2

So, I was reading section 9.6 in Dummit and Foote, and I got a bit confused by the proof below, particularly the part where they claim $f_{i-1}'-f_i'=S(f_{i-1}, f_i)$. This is on page 323.

Dummit_Foote

For example, let $f_1=x^3y$ and let $f_2=x^2y^2$. Then $S(f_1, f_2)=yf_1-xf_2=0$. However, $f_1'=f_1$, and $f_2'=f_2$, since both polynomials are already monic, but $f_1'-f_2'=f_1-f_2\neq 0$. What am I missing here?

Nishant
  • 9,155
  • I don't have Dummit and Foote near me, but I am fairly sure that your two polynomials don't have the same multidegree. (Also, please cite your texts in such a way that the post still makes sense when the externally hosted images expire -- so, say something to the extent of "Lemma 25 in section 9.6 of Dummit and Foote, $x$-th edition" (fill in the $x$ please). Thank you.) – darij grinberg May 15 '14 at 19:41
  • 1
    Ah, that's it. I thought multidegree meant the sum of the degrees of the each variables, when it actually means the ordered tuple of degrees of each variable. Thanks! – Nishant May 15 '14 at 19:44
  • What you meant is called total degree. – darij grinberg May 15 '14 at 19:44

0 Answers0