zw)}$. We write $A = k[\overline{x}, \overline{y}, \overline{z}, \overline{w}]$. Consider the polynomial ring $A[t_0, t_1]$ in two variables over $A$.

My question is the following:

\begin{equation*} \text{Is $(\overline{x}t_1 - \overline{z}t_0)$ a prime ideal in $A[t_0,t_1]$}? \end{equation*}

My approach has been to consider the ring homomorphism $\phi: A[t_0, t_1]\longrightarrow A[t]$ which sends $t_0\longmapsto \overline{x}t$ and $t_1\longmapsto \overline{z}t$. Then $\ker\phi$ is a homogeneous prime ideal of height one. Clearly $\overline{x}t_1-\overline{z}t_0 \in \ker\phi$. It is then sufficient to show that it generates the entire kernel, but I can not proceed further.

I have considered the following more general question: \begin{equation*} \text{Is $(at-b)$ a prime ideal of $R[t]$, where $R$ is a Noetherian domain, $a\in R$ is irreducible and $b\notin aR$?} \end{equation*} Note that we do not assume $R$ to be a UFD, hence we can not apply Gauss' Lemma. I have found the following result (Show that $A[X]/(aX+b)$ is an integral domain) in this direction. Unfortunately it does not answer our question, since in our setting we have the following conditions over the ring $A[t_0]$: \begin{equation*} \overline{x}\overline{y}t_0\in \overline{x}A[t_0]\cap \overline{z}A[t_0] \text{ but } \overline{x}\overline{y}t_0 \notin \overline{x}\overline{z}A[t_0]. \end{equation*} In case the answer to the general statement is negative, does it change if we impose the additional condition that $R$ is a finitely generated $k$-algebra? How about if we impose the condition that $R$ is integrally closed?

UPDATE: In light of Walkar's excellent insights, I now have the following modified question:

\begin{equation*} \text{Is $(\overline{x}t_1 - \overline{z}t_0, \overline{w}t_1 - \overline{y}t_0)$ a prime ideal in $A[t_0,t_1]$}? \end{equation*} We still consider the ring homomorphism $\phi:A[t_0,t_1]\longrightarrow A[t]$ as mentioned above. It is then sufficient to show that $\ker\phi = (\overline{x}t_1 - \overline{z}t_0, \overline{w}t_1 - \overline{y}t_0)$.

Any help is greatly appreciated!

Answer 1

Notice that $A[t_0,t_1]/(xt_1-zt_0)$ is isomorphic to $$R=k[x,y,z,w,t_0,t_1]/(xy-zw,xt_1-zt_0),$$ so it suffices to ask if this ring is an integral domain. However, this is not the case; forgoing the bars, we see: $$x(yt_0-wt_1) = xyt_0-xwt_1 = zwt_0 - wzt_0 = 0.$$

To finish the conclusion, we simply need to see $x$ and $yt_0-wt_1$ are not in the ideal $$I=(xy-zw,xt_1-zt_0).$$ We will do this by producing a Groebner basis for $I$, which is a specific kind of generating set of $I$ for which ideal membership can be checked easily. I am following the conventions in Chapter 2 of Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea.

We order the monomials in $k[x,y,z,w,t_0,t_1]$ with $x>y>z>w>t_0>t_1$ and the lexicographic ordering, e.g. $yz^2w > yzw^2$ since the exponent vectors of the two monomials $(0,1,2,1,0,0)>(0,1,1,2,0,0)$ under the lexicographic ordering.

We start with the generating set $F=\{xy-zw,xt_1-zt_0\} = \{f_1,f_2\}$. We then form the reduced $S$-polynomial $f_3=\overline{S(f_1,f_2)}^F = yzt_0 - zwt_1$ by using the multivariate division algorithm, see 3.6.3 and 3.6.4 of [CLO]. This now updates our generating set to $F=\{f_1,f_2,f_3\}$. We then check that the reduced $S$-polynomials $\overline{S(f_i,f_j)}^F = 0$ for all $i,j$. Thus $F$ is a Groebner basis for $I$.

The defining feature of a Groebner basis is that $g \in I$ if and only if $\overline{g}^F = 0$, that is, if and only if $g$ has remainder $0$ upon application of the multivariate division algorithm for the polynomials in $F$. Letting $g_1 = x$ and $g_2 = yt_0-wt_1$, we compute the remainders: $$\overline{g_1}^F = x \,\,\text{ and }\,\, \overline{g_2}^F=yt_0-wt_1.$$ Since neither remainder is $0$, neither of $x$ nor $yt_0-wt_1$ are in $I$. This shows that $R$ is not an integral domain.

Macaulay2 is a computer algebra system which works with Groebner bases as one of its primary tools. With the code below, you can replicate the details of the story told above, with some additional code as to how I found the example, skipping the output lines that are less relevant. I also worked over the field $\mathbb{Q}$, but I do not think the underlying field should affect the calculations at all.

i1: k=QQ
i2: S=k[x,y,z,w,t_0,t_1]
i3: I=ideal(x*y-z*w,x*t_1-z*t_0)
i4: associatedPrimes I
o4: {ideal(z,x),ideal(z*t_0-x*t_1,x*t_0-w*t_1,x*y-z*w)}

So, $(x,z)$ is an associated prime; maybe we should try something killed by $x$ or $z$ since associated primes of $I$ are exactly annihilators of elements of the ring $R/I$.

i5: R=S/I
i6: annihilator ideal(x)
o6: ideal(y*t_0-w*t_1)

One can then check that $x$ and $yt_0-wt_1$ are nonzero in $R$, for an element $r$ and ideal $I$, the code $r\% I$ computes the remainder upon division by a Groebner basis of $I$.

i7: use S
i8: x%I
o8: x
i9: h=y*t_0-w*t_1
i10: h%I
o10: y*t_0-w*t_1

Since o9 and o10 were not $0$, $x$ and $h$ are not in $I$. You can try this out yourself by going to Macaulay2Web and typing in this or another calculation.

A word of caution if you are new to Macaulay2; it is very proficient at things like Groebner bases, but for other non-core functions, it can be less than 100% trustworthy. I always attempt to verify any calculation Macaulay2 does for me by hand or another source.