2

Let $k$ be a field and $k[x]$ be the ring of polynomials over $k$. Given two polynomis $m_1(x), m_2(x) \in k[x]$, I want to know the relationship between the resultants and the least linear combination.

(In Euclidean rings, the least linear combination is the greatest common divisor (GCD).)

R.P.
  • 4,745
worm
  • 29
  • 1
    What kind of relationship could there possibly be? The resultant is a scalar, not a polynomial. – abx Aug 03 '20 at 12:49
  • If we simplify the sylvester matrix of m1(x) and m2(x) to a row echelon form, we can transform the non-zero row at the bottom of the matrix into polynomial form which is exactly the least linear combination. – worm Aug 03 '20 at 13:00
  • 3
    If the resultant is $0$, then the GCD has positive degree; if the resultant is nonzero, then the GCD is $1$. I recall some more interesting questions about this where $k$ is a ring, but there isn't anything more to say about a field. – David E Speyer Aug 03 '20 at 13:31
  • Let $n\in\mathbb{N}^*$ be a composite number. So $\mathbb{Z}_n[x]$ is not an integral domain. Through the experimental results, the above reply is still valid in $\mathbb{Z}_n[x]$ . – worm Aug 03 '20 at 13:38
  • 1
    Could you please tell me some more interesting questions about this where $k$ is a ring? – worm Aug 03 '20 at 13:41
  • 2
    Related questions, over rings: https://mathoverflow.net/questions/227227/reduced-resultant-of-monic-polynomials and https://mathoverflow.net/questions/17501/the-resultant-and-the-ideal-generated-by-two-polynomials-in-mathbbzx and https://mathoverflow.net/questions/248488/reduced-resultants-and-bezouts-identity and https://mathoverflow.net/questions/248574/ideal-generated-by-two-univariate-coprime-integer-polynomials and probably others. Search reduced resultant site:mathoverflow.net and related questions in the results you get. – Gerry Myerson Aug 04 '20 at 10:22
  • I was remembering the question https://mathoverflow.net/questions/17501 and its many interesting answers. – David E Speyer Aug 04 '20 at 13:23
  • Have you had a look, worm, at those links I gave? – Gerry Myerson Aug 06 '20 at 03:23
  • Thank you very much. I have solved my problem well according to your suggestion. – worm Aug 11 '20 at 03:25

1 Answers1

3

The important notion here is that of subresultant. Suppose $P_1, P_2$ are two polynomials of degrees $d_1,d_2$ and suppose $d_1\ge d_2$. To compute the GCD you would typically use Euclid's division algorithm: you divide $P_1$ by $P_2$ and get a remainder $P_3$ then you divide $P_2$ by $P_3$ and get the remainder $P_4$ etc. The last nonzero remainder is the GCD. Now imagine doing that for generic polynomials $$ P_1(x)=a_{1,d_1}x^{d_1}+a_{1,d_1-1}x^{d_1-1}+\cdots+a_{1,1}x+a_{1,0} $$ and $$ P_2(x)=a_{2,d_2}x^{d_2}+a_{2,d_2-1}x^{d_2-1}+\cdots+a_{2,1}x+a_{2,0}\ . $$ The very first step would be to subtract $\frac{a_{1,d_1}}{a_{2,d_2}}x^{d_1-d_2}P_2(x)$ from $P_1(x)$. Don't do that. Instead multiply $P_1$ by $a_{2,d_2}$ and then subtract $a_{1,d_1}x^{d_1-d_2}P_2(x)$ so as not to produce fractions. Rince and repeat. Generically the degree of the remainder only drops by one. The resultant essentially is the degree zero remainder, i.e., $P_{d_2+2}$. The previous $P$'s are the subresultants (up to one's choice of normalization convention, there may also be extraneous factors to peel off).

A good reference on the subject is the book "Algorithms in Real Algebraic Geometry" by Basu, Pollack and Roy.

Abdelmalek Abdesselam
  • 21,801