By David Cox, Andrew R. Kustin, Claudia Polini, Bernd Ulrich
Ponder a rational projective curve C of measure d over an algebraically closed box kk. There are n homogeneous kinds g1,...,gn of measure d in B=kk[x,y] which parameterise C in a birational, base aspect unfastened, demeanour. The authors examine the singularities of C by way of learning a Hilbert-Burch matrix f for the row vector [g1,...,gn]. within the ""General Lemma"" the authors use the generalised row beliefs of f to spot the singular issues on C, their multiplicities, the variety of branches at every one singular element, and the multiplicity of every department. allow p be a novel aspect at the parameterised planar curve C which corresponds to a generalised 0 of f. within the ""Triple Lemma"" the authors supply a matrix f' whose maximal minors parameterise the closure, in P2, of the blow-up at p of C in a neighbourhood of p. The authors follow the overall Lemma to f' in an effort to know about the singularities of C within the first neighbourhood of p. If C has even measure d=2c and the multiplicity of C at p is the same as c, then he applies the Triple Lemma back to profit concerning the singularities of C within the moment neighbourhood of p. give some thought to rational aircraft curves C of even measure d=2c. The authors classify curves in line with the configuration of multiplicity c singularities on or infinitely close to C. There are 7 attainable configurations of such singularities. They classify the Hilbert-Burch matrix which corresponds to every configuration. The examine of multiplicity c singularities on, or infinitely close to, a hard and fast rational aircraft curve C of measure 2c is such as the learn of the scheme of generalised zeros of the fastened balanced Hilbert-Burch matrix f for a parameterisation of C
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Extra info for A study of singularities on rational curves via syzygies
Assume that the entries of ϕ span a vector space of dimension 4 and ht I2 (ϕ) = 2. Then there exist invertible matrices χ and ξ over k so that χϕξ has one of the following forms: ϕ(∅,μ4 ) = Q1 Q2 Q2 Q3 Q3 Q4 , ϕ(c,μ4 ) = Q1 Q2 Q3 Q1 0 Q4 , ϕc,c = Q1 Q2 Q3 Q3 0 Q4 , or ϕc:c = Q1 Q3 Q2 Q4 0 Q2 , with Q1 , Q2 , Q3 , Q4 linearly independent. Proof. There are two possibilities for the original matrix ϕ. In Case 1, the entries in each column of ϕ span a vector space of dimension 2. In Case 2, the entries of at least one of the columns of ϕ span a vector space of dimension 3.
Proof. Fix ϕ ∈ BalHd . We ﬁrst prove that there exists g ∈ G with gϕ ∈ M Bal for some ∈ ECP. Consider the parameter μ = μ(I1 (ϕ)). The matrix ϕ has six homogeneous entries of degree c, so μ ≤ 6. On the other hand, the hypothesis that ht I2 (ϕ) = 2 guarantees that 2 ≤ μ. Thus, 2 ≤ μ ≤ 6. We treat each possible value for μ separately. If μ = 6, then the entries of ϕ are linearly independent and Bal . Suppose now that μ = 5. 7), then, after row and column operations, ϕ is transformed into gϕ ∈ M(c,μ .
THE BIPROJ LEMMA into one of the following four forms: ⎡ T1 M1 = ⎣T2 T3 ⎤ ∗ ∗⎦ , ∗ ⎡ T1 ⎢ T2 ⎢ M2 = ⎣ T3 0 ⎤ f1 f2 ⎥ ⎥, f3 ⎦ T1 ⎡ T1 ⎢ T2 ⎢ M3 = ⎢ ⎢ T3 ⎣0 0 ⎤ g1 g2 ⎥ ⎥ g3 ⎥ ⎥, T1 ⎦ T2 ⎡ T1 ⎢ T2 ⎢ ⎢ T3 M4 = ⎢ ⎢0 ⎢ ⎣0 0 ⎤ 0 0⎥ ⎥ 0⎥ ⎥, T1 ⎥ ⎥ T2 ⎦ T3 where the fi are linear forms in k [T2 , T3 ] and the gi are linear forms in R = k [T3 ]. In the case C = M1 , the ideal I2 (C) is a perfect ideal of height 2 with a 2-linear resolution; hence e(R/I2 (C)) = 3 = μ(I2 (C)). In the case C = M2 , we have I2 (C) = T1 (T1 , T2 , T3 )+JR, where J is a non-zero ideal of k [T2 , T3 ] generated by quadrics.