Up till now, I’ve been using power series to parametrize branches:
x(t) = a0+a1t+a2t2+ ⋯, y(t) = b0+b1t+b2t2+ ⋯
If the branch passes through the origin, then a0=b0=0. In the last post, we established Facts 4 and 5, assuming that y(t)=t for all branches, so x(t) = a0+a1y+a2y2+ ⋯.
Filed under Algebraic Geometry
At last we come to Kendig’s proof of Bézout’s Theorem. Although not long, it will take me a few posts to appreciate it in full.
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The Resultant, Episode 5: Inside the Episode
The double-product form for the resultant:
implies Fact 3:
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The Resultant, Episode 5 (The Finale)
Recap: The setting is an integral domain R, with fraction field K, and extension field L of K in which E(x) and F(x) split completely. E(x) and F(x) have coefficients in R. E(x) has degree m, F(x) degree n; we assume m,n>0. The main special case for us: R=k[y], K=k(y), so R[x]=k[x,y], and E and F are polynomials in x and y. As always, we assume k is algebraically closed.
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The Resultant, Episode 4
This episode has one sole purpose: to show that the two formulas for the resultant are equivalent. The next episode, the finale, will tie up some loose ends.
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The Resultant, Episode 3: Inside the Episode
So we have, at long last, several expressions for the resultant:
Filed under Algebraic Geometry
The Resultant, Episode 3
Last time the linear operator
Φ: Kn[x]⊕Km[x] → Km+n[x]
Φ(p,q)=pE+qF
made its grand entrance, clothed in the Sylvester matrix. (Recall that Kn[x] is the vector space of all polynomials of degree <n with coefficients in K, likewise for Km[x] and Km+n[x].)
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The Resultant, Episode 2
By now you know the characters: the polynomials E(x) (degree m) and F(x) (degree n) with coefficients in an integral domain R, its fraction field K, and the extension field L of K in which E and F split completely:
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The Resultant, Episode 1: Inside the Episode
In Episode 1 of our miniseries, “The Resultant”, the characters were introduced: integral domain R with fraction field K and extension field L, and polynomials E(x) and F(x) in R[x], factoring completely in L as a(x–u1)···(x–um) and b(x–v1)···(x–vn). (Repeated roots allowed.) We had our first formulas for the resultant:
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The Resultant, Episode 1
Time to discuss the resultant; we’ll need it for Kendig’s proof of Bézout’s theorem, but it has other uses too. The story will take several episodes, plus extras. Like a miniseries!
Filed under Algebraic Geometry
Diagonal Argument 