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 Incoming links: Coefficient normalization in an ideal of a tensor product

Let

J \neq = 0

A \otimes_{k} B

. Choose a nonzero

x \in J

n

, and take a shortest expression

x = \sum_{i = 1}^{n} a_{i} \otimes b_{i} .

Both families

(a_{i})

and

(b_{i})

are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular

a_{1}, b_{1} \neq = 0

Because

A

a_{1}

A

. Thus there are finitely many

u_{j}, v_{j} \in A

with

\sum_{j} u_{j} a_{1} v_{j} = 1

. Form

y = \sum_{j} (u_{j} \otimes 1) x (v_{j} \otimes 1) = \sum_{i = 1}^{n} c_{i} \otimes b_{i} \in J, c_{i} = \sum_{j} u_{j} a_{i} v_{j}, c_{1} = 1.

Since the

b_{i}

b_{1}

shows

y \neq = 0

n

and hence, by the choice of

x

, exactly

n

. Consequently

c_{1} = 1, c_{2}, \dots, c_{n}

Similarly, simplicity of

B

gives finitely many

p_{j}, q_{j} \in B

with

\sum_{j} p_{j} b_{1} q_{j} = 1

. Now

z = \sum_{j} (1 \otimes p_{j}) y (1 \otimes q_{j}) = \sum_{i = 1}^{n} c_{i} \otimes d_{i} \in J, d_{i} = \sum_{j} p_{j} b_{i} q_{j}, d_{1} = 1.

c_{i}

shows

z \neq = 0

, and minimality again makes its two-factor tensor rank exactly

n

. We have obtained

z = 1 \otimes 1 + i = 2 \sum n c_{i} \otimes d_{i} \in J ∖ {0} .

Both coefficient families in this expression are linearly independent. Retaining this additional property is the key to the next part. This is coefficient normalization in an ideal of a tensor product.