3 Tangent spaces, dimension and singulari- ties

n Let H ⊂ A (K) be a hypersurface deﬁned by the equation f(x1, x2, . . . , xn) = 0, where f ∈ K[x1, x2, . . . , xn], and and let P ∈ H. One of the possible ways n of deﬁning that the line {P + tv | t ∈ K} (v = (v1, v2, . . . , vn) ∈ K \{0}) is tangent to H at P is to require that f(P + tv), now simply a function of t, has a stationary point at t = 0. The Taylor expansion of f(P + tv) about t = 0 is

df(P + tv) f(P + tv) = f(P ) + t +(terms of order ≥ 2 in t) dt P n X ∂ f = f(P ) + t vj + (terms of order ≥ 2 in t). ∂ x P j=1 j

(Derivatives of polynomials can be deﬁned purely formally over any ﬁeld without using the concept of limits and Taylor expansion works for polyno- mials over any ﬁeld.) n X ∂f Thus f(P +tv) has a stationary point at t = 0 if and only if vj = 0. ∂x P j=1 j n ∂f ∂f ∂f X ∂f By using the gradient ∇f = , , ··· , , the condition vj = ∂x ∂x ∂x ∂x P 1 2 n j=1 j 0 can be re-written as (∇f)P .v = 0.

Deﬁnition. Let V ⊆ An(K) be an aﬃne algebraic variety and let P ∈ V .A line ` given in parametric form as {P + tv | t ∈ K},(v ∈ Kn \{0}), is said to be tangent to V at P if and only if (∇f)P .v = 0 for every f ∈ I(V ).

Deﬁnition. The tangent space to V at P , denoted by TP V is the union of the tangent lines to V at P and the point P .

n Theorem 3.1 Let V ⊆ A be an aﬃne algebraic variety. Let f1, f2,..., fr be a set of generators for I(V ). Let P ∈ V and let v = (v1, v2, . . . , vn) ∈ Kn \{0}. The line ` = {P + tv | t ∈ K} is tangent to V at P if and only if

33 JP v = 0, where J is the Jacobian matrix, ∂f ∂f ∂f 1 1 ··· 1 ∂x1 ∂x2 ∂xn ∂f2 ∂f2 ∂f2 ··· J = ∂x1 ∂x2 ∂xn . . . . . . .. . ∂f ∂f ∂f r r ··· r ∂x1 ∂x2 ∂xn and JP is J evaluated at P .

Proof. Assume that ` is tangent to V at P . Then (∇f)P .v = 0 for every f ∈ I(V ), in particular, (∇fi)P .v = 0 for every i, 1 ≤ i ≤ r.(∇fi)P .v is exactly the ith component of JP v, therefore JP v = 0. r P Assume now that JP v = 0. Let f ∈ I(V ). Then f = gifi for some i=1 gi ∈ K[x1, x2, . . . , xn], 1 ≤ i ≤ r. Now

r r X X ∇f = ∇(gifi) = ((∇gi)fi + gi∇fi), i=1 i=1 therefore

r r X X (∇f)P = ((∇gi)P fi(P ) + gi(P )(∇fi)P ) = gi(P )(∇fi)P . i=1 i=1

In the last step we used that fi(P ) = 0 for every i, 1 ≤ i ≤ r. Hence r P (∇f)P .v = gi(P )(∇fi)P .v, but (∇fi)P .v = 0, because it is the ith com- i=1 ponent of JP v = 0, therefore (∇f)P .v = 0. n It follows from this theorem that the set W = {v ∈ K | JP v = 0} is a linear subspace, therefore TP V = P + W is an aﬃne subspace. Its dimension is dim TP V = dim W = n − rank JP by the Rank-Nullity Theorem. Example: Let V ⊂ A2(C) be the cuspidal cubic curve deﬁned by the equation y2 − x3 = 0. It can be checked that y2 − x3 is irreducible, so hy2 − x3i is a prime ideal, therefore it is radical and then the Nullstellensatz implies that that I(V ) = hy2 − x3i. Therefore J = (−3x3 2y).

34 y

2

1

x -0.5 0.5 1.0 1.5 2.0 2.5

-1

-2

At P = (1, 1), JP = (−3 2), therefore the line {P + tv | t ∈ C} is tangent to V at (1, 1) if and only if −3v1 + 2v2 = 0. Hence T(1,1)V is the line of slope 3/2 through P with equation y = 3x/2 − 1/2. It agrees with the tangent line obtained by, for example, implicit diﬀerentiation.

At Q = (0, 0), however, something strange happens. JQ = (0, 0), therefore 2 every line through (0, 0) is a tangent line and T(0,0)V = A ! Let v = (α, β) 6= (0, 0) be the direction vector of a line through Q = (0, 0). Then Q + tv = (tα, tβ) and if we substitute x = tα, y = tβ into y2 − x3 we obtain t2β2 − t3α3 = t2(β2 − tα3), which has a zero of multiplicity at least 2 at t = 0, so any line through (0, 0) is a tangent line by our deﬁnition. If β = 0, then t2(β2 − tα3) has a zero of multiplicity 3 at t = 0, so the line y = 0 has a higher order tangency to the curve than a general line through (0, 0).

Proposition–Deﬁnition 3.2 Let V ⊆ An be an irreducible aﬃne algebraic variety. If V 6= ∅, there exists a proper subvariety Z of V (Z 6= V , but Z may be ∅) and an integer d ≥ 0 such that dim TP V = d for every P ∈ V \ Z and dim TP V > d for P ∈ Z. (In other words, dim TP V is constant on a non- empty Zariski open subset of V .) The dimension of V , denoted by dim V , is deﬁned to be d. The points of V \Z, where dim TP V = dim V , are called non- singular, while the points of Z, where dim TP V > dim V , are called singular and the set of singular points of V is denoted by Sing V . (Sometimes it is convenient to deﬁne dim ∅ = −1 and Sing ∅ = ∅.)

Proof. Let Vm = {P ∈ V | dim TP V ≥ m}, where m ≥ 0 is an integer. By the remark after Theorem 3.1, Vm = {P ∈ V | rank JP ≤ n − m}. It is a theorem in linear algebra that a matrix has rank ≤ k if and only if all of its

35 (k + 1) × (k + 1) minors vanish. Therefore

Vm = V ∩ V(h(n − m + 1) × (n − m + 1) minors of J}.

This shows that Vm is the aﬃne algebraic variety deﬁned by the ideal gener- ated by I(V ) and by the (n − m + 1) × (n − m + 1) minors of J. We have V = V0 ⊇ V1 ⊇ V2 ⊇ ... ⊇ Vn ⊇ Vn+1 = ∅. Choose d maximal such that Vd = V . Then dim TP V = d for P ∈ V \Vd+1 and dim TP V > d for P ∈ Vd+1, so dim V = d and Sing V = Vd+1.

n Example: If V is an aﬃne subspace of A , TP V = V at every point P ∈ V , so the dimension of V as a variety agrees with its dimension as an aﬃne space and it has no singular points. In particular, An is non-singular and has dimension n. Worked examples can be found in the separate handout at https://personalpages. manchester.ac.uk/staff/gabor.megyesi/teaching/MATH32062/singularities. pdf. Herwig Hauser’s gallery at http://homepage.univie.ac.at/herwig. hauser/bildergalerie/gallery.html has many nice examples of singular surfaces. The combination of the preceding results gives the following procedure for ﬁnding the dimension and singular points of an irreducible variety V ⊆ An(K).

1. Choose a set of generators f1, f2, . . . , fr for I(V ). (In all the problems you will have to solve, the ﬁeld will be algebraically closed and the deﬁning ideal of V will be a radical ideal, so it will be equal to I(V ).) 2. Calculate the Jacobian matrix J. 3. For k = 0, 1, 2,..., ﬁnd the points where all the (k + 1) × (k + 1) minors of J vanish and also f1 = f2 = ... = fr = 0. These are the points where rank JP ≤ k and therefore dim TP V ≥ n − k. Do this until you ﬁnd the smallest k such that dim TP V ≥ n − k for every point P ∈ V . 4. Then dim V = n−k and Sing V is the set of points where dim TP V > n−k, i. e., the points of V where all the k × k minors of J vanish. Deﬁnition. Let V be an arbitrary (not necessarily irreducible) aﬃne algebraic variety. The dimension of V , dim V , is the maximum of the dimensions of the irreducible components of V . The local dimension of V at P ∈ V is deﬁned to be the maximum of the dimension of the irreducible components of V containing P . P is called non-singular if and only if dim TP V is equal to the local dimension of V at P . P is called singular if and only if dim TP V is greater than the local dimension

36 of V at P . The set of singular points of V is denoted by Sing V , just as in the irreducible case. Fact: Sing V consists of the singular points of the individual irreducible com- ponents and of intersection points of diﬀerent components. This means that in general, one needs to decompose the variety into its irreducible components in order to ﬁnd the its dimension and its singular points, however, by the proposition below and by question 2 on tutorial sheet 7, hypersurfaces over an algebraically closed ﬁeld are an exception.

Proposition 3.3 Let K be an algebraically closed ﬁeld. Let H ⊂ An(K) be a hypersurface, that is, H = V(hfi) for some non-constant polynomial f ∈ K[x1, x2, . . . , xn], then dim H = n − 1.

α1 α2 αr Proof. f can be factorised as f = f1 f2 ··· fr , where the fi, 1 ≤ i ≤ r, are irreducible polynomials such that fi is not a scalar multiple of fj if i 6= j, and αi, 1 ≤ i ≤ r, are positive integers. The irreducible components are V(hfii), 1 ≤ i ≤ r, so if we prove the proposition for f irreducible, then it will imply the general case, because each component of H has dimension n − 1. Let’s assume that f is irreducible, then H is also irreducible and I(H) = hfi. ∂f ∂f ∂f We have J = , ,..., . rank JP can only be 0 or 1 and if ∂x1 ∂x2 ∂xn rank JP = 1 for some P ∈ H, then dim H = n − 1. ∂f ∂f Assume to the contrary that rank JP = 0 for every P ∈ H. Then , , ∂x1 ∂x2 ∂f ..., ∈ I(H) = hfi, that is, all the partial derivatives are multiples of f. ∂xn As their degree is smaller than the degree of f, the only way this can happen is if they are all 0. This is not possible in characteristic 0 since then f would be a constant, which has been excluded. In characteristic p it is possible for all the partial derivatives to be 0, it happens if and only if the exponent of every variable in every term is a dxn multiple of p.( = nxn−1 = 0 if and only if p divides n.) Then dx √ p X pi1 pi2 pin X p i1 i2 in f = αi1,i2,...,in x1 x2 ··· xn = αi1,i2,...,in x1 x2 ··· xn , i1,i2,...,in i1,i2,...,in so f is reducible, contradicting our assumption that it was irreducible.

n Remark. Let K be an algebraically closed ﬁeld. If V1, V2 ⊆ A (K) are aﬃne algebraic varieties of dimension d1 and d2, respectively, then every irreducible

37 component of V1 ∩ V2 has dimension at least d1 + d2 − n, but this is hard to prove with our deﬁnition of dimension. In particular, this implies that if n V = V(hf1, f2, . . . , fri) ⊆ A (K), then every component of V has dimension at least n−r. If f1 and f2 are non-constant polynomials without a non-trivial common factor, then every component of V(hf1, f2i) has dimension exactly n − 2. Theorem 3.4 Let V ⊆ An be an aﬃne algebraic variety and let P ∈ V . Let mP = {F ∈ K[V ] | F (P ) = 0} be the maximal ideal of K[V ] corresponding 2 2 to P . Then dim TP V = dim(mP /mP ). (The quotient ring K[V ]/mP is a 2 ﬁnite dimensional vector space, mP /mP is a subspace in it.) Proof. Let’s introduce the notation f = f + I(V ) for f ∈ K[V ]. Let P = (a1, a2, . . . , an) and let F ∈ mP . F = f for a suitable f ∈ K[V ] and F (P ) = f(P ) = 0. We noted in the proof of the Nullstellensatz (Theorem 1.7) that f(P ) = 0 if and only if f ∈ hx1 − a1, x2 − a2, . . . , xn − ani, therefore n P there exist polynomials gi ∈ K[x1, x2, . . . , xn] such that f = (xi − ai)gi in i=1 n P K[x1, x2, . . . , xn]. Hence f = (xi − ai)gi in K[V ]. xi − ai ∈ mp for every i, i=1 1 ≤ i ≤ n, obviously, therefore mp = hx1 − a1, x2 − a2,..., xn − ani /K[V ]. In the rest of the proof we shall assume P = (0, 0,..., 0) for simplicity, then we have mP = hx1, x2,..., xni /K[V ].

We can also assume that TP V is the subspace x1 = x2 = ... = xk = 0 of n K , where k = n − dim TP V , this can be achieved by a linear change of co-ordinates. This means that the reduced row echelon form of JP is 1 0 ··· 0 0 ··· 0 0 1 ··· 0 0 ··· 0 ...... ...... . . . . . 0 0 ··· 1 0 ··· 0 , 0 0 ··· 0 0 ··· 0 . . . . . . . . . . 0 0 ··· 0 0 ··· 0 with 1s on the diagonal in the ﬁrst k rows and all the other entries are 0. The rows of the reduced row echelon form are linear combinations of the rows of JP . By applying the same linear combinations to the generators of I(V ) we obtain a set of generators f1, f2,..., fr for I(V ) such that the Jacobian of this set of generators is the reduced row echelon form above, therefore fi = xi + (terms of degree ≥ 2) for 1 ≤ i ≤ k, and fi only contains terms of degree ≥ 2 for k + 1 ≤ i ≤ r.

38 n−k Let’s deﬁne a function D : mP → K in the following way. Let F ∈ mP . Then F = f for some f ∈ hx1, x2, . . . , xni /K[x1, x2, . . . , xn]. f can n P be written as f = αixi + (terms of degree ≥ 2) and we set D(F ) = i=1 (αk+1, αk+2, . . . , αn). 2 We shall prove that D is a surjective linear map with kernel mP , then it will 2 ∼ n−k 2 n−k follow that mP /mP = K , so dim(mP /mP ) = dim K = n − k.

Step 1. First we need to show that the vector (αk+1, αk+2, . . . , αn) does not depend on the choice of f, so that D is indeed a function.

Let g ∈ K[x1, x2, . . . , xn] be another polynomial such that F = g. Then f − g ∈ I(V ), so there exist hi ∈ K[x1, x2, . . . , xn], 1 ≤ i ≤ r, such that r P f − g = hifi. As fi = xi + (terms of degree ≥ 2) for 1 ≤ i ≤ k, and fi i=1 r P only contains terms of degree ≥ 2 for k + 1 ≤ i ≤ r, the sum hifi has no i=1 linear terms in xk+1, xk+2,..., xn, so the coeﬃcients αk+1, αk+2,..., αn are uniquely determined and therefore D(F ) is indeed a function.

Step 2. D is clearly linear, D(F + G) = D(F ) + D(G) for any F,G ∈ mP and D(λF ) = λD(F ) for any F ∈ mP and any λ ∈ K. n P Step 3. D is surjective, since D( αixi) = (αk+1, αk+2, . . . , αn) for any i=k+1 αk+1, αk+2,..., αn ∈ K.

Step 4. ker D = {F ∈ mP | D(F ) = 0} by deﬁnition. We shall prove that 2 ker D = mP . n P Let F = f, f = αixi + (terms of degree ≥ 2) as in the deﬁnition of D. i=1 Then F ∈ ker D if and only if αk+1 = αk+2 = ... = αn. This implies that k P f − αifi only contains terms of degree ≥ 2, i. e., it is an element of the i=1 ideal h{xixj | 1 ≤ i ≤ j ≤ n}i /K[x1, x2, . . . , xn]. This is equivalent to the existence of polynomials gij ∈ K[x1, x2, . . . , xn], 1 ≤ i ≤ j ≤ n, such that

k X X f − αifi = gijxixj i=1 1≤i≤j≤n in K[x1, x2, . . . , xn]. Then in K[V ], we have X f = gij xi xj, 1≤i≤j≤n

39 2 i. e., F = f ∈ h{xi xj | 1 ≤ i ≤ j ≤ n}i = mP /K[V ]. This shows that 2 ker D ⊆ mP . 2 Conversely, let’s now assume that F ∈ mP . Then X F = Gijxi xj 1≤i≤j≤n for suitable Gij ∈ K[V ], 1 ≤ i ≤ j ≤ n. Now let gij ∈ K[x1, x2, . . . , xn], P 1 ≤ i ≤ j ≤ n, be such that gij = Gij, then we can take f = gijxixj 1≤i≤j≤n and it satisﬁesf = F . As f has no linear terms, we have D(F ) = (0, 0,..., 0), 2 therefore F ∈ ker D. This shows mP ⊆ ker D, by combining it with the 2 previous result we obtain ker D = mP . n−k We have proved that D is a surjective linear map mP → K with kernel 2 2 ∼ n−k 2 n−k mP , therefore mP /mP = K , in particular dim(mP /mP ) = dim K = n − k = dim TP V . Corollary 3.5 Let ϕ : V → W be an isomorphism between aﬃne algebraic varieties. Then dim TP V = dim Tϕ(P )W for every P ∈ V , therefore dim V = dim W and ϕ(Sing V ) = Sing W . In particular, a singular variety cannot be isomorphic to a non-singular variety. Proof. By Theorem 2.1, ϕ induces a morphism ϕ∗ : K[W ] → K[V ] and by Corollary 2.2, if ϕ is an isomorphism of varieties then ϕ∗ is an isomorphism ∗ of rings. It easy to check ϕ also gives a bijection between mP /K[V ] and 2 2 2 mϕ(P ) /K[W ] and also between mP /K[V ] and mϕ(P ) /K[W ]. Hence mP /mP 2 and mϕ(P )/mϕ(P ) are isomorphic vector spaces, therefore by Theorem 3.4, 2 2 dim TP V = dim mP /mP = dim mϕ(P )/mϕ(P ) = dim Tϕ(P )W . The other assertions follow from this, since dimension and singularity are deﬁned in terms of the dimensions of the tangent spaces at the points of the variety. Examples: 1. The aﬃne line A1(C) and the cuspidal cubic curve deﬁned by the equation y2 − x3 = 0 in A2(C) are not isomorphic, since the former has no singular point, while the latter has a singular point at (0, 0).

3 2 2. Let W1 = V(hxy, xz, yzi) ⊂ A (C) and let W2 = V(hxy(x + y)i) ⊂ A (C).

40 2

1 y 0 2 -1

-2 W1 2 1 W2

1

x 0 -2 -1 1 2

-1 -1

-2 -2

-1 -2 0

1

2

The irreducible components of W1 are the co-ordinate axes, V(hx, yi), V(hx, zi) and V(hy, zi). The irreducible components of W2 are the lines V(hxi), V(hyi) and V(hx + yi). Both varieties consist of three lines meeting at a point, the origin, but we shall prove that W1 and W2 are not isomorphic.

Let’s calculate T(0,0,0)W1 and T(0,0)W2. hxy, xz, yzi) and hxy(x + y)i are radical ideals, so they are equal to I(W1) and I(W2), respectively. y x 0 The Jacobian of W1 is J(W1) = z 0 x. At (0, 0, 0), J(0,0,0)(W1) = 0 z y 0 0 0 0 0 0, so it has rank 0 and therefore dim T(0,0,0)W1 = 3. (T(0,0,0)W1 = 0 0 0 3 A can be proved even without calculating the Jacobian. Any f ∈ I(W1) is identically 0 on the three co-ordinate axes, therefore these axes are tangent lines. If we know that T(0,0,0)W1 is an aﬃne subspace, then it follows that 3 T(0,0,0)W1 = A as no other aﬃne subspace contains all three co-ordinate axes.) 2 2 The Jacobian of W2 is J(W2) = 2xy + y x + 2xy . At (0, 0), J(0,0)(W2) = 0 0 , so it has rank 0 and therefore dim T(0,0)W2 = 2.

If there existed an isomorphism ϕ : W1 → W2, then we would have ϕ(0, 0, 0) = (0, 0), as (0, 0, 0) and (0, 0) are uniquely characterised as the intersection points of the components in W1 and W2, respectively. However, isomor- phisms preserve the dimension of the tangent space and dim T(0,0,0)W1 = 3 6= dim T(0,0)W2 = 2, therefore there does not exist an isomorphism between W1 and W2.

The rest of the material in this section is not examinable.

41 For any vector space X over K, the dual space X∗ is deﬁned to be the vector space of all linear maps X → K with the obvious operations. For any linear map α : X → Y , there is a dual map α∗ : Y ∗ → X∗. What Theorem 3.4 really ∼ 2 ∗ proves, without mentioning dual spaces explicitly, is that TP V = (mP /mP ) . While any two ﬁnite dimensional vector spaces of the same dimension are 2 ∗ isomorphic, the isomorphism between TP V and (mP /mP ) is natural, it can be deﬁned without reference to bases. Given an arbitrary morphism ϕ : V → W , by Theorem 2.1 we have a morphism of rings ϕ∗ : K[W ] → K[V ], which also induces a linear map 2 2 mϕ(P )/mϕ(P ) → mP /mP . Taking duals, we obtain the diﬀerential map ∼ 2 ∗ ∼ 2 ∗ dϕP : TP V = (mP /mP ) → Tϕ(P )W = (mϕ(P )/mϕ(P )) . dϕP is similar to the diﬀerential map betwen tangent spaces in diﬀerential geometry. Isomorphisms can be characterised by using the diﬀerential map, ϕ is an isomorphism if and only if it is bijective and dϕP is an isomorphism for every P ∈ V . The dimension of a variety can be deﬁned in other ways as well, which are more suited for certain purposes, but are less useful for practical calculations. 1. dim V is equal to the maximum d for which there exists a chain of irre- ducible subvarieties V0 ⊂ V1 ⊂ ...Vd ⊆ V . Moreover, if V is irreducible, any such chain that cannot be extended by inserting further subvarieties has the same length. From this deﬁnition it is clear that the dimension of a proper subvariety of an irreducible variety is smaller than the dimension of the whole variety, which is hard to prove by using tangent spaces. 2. If V is irreducible, then dim V is transcendence degree of K(V ) over K, i. e., the maximal number of elements x1, x2,..., xd ∈ K(V ) which do not satisfy any non-zero polynomial with coeﬃcients in K. As a consequence of an algebraic result called the Noether Normalisation Lemma, any variety is birationally equivalent to a hypersurface, therefore dim V = d if and only if V is birationally equivalent to a hypersurface in Ad+1. This interpretation of the dimension shows that the dimension of irreducible varieties is invariant under birational equivalence, not just under isomorphism. Fact: For any irreducible variety V in characteristic 0, there exist a non- singular variety V˜ and a birational morphism ϕ : V˜ → V such that ϕ is an isomorphism outside Sing V . This is called the desingularisation of V . The existence of a desingularisation is a very hard theorem, for which the Japanese mathematician Heisuke Hironaka received the Fields Medal in 1970. The problem is still open in prime characteristic in dimension ≥ 3.

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