Algebraic Geometry, PS 1

Bryden Cais

September 18, 2002

Exercise 1. Recall that for any field k, two systems of algebraic equations S, S′ define the same algebraic variety if and only if (S) = (S′). Let (S) be the k[T_{1}, …, T_{n}] := k[T] ideal generated by the elementary symmetric polynomials S_{i} of degree at most n in the T_{i}, and let (S′) be the ideal generated by the set of polynomials

S′ = {X_{i} := ∑_{j=1}^{n} T_{j}^{i} : 1 ≤ i ≤ n} .

It is an elementary result from algebra that (S) consists of all symmetric polynomials in the T_{i}, (see, for example [1, p. 550]) and therefore that (S′) ⊂ (S). The formula

j!S_{j} = X_{1}^{j} − X_{j} + ∑_{i=1}^{⌊ \frac{j}{2} ⌋} \binom{j}{i} (X_{j} − X_{i}X_{j−i})

may be proved by induction and shows that j!S_{j} ∈ (S′) for each 1 ≤ j ≤ n. Now if char(k) |̸n! then j! is invertible in k for each 1 ≤ j ≤ n and therefore S_{j} ∈ (S′). Moreover, the representation of S_{j} given above is unique so we see that (S) = (S′) if and only if char(k) > n.

Exercise 2. Consider the systems

S = {T_{1}^{2} + T_{2}, T_{1}}
S′ = {T_{2}^{2}T_{1}^{2} + T_{2}^{3} + T_{2} + T_{1}T_{2}, T_{2}T_{1}^{2} + T_{2}^{2} + T_{1}}

of algebraic equations over ℚ. The system S′ is equal to

{T_{2}(T_{2}T_{1}^{2} + T_{2}^{2} + T_{1}) + T_{2}, T_{2}T_{1}^{2} + T_{2}^{2} + T_{1}} ,

which is equivalent to the system

{T_{2}, T_{2}T_{1}^{2} + T_{2}^{2} + T_{1}} ,

which is in turn equivalent to

{T_{2}, T_{1}} .

On the other hand, S is equivalent to

{T_{2}, T_{1}} ,

since if T_{1} and T_{1}^{2} + T_{2} vanish on any variety V then so does T_{2} and conversely (because we are working over a field). Thus, S and S′ define the same ℚ varieties.

Exercise 3. Let X ⊂ 𝔸_{k}^{n}, Y ⊂ 𝔸_{k}^{m} be two affine algebraic k varieties. Let X, Y be generated by the systems S = {f_{1}, …, f_{r}} ⊂ k[T_{1}, …, T_{n}] and S = {g_{1}, …, g_{s}} ⊂ K[U_{1}, …, U_{m}]. Let

I = {f_{1}, …, f_{r}, g_{1}, … g_{s}} ⊂ k[T_{1}, …, T_{n}, U_{1}, …, U_{m}].

We then define the cartesian product of X, Y as

X × Y (K) := Sol(I; K).

Clearly, X × Y (K) = X(K) × Y (K) for any k algebra K (we just concatenate solutions).

Exercise 4. Let us show that an arbitrary intersection of algebraic varieties is again an algebraic variety and that any finite union of algebraic varieties is an algebraic variety.

1. Let S be a finite index set and let V_{s} for s ∈ S be algebraic k sets and let I_{s} = I(V_{s}). Define

I = ∏_{s∈S} I_{s}. Then we claim that

V (I) = ⋃_{s∈S} V_{s}.

To see this, observe that if x ∈ V_{s} for some s then every f ∈ I_{s} vanishes at x, so any finite sum of products of elements of all the I_{r} vanishes at x (since one of the factors must be in I_{s}); i.e. x ∈ V (I). Conversely, let x ∈ V (I) and suppose that x ∉ V_{s} for all s. Then for each s there exists f_{s} such that f_{s}(x) ≠ 0, whence f = ∏_{s∈S} f_{s}(x) ≠ 0, but f ∈ I, a contradiction.

2. Now let S be an arbitrary index set and let V_{s}, I_{s} be as above. Set I = ∑_{s∈S} I_{s}. Now we

claim that

⋂_{s∈S} V_{s} = V (I).

Let x ∈ V_{s} for all s. Then I_{S} vanishes at x for each s, so that I also vanishes at x; that is, x ∈ V (I). Conversely, since I_{s} ⊂ ∑_{s∈S} I_{s}, we see that V (I) ⊂ V (I_{s}) (since the V operator is inclusion reversing) and hence that V (I) ⊂ ⋂ V_{s}.

Exercise 5. Let

S, S′ ⊂ k[T_{1}, …, T_{n}]

be two systems of linear equations and suppose that Sol((S); k) = Sol((S′); k). Since S is linear, we can identify S with a subset of k^{n} via

∑_{i=1}^{n} a_{i}T_{i} ↦ (a_{1}, …, a_{n}).

Indeed, we can in fact view S as a subspace of k^{n} (as a vector space) since if f(c_{1}, …, c_{n}) = 0 for some f ∈ S and c = (c_{1}, …, c_{n}) ∈ k^{n} then f(λc_{1}, …, λc_{n}) = 0 also since f is linear. It is not difficult to see that under this identification, Sol((S); k) is just the orthogonal complement of S in k^{n} with respect to the usual inner product. Now observe that any space is determined uniquely by its orthogonal complement, so that the vector spaces S, S′ must be equal. This implies that the ideals (S), (S′) are equal. The converse of this result is trivial.

Exercise 6. We show that if a ring A is Noetherian then so is the polynomial ring A[X]. For suppose that A[X] is not Noetherian. Then there exists a nonstabilizing chain of ideals I_{1} ⊂ I_{2} ⊂ … ⊂ A[X]. Let f_{1} ∈ I be a polynomial of minimal degree in X. Define f_{i} for i ≥ 2 inductively to be a polynomial of minimal degree in I \ (f_{1}, f_{2}, …, f_{i−1}). Let f_{i} have degree d(i) in X and leading coefficient a_{i}. Then we claim that (a_{1}) ⊂ (a_{1}, a_{2}) ⊂ … ⊂ A is a nonstabilizing chain of ideals. For suppose that it stabilizes: then there exists some n such that (a_{1}, … a_{n}) = (a_{1}, …, a_{m}) for all m ≥ n, so that we can write

a_{n+1} = ∑_{i=1}^{n} b_{i}a_{i}

where b_{i} ∈ A. Then the polynomial f_{n+1} − ∑_{i=1}^{n} b_{i}X^{d(n)−d(i)}f_{i} is in I \ (f_{1}, …, f_{n}) (since f_{n+1} ∉ (f_{1}, …, f_{n})), but is of lower degree than f_{n+1}. This is a contradiction and our chain of ideals in A does not terminate. The contrapositive then shows that if A is Noetherian, so is A[X], and by induction, this implies that A[T_{1}, …, T_{n}] is Noetherian.

Exercise 7. Let A = k[T_{1}, T_{2}]/(T_{1}^{2} − T_{2}^{3}). Let K denote the field of fractions of A. Then T_{1}/T_{2} ∈ K \ A and

(\frac{T_{1}}{T_{2}} )^{2} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{T_{2}^{3}}{T_{2}^{2}} = T_{2}.

Therefore, T_{1}/T_{2} is a root of the polynomial X^{2} − T_{2}. It is therefore integral over A.

Exercise 8. Let X, Y be two affine algebraic sets in K^{n}. Suppose that f ∈ I(X) ∩ I(Y). Then f vanishes on X and on Y so that f vanishes on X ∪ Y and hence f ∈ I(X ∪ Y). Conversely, if f ∈ I(X ∪ Y) then f vanishes on X and on Y and therefore f ∈ I(X) and f ∈ I(Y). In short, this shows that

I(X ∪ Y) = I(X) ∩ I(Y).

We do not always have I(X) ∩ I(Y) = I(X)I(Y). For example, let Y = X = Sol(T_{1}; K). Then I(X) = I(Y) = (T_{1}) (since this is a prime ideal it is already radical) and hence I(X)∩I(Y) = (T_{1}). On the other hand, I(X)I(Y) = (T_{1}^{2}).

Exercise 9. Let I ⊂ k[T_{1}, T_{2}] be generated by T_{1}^{2}T_{2}, T_{1}T_{2}^{3} and suppose that f ∈ rad(I). Then f^{N} = aT_{1}^{2}T_{2} + bT_{1}T_{2}^{3} for some N ∈ ℕ and a, b ∈ k[T_{1}, T_{2}]. This shows that f^{N} is divisible by T_{1}T_{2} and therefore that f is divisible by T_{1}T_{2} (since T_{1} and T_{2} are both irreducible, clearly). Hence f ∈ (T_{1}T_{2}) and therefore that rad(I) ⊂ (T_{1}T_{2}). On the other hand, (T_{1}T_{2})^{2} = T_{2}(T_{1}^{2}T_{2}) ∈ I so that (T_{1}T_{2}) ⊂ rad(I) and therefore rad(I) = (T_{1}T_{2}).

Exercise 10. We claim that the Zariski topology on 𝔸^{n}(K) is not Hausdorff. For let x, y ∈ 𝔸^{n}(K) be distinct points. Then if the Zariski topology were Hausdorff, we could find varieties V, U such that x ∈ 𝔸^{n}(K) \ V and y ∈ 𝔸^{n}(K) \ U and

𝔸^{n}(K) \ V ∩ 𝔸^{n}(K) \ U = 𝔸^{n}(K) \ (V ∪ U) = ∅,

or equivalently, V ∪ U = 𝔸^{n}(K). But observe that 𝔸^{n}(K) is an irreducible variety and therefore we must have V = 𝔸^{n}(K) or U = 𝔸^{n}(K) and consequently we cannot separate x, y. The Zariski topology on 𝔸^{n}(K) is, however separable: Let x, y be distinct points of 𝔸^{n}(K). Then the variety V = Sol(T − x; K) contains x and not y; consequently O = 𝔸^{n}(K) \ V is an open set containing y and not x. The same is not true in general for the Zariski topology on 𝔸_{k}^{n}(K) is k ≠ K. As an example, let k = ℚ and K = ℚ(\sqrt{2}). Let x = \sqrt{2} and y = −\sqrt{2}. Then any algebraic ℚ set containing x must consist of polynomials all divisible by T_{2} − 2 and therefore contains y also. I conjecture that if x ∉ G_{K/k}y then x, y can be separated in the Zariski topology, where G_{K/k} denotes the galois group of K/k, which has a natural action on K^{n} and hence on any variety.

Exercise 11. Observe that any solution in K^{n} to the system of algebraic equations

T_{1}^{3} = T_{2}^{3} = T_{1}T_{2}(T_{1} + T_{2}) = 0

over any field K must have T_{1} = T_{2} = 0 (this is already implied by the first two equations over a field). Any polynomial f ≢ 0 vanishing at T_{1} = T_{2} = 0 is certainly in the ideal (T_{1}, T_{2}): if some term of f is not divisible by either T_{1} or T_{2} then there exists some point of K^{n} where f does not vanish. Conversely, any polynomial in this ideal vanishes at T_{1} = T_{2} = 0. Hence, I(V) = (T_{1}, T_{2}) ⊂ k[T_{1}, … T_{n}].

Exercise 12. Observe that the set S = {(z_{1}, z_{2}) ∈ ℂ^{2} : |z_{1}|^{2} + |z_{2}|^{2} = 1} has a limit point in ℂ^{2}. The closure of S in the Zariski ℂ^{2} topology is contained in some closed set, that is, some variety V. Now let f ∈ I(V). Then f is a polynomial in z_{1}, z_{2} that vanishes on a subset of ℂ^{2} having a limit point in ℂ^{2}. From several complex variables we know that any holomorphic function that vanishes on a set having a limit point vanishes identically. Therefore, f ≡ 0 and I(V) = (0) so that V = C^{2}. This shows that S is dense in ℂ^{2}.

References

[1] Artin, M. Algebra. Prentice Hall, N.J. 1991.