J= (a_{1}, …, a_{r})⊂ I.
Now suppose x∈ I. Write x=a/b with a, b∈ R and b∉ P. Since a = bx ∈ I∩ R, we have a∈ I^{-}, and so a/b∈ J. In particular, the same set of generators works.Θ_{X,x} = Hom_{k(x)}(m/m^{2}, k(x)).
D_{i}(F) = [∂ F / ∂ x_{i}](0).
Now let Θ = ∑_{i=1}^{n} t_{i} D_{i} be an arbitrary tangent vector. If F∈ M, thenΘ(F) = ∑_{i=1}^{n} t_{i} [∂ F / ∂ x_{i}](0).
After identifying the tangent space with A^{n}, the previous equation allows us to interpret F as a linear function on Θ, and hence as an element of the dual vector space Θ^{*} = Hom(Θ, k).Now consider the case of an algebraic set X∈A^{n}, with x=(0, …,0). Let L⊂A^{n} be a line through the origin, given parametrically as {ta : t∈ k} for some fixed point a distinct from the origin. If the ideal of X is generated by {f_{1},…,f_{r}}, then the intersection X∩ L is defined by
f_{1}(ta) = … = f_{r}(ta) = 0.
This is a system of polynomials in the single variable t (picking out a structured algebraic set on the line L≈A^{1}). So, we can replace the set of polynomials by its greatest common divisorf(t) = gcd(f_{1}(ta), …, f_{r}(ta)) = c∏(t-α_{i})^{ni}.
dF = ∑_{i=1}^{n} [∂ F / ∂ x_{i}](x) t_{i}.
It is easy to check thatd(F+G) = dF + dG
andd(FG) = F(x)d(G) + G(x)dF.
Now write I(X) = (F_{1}, …, F_{r}). Then the tangent space to X at x is defined by the equationsdF_{1} = … = dF_{r} = 0.
To see this, let M=(x_{1},…,x_{n})⊂ k[x_{1},…,x_{n}] and let m be the image of this ideal in A(X). Then there is an exact sequenceI(X) → M/M^{2} → m/m^{2} → 0.
By duality, the tangent space to X at the origin is the subspace to the tangent space to A^{n} at the origin consisting of those linear forms Θ=∑ t_{i}D_{i} that vanish on functions in I(X). However, the differential dF is just Θ(F) under the natural pairing; the claim follows.Now let L={ta : t∈ k } be a (parametric) line through the origin. Write
F_{j}(x) = ∑_{i=1}^{n} a_{i} x_{i} + h.o.t. = dF_{j} + G_{j}.
ThenF_{j}(at) = (∑_{i=1}^{n} a_{i}a)t + t^{2}*stuff.
So, t^{2} divides F_{j}(ta) if and only if dF_{j}(a) = 0. In other words, the line L touches X at 0 if and only if the point a defining the line lies in the tangent space to X at 0.dim(m_{x}/m_{x}^{2}) = n - rank(J_{X}(x)).
∑_{i=1}^{n} [∂ F / ∂ x_{i}](x) t_{i} = 0.
Thus, the tangent space is either A^{n-1} (if at least one coefficient is nonzero) or A^{n} (if all the coefficients xs vanish). One can show that if X is a variety, then there exists at least one point (and hence an entire open set) where the tangent space equals A^{n-1}.m/m^{2} ≈ M/M^{2} ≈ Θ^{*}_{X,x}
and Θ_{X,x} is isomorphic to the null space of the Jacobian. So,dim(m/m^{2}) + rank(J_{X}(x)) = n.
One also has A(X)⊂O_{X,x}⊂ k(X). Thus, all three rings have the same transcendence degree d=dim(A(X))=dim(O_{X,x}). So, O_{X,x} is regular if and only if d=dim(m/m^{2}) if and only if rank(J_{X}(x)) = n-d if and only if x is a nonsingular point on X.Next, we show that the set of singular points is proper. Since birationally isomorphic varieties have isomorphic open subsets, this property is birational. So, we can assume that X=Z(F) is an irreducible hypersurface in A^{n}. If the singular set is all of X, then one has an equality of ideals (F) = I(X) = (F,∂ F/∂ x_{1}, …, ∂ F/∂ x_{n}). But the partial derivatives have degree smaller than the degree of F. The equality of ideals can only hold if all the partials vanish. In characteristic 0, this can only happen if F is a constant. In characteristic p, it can only happen if F is a p-th power. Both cases are impossible, since X is an irreducible hypersurface.
…→ A/M^{i+1} → A/M^{i} → A/M^{i-1} → … A/M.
The completion will be denoted by I^{^}. The usual category theoretic notation is to writeI^{^} = lim_{←} A/M^{i};
this is an example of an inverse limit.∑_{i=0}^{@}[infty]@ a_{i} p^{i},
where a_{i} ∈ { 0,1,…,p-1}. Actual integers are identified with their expansions in base p. You add and multiply using the usual formulas; the only essential difference is that the sum or product of two "monomials" of the same degree may not be another monomial. (This occurs because a_{i}+b_{j} or a_{i}b_{j} can be bigger than p.)We would like to try to generalize these examples, in order to think of elements in completions as power series, but in as coordinate-free a way as possible. The basic tool form commutative algebra is the following.
Let w_{1},…,w_{s} be generators for N. If s=1, then
N = MN = M* Aw_{s} = M* w_{s}.
In particular, we can write w_{s} = mw_{s} for some element m∈ M. Hence, (1-m)w_{s}=0. Since 1-m ∉ M, it must be a unit in the local ring A. Therefore, w_{s}=0, and N=0.Now suppose s>1, and use induction. Write N'=N/Aw_{s}. Then N' is generated by s-1 elements, so N'=0. Thus, N=Aw_{s} is generated by one element, and we're done.
O^{^}_{P} = k[[x,y]]/(y^{2}-x^{2}-x^{3}) and O^{^}_{Q} = k[[x,y]]/(xy)
So, we have to find power series g, h∈ k[[x,y]] such thatg | = (y+x) + g_{2} + g_{3} + … |
h | = (y-x) + h_{2} + h_{3} + … |
gh | = y^{2} - x^{2} - x^{3}. |
gh = y^{2}-x^{2} +g_{2}(y-x)+h_{2}(y+x) + (degree≥3).
So,g_{2} (y-x) + h_{2}(y+x) = x^{3}.
We can take g_{2} = xy+[1 / 2]y^{2} and h_{2}=x^{2}-[1 / 2]y^{2}. Next, we need to solve(y-x)g_{3} + (y+x)h_{3} = -g_{2}h_{2}.
We can always find such a solution (because y-x and (y+x) generate the maximal ideal). Continue until you build the entire power series. In this example, it is interesting to note that the ring O_{P} is an integral domain, but its completion O^{^}_{P} is not.f = f_{r} + f_{r+1} + …,
where each f_{i} is a homogeneous polynomial of degree i. In this case, we say that P is an r-fold point of X. If f_{r} is a product of r distinct linear factors, then we say that P is ordinary. When r=1, the point P is nonsingular. Any two ordinary double points are analytically isomorphic. Any two ordinary triple points are analytically isomorphic. However, there is a one-parameter family of different ordinary fourfold points!Comments on this web site should be addressed to the author:
Kevin R. Coombes