Math 593 Assignment # 9 Solutions

Bryden R. Cais

December 2, 2002

1. Let R = ℂ[x, y] and I = (x, y) ⊂ R. Define ϕ : I ⊗_{R} I → ℂ by

ϕ(f ⊗ g) = f_{x}(0, 0)g_{y}(0, 0) − f_{y}(0, 0)g_{x}(0, 0).

We claim that ψ is an R module homomorphism, where ℂ is considered as an R module by

f · z := f(0, 0)z

for any f ∈ R and z ∈ ℂ. We first show that the map ψ : I × I → ℂ given by

ψ(f, g) = f_{x}(0, 0)g_{y}(0, 0) − f_{y}(0, 0)g_{x}(0, 0)

is R bilinear. This is essentially a sequence of straightforward calculations. For example, for any r ∈ R and f, g, h ∈ I we have

ψ(rf + g, h)

= (rf + g)_{x}(0, 0)h_{y}(0, 0) − h_{x}(0, 0)(rf + g)_{y}(0, 0)
= (r_{x}(0, 0)f(0, 0) + f_{x}(0, 0)r(0, 0) + g_{x}(0, 0))h_{y}(0, 0) − h_{x}(0, 0)(r_{y}(0, 0)f(0, 0) + f_{y}(0, 0)r(0, 0) + g_{y}(0, 0))
= (f_{x}(0, 0)r(0, 0) + g_{x}(0, 0))h_{y}(0, 0) − h_{x}(0, 0)(f_{y}(0, 0)r(0, 0) + g_{y}(0, 0))

since f ∈ I and hence f(0, 0) = 0,

= r(0, 0)(f_{x}(0, 0)h_{y}(0, 0) − h_{x}(0, 0)f_{y}(0, 0)) + g_{x}(0, 0)h_{y}(0, 0) − h_{x}(0, 0)g_{y}(0, 0)
= r · ψ(f, h) + ψ(g, h).

The calculations to show R linearity in the second component are similar, so we omit them. By the universal property of the tensor product, we have that ϕ is R linear, as required.

Now if x ⊗ y − y ⊗ x = 0 then since ϕ is an R module homomorphism we have

0 = ϕ(0)
= ϕ(x ⊗ y − y ⊗ x)
= ϕ(x ⊗ y) − ϕ(y ⊗ x)
= (1 · 1 − 0 · 0) − (0 · 0 − 1 · 1)
= 2,

a contradiction.

2. For p, q ∈ ℚ write p = a/b, q = c/d for integers a, b, c, d. Then for p ⊗ q ∈ ℚ ⊗_{ℤ} ℚ we have

p ⊗ q = \frac{a}{b} ⊗ \frac{c}{d}

= (\frac{acd}{bd} ⊗ \frac{1}{d})

= d (\frac{ac}{bd} ⊗ \frac{1}{d})

= (\frac{ac}{bd} ⊗ \frac{d}{d})

= (\frac{ac}{bd} ⊗ 1)

= (pq ⊗ 1),

where we have only used ℤ-bilinearity of the tensor. This suggests that we consider the map ϕ : Q ⊗_{ℤ} ℚ → ℚ given by p ⊗ q ↦ pq. Observe that from our calculations above, it is clear that this is a ℚ module homomorphism. Indeed, we have

ϕ(p ⊗ q + r ⊗ s) = ϕ((pq ⊗ 1) + (rs ⊗ 1))
= ϕ(((pq + rs) ⊗ 1))
= pq + rs
= ϕ(p ⊗ q) + ϕ(r ⊗ s),

and

ϕ(\frac{x}{y}(p ⊗ q)) = ϕ(\frac{x}{y}(\frac{y}{y} pq ⊗ 1))

= ϕ((\frac{x}{y}pq ⊗ 1))

= \frac{x}{y}pq

= \frac{x}{y}ϕ(p ⊗ q),

for any x/y ∈ ℚ. Surjectivity of ϕ is clear since ϕ(p ⊗ 1) = p for any p ∈ ℚ. Injectivity is evident from the above calculation since if ϕ(p ⊗ q) = ϕ(r ⊗ s) then pq = rs so that

p ⊗ q = pq ⊗ 1 = rs ⊗ 1 = r ⊗ s.

It follows that ϕ is an isomorphism of ℚ modules, as required.

3. Define

ϕ : V^{∗} × W → Hom(V, W)

by

(f, w) ↦ (v ↦ f(v) · w) .

It is not difficult to see that ϕ is bilinear. For every v ∈ V we have

ϕ(af + g, bw + u)(v) = (af + g)(v)(bw + u)
= (af(v) + g(v))(bw + u)
= abf(v)w + af(v)u + bg(v)w + bg(v)u
= abϕ(f, w)(v) + aϕ(f, u)(v) + bϕ(g, w)(v) + bϕ(g, u)(v).

Thus we obtain a homomorphism ϕ̃ : V^{∗} ⊗ W → Hom(V, W) given by ϕ(f, w)(v) = f(v)w. To check that this is an isomorphism, pick any basis {v_{i}} of V and {w_{j}} of W. Let {v_{i}^{∗}} be the dual basis to {v_{i}}. Then we know that {v_{i}^{∗} ⊗ w_{j}} is a basis for V^{∗} ⊗ W. Suppose that f ∈ Hom(V, W) is given by

f(v_{i}) = ∑_{j} a_{ij}w_{j}.

Let

x = ∑_{k,j} a_{kj}(v_{k}^{∗} ⊗ w_{j}) ∈ V^{∗} ⊗ W.

Then clearly we have

ϕ(x)(v_{i}) = ∑_{k,j} a_{kj}(v_{k}^{∗}(v_{i})w_{j}) = ∑_{j} a_{ij}w_{j} = f(v_{i}),

so that ϕ is surjective. Since dim Hom(V, W) = dim V^{∗} ⊗ W, we must have that ϕ is an isomorphism.

Observe that although we chose a basis to show that ϕ was an isomorphism, the definition of ϕ is independent of any choice of basis. This is what it means for an isomorphism of vector spaces to be “canonical.”

4. Let M be an R module and I ⊂ R an ideal. Recall that we have an exact sequence

0 ⟶ I ⟶^{u} R ⟶^{v} R/I ⟶ 0.

By right exactness of the tensor product, this gives an exact sequence

I ⊗_{R} M ⟶^{u⊗id} R ⊗_{R} M ⟶^{v⊗id} R/I ⊗_{R} M ⟶ 0.

We have

u ⊗ id (∑_{l} i_{l} ⊗ m_{l}) = u ⊗ id (∑_{l} 1 ⊗ i_{l}m_{l})

= ∑_{l} u(1) ⊗ i_{l}m_{l}

= u(1) ⊗ (∑_{l} i_{l}m_{l})

so that if u ⊗ id(x) = 0 then since u is injective and u(1) ≠ 0 we have x = 0 so that u ⊗ id is injective. We claim that I ⊗_{R} M ≃ IM. For we may define

f :I ⊗ M → IM,
g :IM → I ⊗ M

by

f (∑_{l} i_{l} ⊗ m_{l}) = ∑_{l} i_{l}m_{l},

g (x) = 1 ⊗ x.

A priori, it is unclear that the image of g is contained in IM. However, we have

g ∘ f (∑_{l} i_{l} ⊗ m_{l}) = 1 ⊗ ∑_{l} i_{l}m_{l}

= ∑_{l} 1 ⊗ i_{l}m_{l}

= ∑_{l} i_{l} ⊗ m_{l}

since the tensor product is over R and I ⊂ R. Moreover, for x = ∑_{l} i_{l}m_{l} ∈ IM we have

f ∘ g(x) = f (1 ⊗ ∑_{l} i_{l}m_{l})

= ∑_{l} i_{l}m_{l}

= x,

from which it follows that f is an isomorphism, since it is obviously a homomorphism. Similarly, one shows that R ⊗_{R} M ≃ M (the isomorphism is the same as above). It follows that

M ⊗_{R} R/I ≃ R/I ⊗_{R} M ≃ M/IM,

as required.

5. Let {e_{1}, e_{2}} be a basis of ℝ^{2}. Let v = ae_{1} + be_{2} and w = ce_{1} + de_{2}. Then

v ⊗ w = ace_{1} ⊗ e_{1} + ade_{1} ⊗ e_{2} + bce_{2} ⊗ e_{1} + bde_{2} ⊗ e_{2}
= αe_{1} ⊗ e_{1} + βe_{1} ⊗ e_{2} + γe_{2} ⊗ e_{1} + δe_{2} ⊗ e_{2},

with

αδ − βγ = 0.

This shows that any simple tensor corresponds to a point of the quadric hypersurface given by

αδ − βγ = 0.

Conversely, suppose that we have a point (α, β, γ, δ) of the above hypersurface. If α, β, γ, δ are all 0, it is clear that this corresponds to the simple tensor 0 ⊗ 0. Otherwise, without loss of generality, suppose that α ≠ 0 and that αδ − βγ = 0. Then

δ = \frac{βγ}{α}

and

αe_{1} ⊗ e_{1} + βe_{1} ⊗ e_{2} + γe_{2} ⊗ e_{1} + δe_{2} ⊗ e_{2} = (αe_{1} + γe_{2}) ⊗ (e_{1} + \frac{β}{α}e_{2})

is a simple tensor.