The theory of F-rational signature

We shall say that there are 𝑛 simultaneous injections from 𝑉 to 𝑊 in 𝐻 provided there exist ϕ 1 , … , ϕ n ∈ H such that the induced map

is an injection. We will use MaxInj ⁡ ( H ) to denote the maximal non-negative integer 𝑛 such that there are 𝑛 simultaneous injections from 𝑉 to 𝑊 in 𝐻.

Let 𝑘 be a field and suppose 𝑉 is a finite-dimensional vector space over 𝑘. Then there exists a positive constant 𝐶 with the following property: for any finite-dimensional vector space 𝑊 and vector subspace H ⊆ Hom k ⁡ ( V , W ) , we have

where 𝑈 varies over all non-zero subspaces of 𝑉.

Our proof will show that the constant 𝐶 appearing in Theorem A.2 can be taken to be

independently of the ground field 𝑘. However, we believe this bound to be far from optimal. In particular, when working over an infinite field 𝑘 and using general 𝑘 linear combinations of maps in 𝐻 appropriately, we believe it is possible to exhibit a quadratic bound in terms of dim ( V ) . This should not be a surprise because new injections can appear in 𝐻 after extending from a finite field. For example, over the field F 2 = Z / 2 ⁢ Z , let V = W = F 2 ⊕ 3 and consider the subspace

of Hom F 2 ⁡ ( V , W ) . There is no injection in 𝐻, but x ⁢ A + B is an injection over

Let 𝑘 be a field and suppose 𝑉 is a two-dimensional vector space over 𝑘. For any finite-dimensional vector space 𝑊 and vector subspace H ⊆ Hom k ⁡ ( V , W ) , we have

where 𝑈 varies over all non-zero subspaces of 𝑉.

Let 𝑘 be an arbitrary field and set V = W = k ⊕ 3 with standard basis vectors e ⃗ 1 , e ⃗ 2 , e ⃗ 3 . Consider the linear transformations from 𝑉 to 𝑊 given by the matrices

and let 𝐻 be the linear span of f , g , h in Hom k ⁡ ( V , W ) . By verifying that

for all values of λ , μ , ν , we see that there are no injections in 𝐻. Yet

as it is easy to verify that H ⁢ ( V ) = W and that dim H ⁢ ( k ⁢ ⟨ v ⃗ ⟩ ) = 2 for any 0 ≠ v ⃗ ∈ V .

Let 𝑉 and 𝑊 be two finite-dimensional vector spaces over a field 𝑘. If ϕ , ξ ∈ Hom k ⁡ ( V , W ) are such that Im ⁡ ϕ ∩ Im ⁡ ξ = 0 , then rank ⁡ ( ϕ + ξ ) ≥ rank ⁡ ϕ and the inequality is strict provided there exists v ⃗ ∈ ker ⁡ ϕ such that ξ ⁢ ( v ⃗ ) ≠ 0 .

Proof

If v ⃗ ∈ ker ⁡ ( ϕ + ξ ) , then ϕ ⁢ ( v ⃗ ) = − ξ ⁢ ( v ⃗ ) ∈ Im ⁡ ϕ ∩ Im ⁡ ψ = 0 . Thus we see that ker ⁡ ( ϕ + ξ ) = ( ker ⁡ ϕ ∩ ker ⁡ ξ ) ⊆ ker ⁡ ϕ , and the desired inequality rank ⁡ ( ϕ + ξ ) ≥ rank ⁡ ϕ follows and is strict provided ker ⁡ ϕ ∩ ker ⁡ ξ ⊊ ker ⁡ ϕ . ∎

Let 𝑘 be a field, 𝑉 a two-dimensional 𝑘-vector space, and n ≥ 1 an integer. Let 𝑊 be a finite-dimensional 𝑘-vector space and 𝐻 a subspace of Hom ⁡ ( V , W ) such that, for every 0 ≠ U ⊆ V , we have dim H ⁢ ( U ) ≥ n ⁢ dim U . Then MaxInj ⁡ ( H ) ≥ n , i.e., there is an injection ⨁ n V → W , where each component is in 𝐻.

Proof

We proceed by induction starting with n = 1 . Suppose all maps in 𝐻 have rank at most 1. Take any 0 ≠ h ∈ H and let U = ker ⁡ h . By the assumption, there is g ∈ H such that g ⁢ ( U ) ≠ 0 . We must have Im ⁡ g = Im ⁡ h , or g + h has rank 2 by Lemma A.6. Since dim H ⁢ ( V ) ≥ 2 , there is f ∈ H such that Im ⁡ f ⊈ Im ⁡ g . This gives a contradiction since either f + g or f + h must have rank 2 by Lemma A.6.

Now, assume that the n + 1 -level condition holds, i.e., dim H ⁢ ( U ) ≥ ( n + 1 ) ⁢ dim U for all 0 ≠ U ⊆ V . By induction, we find independent injections ϕ 1 , … , ϕ n . Set W ′ = W / Im ⁡ ϕ 1 . If the 𝑛-level condition holds for H ′ ⊆ Hom ⁡ ( V , W ′ ) , then there is an injection ⨁ n V → W ′ by induction which then lifts to the required injection ⨁ n + 1 V → W .

Thus we assume that H ′ does not satisfy the 𝑛-level condition, i.e., there is a one-dimensional subspace 𝑈 such that dim H ′ ⁢ ( U ) ≤ n − 1 , forcing that dim H ⁢ ( U ) = n + 1 and ϕ 1 ⁢ ( V ) ⊆ H ⁢ ( U ) . This can only happen if ϕ 1 ⁢ ( V ) + ϕ 2 ⁢ ( U ) + … + ϕ n ⁢ ( U ) = H ⁢ ( U ) since the dimensions are equal. We now pass to W ̄ = W / ∑ i = 2 n ϕ i ⁢ ( V ) . For any ϕ ∈ H , we denote by ϕ ̄ the induced map V → W ̄ and define H ̄ analogously. Since the original ϕ 1 , … , ϕ n are independent injections, we still have that ϕ ̄ 1 ⁢ ( V ) = H ̄ ⁢ ( U ) . Hence dim H ̄ ⁢ ( U ) = 2 and dim H ̄ ⁢ ( V ) ≥ 4 . It remains to build two independent injections U → W ̄ because their lifts will be independent with ϕ 2 , … , ϕ n .

Let ψ ̄ be such that ϕ ̄ 1 ⁢ ( V ) = H ̄ ⁢ ( U ) = ψ ̄ ⁢ ( U ) + ϕ ̄ 1 ⁢ ( U ) . First, assume that ψ ̄ is an injection, and let v ⃗ ∉ U be such that ϕ ̄ 1 ⁢ ( v ⃗ ) ∈ ψ ̄ ⁢ ( U ) . Since dim H ̄ ⁢ ( V ) ≥ 4 , there is g ̄ such that g ̄ ⁢ ( V ) ⊈ ψ ̄ ⁢ ( V ) + ϕ ̄ 1 ⁢ ( V ) . If g ̄ , ψ ̄ are independent injections, then we are done. Otherwise, clearly, g ̄ ⁢ ( v ⃗ ) ∉ ψ ̄ ⁢ ( V ) + ϕ ̄ 1 ⁢ ( V ) , so we may apply Lemma A.6 in W ̄ / ψ ⁢ ( V ) to show that g ̄ + ϕ ̄ 1 and ψ ̄ are independent injections.

Last, suppose that ψ ̄ is not an injection and fix 0 ≠ e ⃗ ∈ ker ⁡ ψ ̄ . As dim H ̄ ⁢ ( V ) ≥ 4 , there is h ̄ such that h ̄ ⁢ ( V ) ⊈ Im ⁡ ϕ 1 = H ̄ ⁢ ( U ) . By the choice of ψ ̄ , we have e ⃗ ∉ U , so h ̄ ⁢ ( e ⃗ ) ∉ H ̄ ⁢ ( U ) . If h ̄ ⁢ ( U ) ⊈ ϕ ̄ 1 ⁢ ( U ) , then h ̄ is injective and we reduce to the previous case by replacing 𝜓 with ℎ. Otherwise, if h ̄ ⁢ ( U ) ⊆ ϕ ̄ 1 ⁢ ( U ) , then h ̄ ⁢ ( V ) ∩ ψ ̄ ⁢ ( V ) = 0 , so h ̄ + ψ ̄ is injective by Lemma A.6 and we reduce to the previous case since ϕ ̄ 1 ⁢ ( V ) = ( ψ ̄ + h ̄ ) ⁢ ( U ) + ϕ ̄ 1 ⁢ ( U ) . ∎

Proof of Theorem A.4

From (A.1), we have that n ⁢ dim ( U ) ≤ dim H ⁢ ( U ) for all subspaces 0 ≠ U ⊆ V , and hence

Moreover, we must have dim H ⁢ ( U 0 ) < ( n + 1 ) ⁢ dim ( U 0 ) for some 0 ≠ U 0 ⊆ V by Proposition A.7. Altogether this gives

and so equality must hold throughout completing the proof. ∎

Let U , W be vector spaces over a field 𝑘. Suppose that

are such that Φ = ( ϕ 1 , … , ϕ N ) : ⨁ N U → W is an injection. If 𝑍 is a subspace of 𝑊 such that dim ( Z ∩ Im ⁡ Φ ) = d ≤ N , then omitting some 𝑑 of the ϕ 1 , … , ϕ N will yield an injection ⨁ N − d U → W with image disjoint from 𝑍. In other words, after reordering ϕ 1 , … , ϕ N , one can ensure that Z ∩ ( ∑ i = d + 1 N Im ⁡ ϕ i ) = 0 .

Proof

We proceed by induction on 𝑑, noting first that the lemma is a tautology when d = 0 . Now, assume the statement holds for all 0 ≤ n < d and we have an injection

and a subspace Z ⊆ W with dim ( Z ∩ Im ⁡ Φ ) = d ≤ N . Let 0 ≠ v ⃗ ∈ ( Z ∩ Im ⁡ Φ ) and denote

Since ⋂ i = 1 N Im ⁡ Φ j = 0 , it follows that v ⃗ ∉ Im ⁡ Φ j for some 𝑗 which we may assume to be 1. In particular, Φ 1 : ⨁ N − 1 U → W is an injection with Z ∩ Im ⁡ Φ 1 ⊊ Z ∩ Im ⁡ Φ so that

Using the induction assumption on Φ 1 and 𝑍, it follows that we can reorder ϕ 2 , … , ϕ N to achieve 0 = ( Z ∩ ( ∑ i = n + 2 N Im ⁡ ϕ i ) ) ⊇ ( Z ∩ ( ∑ i = d + 1 N Im ⁡ ϕ i ) ) as desired. ∎

Let 𝑉 and 𝑊 be finite-dimensional vector spaces over a field 𝑘, and 𝐻 a subspace of Hom k ⁡ ( V , W ) . Suppose n ≥ 0 and 1 ≤ d < dim V are integers and assume the following conditions are satisfied.

1. There exist ϕ 1 , … , ϕ n ∈ H giving an injection Φ = ( ϕ 1 , … , ϕ n ) : ⨁ i = 1 n V → W .

2. We have dim ( H ⁢ ( U ) ) > n ⁢ dim ( U ) for any non-zero subspace 0 ≠ U ⊆ V .

3. Writing m : = ( dim ( V ) − d ) ⋅ dim ( V ) + 1 , there exist ψ 1 , … , ψ m ∈ H so that

   dim ( Im ⁡ Φ + ∑ j = 1 ℓ Im ⁡ ψ j ) ≥ d + dim ( Im ⁡ Φ + ∑ j = 1 ℓ − 1 Im ⁡ ψ j )

   for ℓ = 1 , … , m . In other words, we have that each ψ ℓ has rank at least 𝑑 modulo Im ⁡ Φ + ∑ j = 1 ℓ − 1 Im ⁡ ψ j .

Proof

In order to have the rank at least 𝑑 modulo Im ⁡ Φ + ∑ j = 1 ℓ − 1 Im ⁡ ψ j , each ψ ℓ must have rank at least 𝑑 modulo Im ⁡ Φ . The assertion follows trivially if any ψ ℓ has rank at least d + 1 modulo Im ⁡ Φ , so we may assume each ψ ℓ has rank exactly 𝑑 modulo either Im ⁡ Φ or Im ⁡ Φ + ∑ j = 1 ℓ − 1 Im ⁡ ψ j .

Let W ′ = W / Im ⁡ Φ , and for any ϕ ∈ Hom k ⁡ ( V , W ) , we denote by ϕ ′ ∈ Hom k ⁡ ( V , W ′ ) the map V → ϕ W → W ′ induced by the quotient. Since the rank of ψ ℓ ′ does not change after going modulo ∑ j = 1 ℓ − 1 Im ⁡ ψ j ′ , we have Im ⁡ ψ ℓ ′ ∩ ( ∑ j = 1 ℓ − 1 Im ⁡ ψ j ′ ) = 0 for any 1 ≤ ℓ ≤ m , and in particular, Im ⁡ ψ i ′ ∩ Im ⁡ ψ j ′ = 0 for any i ≠ j . If ever ker ⁡ ψ i ′ ≠ ker ⁡ ψ j ′ for some i ≠ j , then rank ⁡ ( ψ i ′ + ψ j ′ ) > d by Lemma A.6, and the assertion follows. Hence assume now that all of these kernels are equal, and set K = ker ⁡ ψ ℓ ′ for all 1 ≤ ℓ ≤ m .

Let 𝑈 be a vector space complement of 𝐾 in 𝑉 so that dim ( U ) = d = dim ( V ) − dim ( K ) and V = U + K with U ∩ K = 0 . Observe that Im ⁡ ψ ℓ ′ = ψ ℓ ′ ⁢ ( U ) , so the restriction of each ψ ℓ ′ to 𝑈 is injective as rank ⁡ ψ ℓ ′ = dim ( U ) = d . Moreover, setting

we similarly have that Ψ ′ | ⊕ m U is an injection as dim ( ∑ j = 1 m ψ j ′ ⁢ ( U ) ) = d ⁢ m . In particular, it follows that each ψ ℓ ′ has rank 𝑑 modulo ∑ j = 1 , j ≠ ℓ m Im ⁡ ψ j ′ , i.e., Im ⁡ ψ ℓ ∩ ( ∑ j = 1 , j ≠ ℓ m Im ⁡ ψ j ′ ) = 0 . Note that we may permute ψ 1 , … , ψ m as needed below while preserving our setup.

If H ⁢ ( K ) ⊈ Im ⁡ Φ , then we will find h ∈ H and v ⃗ ∈ K such that h ′ ⁢ ( v ⃗ ) ≠ 0 . Since Im ⁡ ψ ℓ are disjoint from each other, at most rank ⁡ h ′ ≤ dim ( V ) < m of the Im ⁡ ψ ℓ ′ can intersect Im ⁡ h ′ nontrivially, so after reordering, we may assume Im ⁡ ψ 1 ′ ∩ Im ⁡ h ′ = 0 (as in the proof of Lemma A.8). By Lemma A.6, rank ⁡ ( ψ 1 ′ + h ′ ) > rank ⁡ ψ 1 ′ = d and the assertion follows. Thus we may assume going forward that H ⁢ ( K ) ⊆ Im ⁡ Φ . In particular, note that this implies Im ⁡ Φ ≠ 0 so we must have n ≥ 1 .

Since dim ( H ⁢ ( K ) ) > n ⁢ dim ( K ) , there is some h ∈ H with h ⁢ ( K ) ⊈ Φ ⁢ ( K ) . Because

and an element of ⨁ n V is in ⨁ n K if and only if each of the components is in 𝐾, we may reorder the ϕ i so that

and choose v ⃗ ∈ K with

Let

and for any ϕ ∈ Hom k ⁡ ( V , W ) , we shall denote by ϕ ̄ ∈ Hom k ⁡ ( V , W ̄ ) the map V → ϕ W → W ̄ induced by the quotient by ∑ i = 2 n Im ⁡ ϕ i . It now suffices to find an injection ϕ ̄ and a map ψ ̄ that has rank d + 1 modulo ϕ ̄ . We break the rest of the proof up into two cases depending on whether we could assume that this ℎ is one of ψ i .

Since dim ( Hom k ⁡ ( K , Im ⁡ ϕ ̄ 1 ) ) = dim ( K ) ⋅ dim ( V ) = ( dim ( V ) − d ) ⋅ dim ( V ) < m , there must be a nontrivial linear combination ξ = ∑ i = 1 m α i ⁢ ψ i with α 1 , … , α m ∈ k not all zero and K ⊆ ker ⁡ ξ ̄ . If 0 ≠ u ⃗ ∈ U , then ξ ′ ⁢ ( u ⃗ ) = Ψ ′ ⁢ ( α 1 ⁢ u ⃗ , … , α m ⁢ u ⃗ ) ≠ 0 as Ψ ′ | ⊕ m U is an injection and α j ≠ 0 for some 𝑗. Thus ξ ̄ ⁢ ( u ⃗ ) ≠ 0 as well, so it follows that ξ ̄ | U is an injection, rank ⁡ ξ ̄ = d , and K = ker ⁡ ξ ̄ . Moreover, since ker ⁡ ψ ̄ ℓ ≠ K , 𝜉 cannot be a scalar multiple of ψ ℓ , and we must have that α j ≠ 0 for some j ≠ ℓ .

We will now show that ϕ : = ϕ 1 + ξ and ψ : = ψ ℓ are the required maps. Let us first check that Im ⁡ ξ ̄ ∩ ( Im ⁡ ϕ ̄ 1 + Im ⁡ ψ ̄ ℓ ) = 0 . We already know that

Thus if ξ ̄ ⁢ ( u ⃗ ) = ϕ ̄ 1 ⁢ ( a ⃗ ) + ψ ̄ ℓ ⁢ ( v ⃗ ) , then u ⃗ , v ⃗ ∈ U and ξ ̄ ⁢ ( u ⃗ ) − ψ ̄ ℓ ⁢ ( v ⃗ ) ∈ Im ⁡ Ψ ̄ ⁢ ( ⊕ m U ) ∩ Im ⁡ ϕ ̄ 1 . As Ψ ′ | ⊕ m U is injective, it follows that u ⃗ , v ⃗ = 0 as ξ ′ is an injection modulo ψ ℓ ′ . Thus we conclude Im ⁡ ξ ̄ ∩ ( Im ⁡ ϕ ̄ 1 + Im ⁡ ψ ̄ ℓ ) = 0 . In particular, we have 0 = Im ⁡ ξ ̄ ∩ Im ⁡ ϕ ̄ 1 , giving that ϕ ̄ = ϕ ̄ 1 + ξ ̄ is injective by applying Lemma A.6 and using that ϕ ̄ 1 is injective.

It remains to show that ψ ̄ ℓ has rank at least d + 1 modulo Im ⁡ ( ϕ ̄ 1 + ξ ̄ ) . To that end, let us first check that ψ ̄ ℓ has rank at least d + 1 modulo ϕ ̄ 1 ⁢ ( K ) . By our choice of v ⃗ ∈ K above, we have that ψ ̄ ℓ ⁢ ( v ⃗ ) ∈ Im ⁡ ϕ ̄ 1 ∖ ϕ ̄ 1 ⁢ ( K ) . Put T = U + k ⁢ v ⃗ , which has dimension d + 1 . Suppose we have u ⃗ ∈ U and λ ∈ k with ψ ̄ ℓ ⁢ ( u ⃗ + λ ⁢ v ⃗ ) ∈ ϕ ̄ 1 ⁢ ( K ) . It follows that ψ ̄ ℓ ⁢ ( u ⃗ ) ∈ Im ⁡ ϕ ̄ 1 , and so also ψ ℓ ′ ⁢ ( u ⃗ ) = 0 , which gives u ⃗ = 0 as ker ⁡ ψ ℓ ′ = K . Thus ψ ̄ ℓ ⁢ ( λ ⁢ v ⃗ ) = λ ⁢ ψ ̄ ℓ ⁢ ( v ⃗ ) ∈ ϕ ̄ 1 ⁢ ( K ) , which yields λ = 0 as ψ ̄ ℓ ⁢ ( v ⃗ ) ∉ ϕ ̄ 1 ⁢ ( K ) . It follows that ψ ̄ ℓ | T is injective modulo ϕ ̄ 1 ⁢ ( K ) , i.e., ψ ̄ ℓ is injective on 𝑇 and ψ ̄ ℓ ⁢ ( T ) ∩ ϕ ̄ 1 ⁢ ( K ) = 0 . To conclude the stronger statement that ψ ̄ ℓ has rank at least d + 1 modulo Im ⁡ ( ϕ ̄ 1 + ξ ̄ ) , it suffices verify Im ⁡ ( ϕ ̄ 1 + ξ ̄ ) ∩ ψ ̄ ℓ ⁢ ( T ) = 0 . Suppose we have some w ⃗ ∈ V with ϕ ̄ 1 ⁢ ( w ⃗ ) + ξ ̄ ⁢ ( w ⃗ ) ∈ ψ ̄ ℓ ⁢ ( T ) . It follows that ξ ̄ ⁢ ( w ⃗ ) ∈ Im ⁡ ξ ̄ ∩ ( Im ⁡ ϕ ̄ 1 + Im ⁡ ψ ̄ ℓ ) = 0 and w ⃗ ∈ ker ⁡ ξ ̄ = K . Thus we must have ϕ ̄ 1 ⁢ ( w ⃗ ) ∈ ψ ̄ ℓ ⁢ ( T ) ∩ ϕ ̄ 1 ⁢ ( K ) = 0 , so that ϕ ̄ 1 ⁢ ( w ⃗ ) + ξ ̄ ⁢ ( w ⃗ ) = 0 and hence Im ⁡ ( ϕ ̄ 1 + ξ ̄ ) ∩ ψ ̄ ℓ ⁢ ( T ) = 0 .

Since

we see that dim ( ∑ j = 1 m ψ ̄ j ⁢ ( U ) ) remains unchanged modulo Im ⁡ ϕ ̄ 1 , and thus

Set Z = h ̄ ⁢ ( U ) + Im ⁡ ϕ ̄ 1 . Then

and in particular,

Applying Lemma A.8 to ψ ̄ 1 | U , … , ψ ̄ m | U ∈ Hom ⁡ ( U , W ̄ ) and 𝑍, it follows that, after reordering ψ 1 , … , ψ m , we may assume

Since ψ ̄ d + 1 | K , … , ψ ̄ m | K ∈ Hom k ⁡ ( K , ϕ ̄ 1 ⁢ ( K ) ) and

there must be a nontrivial linear combination ξ = ∑ i = d + 1 m α i ⁢ ψ i with α d + 1 , … , α m ∈ k not all zero and K ⊆ ker ⁡ ξ ̄ . If 0 ≠ u ⃗ ∈ U , then ξ ′ ⁢ ( u ⃗ ) = Ψ ′ ⁢ ( 0 , … , 0 , α d + 1 ⁢ u ⃗ , … , α m ⁢ u ⃗ ) ≠ 0 as Ψ ′ | ⊕ m U is an injection and α j ≠ 0 for some d + 1 ≤ j ≤ m . Thus ξ ̄ ⁢ ( u ⃗ ) ≠ 0 as well, so it follows ξ ̄ | U is an injection, rank ⁡ ξ ̄ = d , and K = ker ⁡ ξ ̄ . Since m ≥ dim ( V ) + 1 ≥ d + 2 , we may choose ψ ℓ with d + 1 ≤ ℓ ≤ m so that α j ≠ 0 for some d + 1 ≤ j ≤ m and j ≠ ℓ . We know ψ ̄ ℓ | U is injective as ψ ℓ ′ | U is injective, and it follows from (A.3) that ψ ̄ ℓ ⁢ ( U ) ∩ h ̄ ⁢ ( U ) = 0 so that ψ ̄ ℓ + h ̄ restricts to an injection on 𝑈 by Lemma A.6. Furthermore, let us argue that

Suppose u ⃗ , u ⃗ ′ , u ⃗ ′′ ∈ U and w ⃗ ∈ V with ξ ̄ ⁢ ( u ⃗ ) = ψ ̄ ℓ ⁢ ( u ⃗ ′ ) + h ̄ ⁢ ( u ⃗ ′′ ) + ϕ ̄ 1 ⁢ ( w ⃗ ) . Then

by (A.3), which implies

giving that α j ⁢ u ⃗ = 0 by the injectivity of Ψ ′ | ⊕ m U as j ≠ ℓ ; thus we must have u ⃗ = 0 , and (A.4) follows.

We will now show that ϕ : = ϕ 1 + ξ and ψ : = ψ ℓ + h are the two required maps. Notice first that, since Im ⁡ ξ ̄ ∩ Im ⁡ ϕ ̄ 1 = ξ ̄ ⁢ ( U ) ∩ Im ⁡ ϕ ̄ 1 = 0 by (A.4) and ϕ ̄ 1 is injective, Lemma A.6 implies that ϕ ̄ 1 + ξ ̄ remains injective. To finish, we need to show that ψ ̄ ℓ + h ̄ has rank at least d + 1 modulo Im ⁡ ( ϕ ̄ 1 + ξ ̄ ) . By our choice of v ⃗ ∈ K above, we have that h ̄ ⁢ ( v ⃗ ) ∈ Im ⁡ ϕ ̄ 1 ∖ ϕ ̄ 1 ⁢ ( K ) . Put T = U + k ⁢ v ⃗ , which has dimension d + 1 . Suppose we have u ⃗ ∈ U and λ ∈ K with

As V = U + K , suppose u ⃗ ′ ∈ U and w ⃗ ∈ K with

Then, as v ⃗ , w ⃗ ∈ K = ker ⁡ ξ ̄ and H ⁢ ( K ) ⊆ Im ⁡ Φ , we see

and it follows that ξ ̄ ⁢ ( u ⃗ ′ ) = 0 by (A.4), and also u ⃗ ′ = 0 because ξ ̄ | U is an injection. Rearranging once again, we have