I was asked the following question by a colleague and was embarrassed not to know the answer.

Let $f(x), g(x) \in \mathbb{Z}[x]$ with no root in common. Let $I = (f(x),g(x))\cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which are linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. Then $I$ is clearly an ideal in $\mathbb{Z}$. Let $D>0$ be a generator of this ideal. The question is: what is $D$?

Now, we do have the standard resultant $R$ of $f,g$, which under our hypotheses, is a non-zero integer. We know that $R \in I$ and it's not hard to show that a prime divides $R$ if and only if it divides $D$. I thought $R = \pm D$ but examples show that this is not the case.


This quantity $D$ is known as the "congruence number" or "reduced resultant" of the polynomials f and g. I first saw this in a preprint by Wiese and Taixes i Ventosa, http://arxiv.org/abs/0909.2724. They ascribe the concept to a paper which I don't have a copy of:

M. Pohst. A note on index divisors. In Computational number theory (Debrecen, 1989), 173– 182, de Gruyter, Berlin, 1991.

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    $\begingroup$ I wrote a paper on this, On resultants, Proc Amer Math Soc 89 (1983) 419-420, MR84j:13004. The "norm = resultant" thing that Felipe and Kevin discuss elsewhere in this thread has been independently rediscovered many times. I talk about this a bit and give a generalization to several variables in Norms in polynomial rings, Bull Austral Math Soc 41 (1990) 381-386, MR91m:13011. $\endgroup$ – Gerry Myerson Mar 8 '10 at 22:58

This difference was well-known in the 19th century when people a) Knew about invariants, and b) Calculated by hand. I believe a lot of the confusion today stems from Lang's Algebra book which is at best misleading about how to interpret what the resultant and discriminant are (and the ideas of famous books, right or wrong, tend to be perpetuated in other people's books!).

As an example, the resultant of the two polynomials $3x+1$ and $3x+2$ is, according to Sylvester's matrix definition, equal to $3$. Here Voloch's $D=1$. Surely this makes no sense according to the well-known theory, that a prime $p$ dividing the resultant of two polynomials should be interpreted to mean that these polynomials share a root when reduced mod $p$? This is evidently nonsense in this example ... unless one re-interprets these polynomials projectively (which is what one should do).

But now if we look at the pair of polynomials $y+2$ and $3y+2$ then the resultant is $4$, and here $D=4$, but how do you interpret here $2^2$ dividing the resultant? It is not immediate from the interpretation of modern algebra books!

There are all sorts of reasons that prime powers can divide a resultant (and discriminant) and it is complicated to understand all the cases when you wish to interpret higher power divisibility.

In Bhargava's work, he needs to understand squarefree values of a multivariable polynomial which is the the value of a discriminant of a class of parametrized polynomials. In other words he needs to parametrize when $p^2$ divides terms in this particular class of discriminants. Even this relatively simple request breaks down into several non-trivial cases, which he handles so beautifully as if to make it look trivial, but it's not.

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    $\begingroup$ A prime $p$ divides the resultant if they have a common root mod $p$ or if both of their degrees strictly decrease by reduction mod $p$. In some sense, the resultant behaves better if it is understood as an invariant of pairs of homogeneous polynomials (of given degrees). So there is a good reason that the resultant of $3x+1$ and $3x+2$ vanishes mod $3$. $\endgroup$ – ACL Apr 12 '15 at 12:08
  • $\begingroup$ Regarding your argument at the second paragraph, you have to take into account that: a prime $p$ dividing the resultant of two polynomials $f(x), g(x)$, is a necessary but not sufficient condition so that these polynomials share a root when reduced mod $p$ or that in other words: for some integer $n$ to exist, such that $p | \gcd(f(n),g(n))$. $\endgroup$ – Konstantinos Kanakoglou Jul 31 '16 at 22:41

An interesting example where the resultant and the "reduced resultant" differ comes from the theory of elliptic curves. Take an elliptic curve $$E:\qquad y^2=x^3+ax+b$$ where $a$ and $b$ are integers. The duplication formula for $E$ states that on the elliptic curve $[2] (x_1,y_1)=(x_2,y_2)$ where $x_2=g(x_1)/4f(x_1)$, $$f(x)=x^3+ax+b\qquad{\rm and}\qquad g(x)=x^4-2ax^2-8bx+a^2.$$ The resultant of $f$ and $g$ is $(4a^3+27b^2)^2$ (the square of the discriminant of $E$ so nonzero). But the reduced resultant is $|4a^3+27b^2|$ which one sees by noting that this divides all entries of the adjugate of the "resultant matrix" of $f$ and $g$.

The fact that this resultant is nonzero is used in a standard proof of the inequality $$h([2] P)\ge 4 h(P)-O(1)$$ for the naive height on $E$ (see Silverman's book).

  • $\begingroup$ Robin: I noticed this when preparing some 4th year UG lectures on elliptic curves a few years ago, and emailed Cassels asking him why this phenomenon was occurring (it was he who had taught me this trick). Cassels replied that he was "out of the loop" as far as mathematics was concerned and I should ask someone else, and I never did, and indeed I never got to the bottom of this until now! See footnote 20 in his elliptic curves book, p80. $\endgroup$ – Kevin Buzzard Mar 11 '10 at 14:39
  • $\begingroup$ Kevin: I only noticed this recently when pondering Miles Reid's comments included in the errata for the second edition of Silverman's Arithmetic of Elliptic Curves. $\endgroup$ – Robin Chapman Mar 11 '10 at 15:00

$D=D(f,g)$ is not particularly well-behaved, is it. For example it's not multiplicative in the variables: if $g=x^2+1$ (nothing special about this example, I'm just fixing ideas) then $D(f,g)$ is the intersection of $\mathbf{Z}$ and the ideal generated by $f(i)$ in $\mathbf{Z}[i]$. So, for example, if $f$ is $x+2$ you get 5, if $f$ is $x-2$ you get 5 too (from the other prime ideal above 5) and if $f$ is $x^2-4$, the product, you still only get 5 (which of course provides a proof that $D$ isn't the resultant in general).

Seems to me that the norm of $f(i)$ would be a much better invariant, which would generalise to the size of the ring $\mathbf{Z}[x]/(f,g)$. It wouldn't surprise me if that were the resultant (or closely related to it), and if it is then that's at least some sort of link. Maybe you knew all that already though ;-)

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    $\begingroup$ The norm is the resultant. This is one of those things which is "well-known" but not really. $\endgroup$ – Felipe Voloch Mar 8 '10 at 20:09
  • $\begingroup$ Agreed! Had I wanted to follow it up I would have found a proof for myself and then just claimed it was "well-known" ;-) $\endgroup$ – Kevin Buzzard Mar 8 '10 at 20:52

This issue came to my attention for the first (and only, up until now) time when Gregory Dresden gave a talk about resultants of cyclotomic polynomials in the UGA number theory seminar last spring. I am pretty sure that the distinction between these two notions of the resultant figured prominently in his work: see


So far as I know the question you ask is unsolved in general, but in this case I don't know very far at all. You might want to ask Greg...


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