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Let $f,g$ be polynomials in $k[y_1,\dots,y_n][x]$ over a field $k$. Assume that at least one of $f$ and $g$ is of positive degree in $x$. Denote by $\operatorname{res}_x(f,g)$ the resultant of $f$ and $g$ with respect to $x$.

On the Wikipedia page for Resultants, under Elimination properties it is stated that if $R$ is a ring of polynomials, $f,g\in R[x]$ at least one of positive degree in $x$ and $I=(f,g)$, then

  1. $I\cap R$ is a principal ideal of $R$, generated by some $r\in R$
  2. $\operatorname{res}_x(f,g)$ is in the principal ideal $(r)$ of $R$
  3. There exists a positive integer $k$ such that $r^k$ is in the principal ideal $(\operatorname{res}_x(f,g))$ of $R$.

I am kindly asking for a reference on all three claims. I was not able to find any mention of these results in the refereces listed on the Wikipedia page (Gelfand Kapranov Zelevinski; Cox Little O'Shea; Macaulay; Salmon).

Randy Marsh
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  • These are not all correct as written. For instance, if $n=2,f=y_1,g=y_2$, then $I\cap R =(y_1,y_2)$ which is not principal, contradicting statement 1. Further, $res_x(f,g)=1$, so statement 2 is wrong as well even after replacing $(r)$ with $I\cap R$. Would you be interested in non-reference answers which prove "fixed" versions of these, or are you only looking for references? – KReiser Aug 21 '20 at 08:28
  • @KReiser, $f$ and $g$ are assumed to have positive degree in $x$, otherwise there's nothing to eliminate. The assumption isn't stated explicitly but is clear from the context of elimination. I'll edit the question. – Randy Marsh Aug 21 '20 at 08:32
  • Thanks for the clarification (perhaps wikipedia should get an edit too). You didn't say anything about the other portion of the comment, though - are you interested in answers which aren't references but prove these statements? – KReiser Aug 21 '20 at 08:35
  • @KReiser Sorry, yes, I'd be interested in answers proving these statements under the assumption that $f$ and $g$ have positive degree in $x$. – Randy Marsh Aug 21 '20 at 08:37

1 Answers1

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Summary: There are problems. (1) seems quite challenging to get a handle on in the case $n>1$, (2) is true modulo the statement of (1) (you can say that the resultant is in $I\cap R$ with not much trouble via some results from Cox, Little and O'Shea), and (3) is missing conditions even in the case $n=1$. I suspect that the statement "... $R[x]$, where $R$ is itself a polynomial ring over a field" on the wikipedia page may be intended to capture $R=k[y]$, or $n=1$ as in your post, but it's not clear to me and I would not feel comfortable reading it as that.

I'll do my bet to provide proofs of some correct statements that resemble those in your post and point out counterexamples to the false portions. I'll be using some of the material from chapter 3 of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms (abbreviated CLO for convenience).

First we'll get a handle on the geometry of $V(I\cap R)$ and $V(res_x(f_1,f_2))$.

Denote $\pi:\Bbb A^{n+1}_k\to\Bbb A^n_k$ the projection, and for any ideal $I=(f_1,\cdots,f_m)\subset k[x,y_1,\cdots,y_n]$, let $I_1$ be the ideal $I\cap k[y_1,\cdots,y_n]$. Write $f_i=l_i(y_1,\cdots,y_n)x^{n_i}+\cdots$, where $n_i$ is the highest power of $x$ appearing in $f_i$. Then by theorem 3.2.2 of CLO, we have that $$V(I_1)=\pi(V(I)) \cup (V(l_1,\cdots,l_m)\cap V(I_1))$$ and by theorem 3 of the same section, we have $$V(I_1)=\overline{\pi(V(I))}.$$

Now I claim that $V(res_x(f_1,f_2))=V(l_1,l_2)\cup \pi(V(I))$. We'll do this in three steps:

  • First, one can see from the resultant as the determinant of the Sylvester matrix that if $l_1,l_2$ both vanish, then the matrix has a row of zeros and thus $res_x(f_1,f_2)=0$ and so $V(l_1,l_2)\subset V(res_x(f_1,f_2))$.
  • Next, if at least one of $f_1,f_2$ is positive-degree in $x$, we have that $\pi(V(I))\subset V(res_x(f_1,f_2))$: if $f_1,f_2$ are both of positive degree with respect to $x$, then we can write $res_x(f_1,f_2)=Af_1+Bf_2$ for two polynomials $A,B$ by CLO proposition 3.5.9. If only one of $f_1,f_2$ is of positive degree with respect to $x$, then the resultant is a power of the other one. Either way, $res_x(f_1,f_2)\in I_1$ and so $V(res_x(f_1,f_2))\supset V(I_1) \supset \pi(V(I))$. (Note that this proves $res_x(f_1,f_2)\in R\cap I$, which immediately implies item 2 if we know $R\cap I$ is principal.)
  • Finally, $V(res_x(f_1,f_2))\setminus V(l_1,l_2)\subset \pi(V(I))$. Pick $c\in V(res_x(f_1,f_2))\setminus V(l_1,l_2)$. If $l_1(c),l_2(c)\neq 0$, then $res_x(f_1,f_2)(c)=0$ is equivalent to $res_x(f_1(x,c),f_2(x,c))=0$. If $l_1(c)\neq 0$ but $l_2(c)=0$, then $res_x(f_1,f_2)(c)=l_1(c)^{\cdots}res_x(f_1(x,c),f_2(x,c))$ by CLO proposition 3.6.3, and so again $res_x(f_1,f_2)(c)=0$ iff $res_x(f_1(x,c),f_2(x,c))=0$. On the other hand, $c\in \pi(V(I))$ iff there's a $c_1$ so that $f_1(c_1,c)=f_2(c_1,c)=0$, which is equivalent to $res_x(f_1(x,c),f_2(x,c))=0$.

These three statements give $V(res_x(f_1,f_2))=V(l_1,l_2)\cup \pi(V(I))$ and we're done.

We note that as $V(res_x(f_1,f_2))$ is a closed set containing $\pi(V(I))$, it must contain its closure, which by our earlier work is $V(I_1)$. If $V(I_1)\supset V(l_1,l_2)$ (this is the case when $l_1$ and $l_2$ have no common factors by a few applications of Krull's Height Theorem and a dimension argument, for instance), then we have item 3 under the additional assumption that $I_1$ is principal: $V(I_1)\supset V(l_1,l_2)$ means $V(res_x(f_1,f_2))=V(I_1)$, which means the radicals of the ideals $(res_x(f_1,f_2))$ and $I_1$ are equal. So $r^k\in (res_x(f_1,f_2))$ for some $k$.

Thus under the conditions in your post and in the presence of item 1, we've proven item 2, and under the additional assumption that $V(l_1,l_2)\subset V(I_1)$ we've proven item 3. If it's not the case that $V(l_1,l_2)\subset V(I_1)$, item 3 could fail: even if $V(I_1)$ is principal, if $V(l_1,l_2)$ is codimension 1 in $\Bbb A^n$ and not a subset of $V(I_1)$, then $res_x(f_1,f_2)$ vanishes on it but $r$ doesn't. This happens when $n=1$, $f_1=yx$ and $f_2=yx^2+y-1$, for instance: $res_x(f_1,f_2)=y^2(y-1)$, while $I_1=(y-1)$ and no power of $y-1$ is in the ideal generated by $res_x(f_1,f_2)=y^2(y-1)$.

Determining when $I_1$ is principal seems to be rather difficult, and I don't have a full characterization to offer you (nor have I found anything in any of the references I consulted on this). For instance, consider $f_1=z^2+xyz$ and $f_2=yz$. We have $(f_1,f_2)\cap R = (yz,z^2)$ which is nonprincipal, though I imagine there are other examples where the coefficients of the highest powers of $x$ don't have any common factor. One potential (silly) answer is that if $n=1$, that is, we're working in $k[x,y]$, then $I\cap k[y]$ is principal because $k[y]$ is a PID, and perhaps this is what the text "... $R[x]$, where $R$ is itself a polynomial ring over a field" means. If "polynomial ring over a field" means any ring of the form $k[x_1,\cdots,x_n]$, then I think this question is rather difficult.

KReiser
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