Beautiful K-box and Their Partitions: Approach via tensor product

Definition. Let $F$$F$ be a field. We identify a box in $F^n$$F^n\text{}$ as a point $\begin{array}{c}(a_0,...,a_n)=\sum _{i=1}^n{a}_i{e}_i\end{array}$$(a_0,...,a_n)=\sum_{i=1}^n a_ie_i\\$. The $a_i$$a_i$ as the side-lengths of the box.

Remark. Let $F=\mathbb{R}$$F=\R$ then we get the traditional box.

Definition. Let $K\subset F$$K\subset F$ be a field. A box in $F^n$$F^n$ is called $K$$K$-beautiful if at least one of its side-lengths blong to $K$$K$ .

Proposition. Let $R$$R$ be a box that is partitioned into finitely many smaller box $R_1,...,R_n$$R_1,...,R_n$. If each $R_i$$R_i$ is $K$$K$-beautiful, then $R$$R$ itself is beautiful.

Proof. Consider the map $f:R\mapsto (a_0,...,a_n)\in F^n$$f: R ↦ (a_0,...,a_n) ∈ F^n$, that is, we only care about the side-lengths of a box.

Then we could consider the canonical map

$$\pi :F^n\to U^n,\text{where }U:=F/K\text{ as a quotient of vector space over }K$$

Then consider the canonical map from proudct to tensor product over $K$$K$.

$$u:U^n\to U^{\otimes n}$$

Since it is tensor product of vector space, we have $[a_1]\otimes ...\otimes [a_n]=0⟺$$[a_1]\otimes ...\otimes[a_n]=0\iff$ there exists a $a_i\in K$$a_i\in K$.

Hence a box $(a_1,...,a_n)$$(a_1,...,a_n)$ is $K$$K$-beautiful iff $[a_1]\otimes ...\otimes [a_n]=0$$[a_1]\otimes...\otimes[a_n]=0$.

The partition of a box, for example, rectangle, corresponds to

$$(a,b)=(a_1+a_2,b)=(a_1,b)+(a_2,b),(a,b)=(a,b_1+b_2)=(a,b_1)+(a,b_2)$$

Compose with $u\circ \pi$$u\circ \pi$ we have

$$(a_1+a_2)\otimes b=a_1\otimes b+a_2\otimes b,a\otimes (b_1+b_2)=a\otimes b_1+a\otimes b_2$$

Now assume that $R$$R$ could be partitioned into finitely many smaller box $R_1,...,R_n$$R_1,...,R_n$ and each $R_k$$R_k$ is $K$$K$-beautiful,

Then we have $u\circ \pi \circ f(R_k)=0$$u \circ \pi \circ f(R_k) = 0$ for all $1\le k\le n$$1 \leq k \leq n$, then

$$u\circ \pi (R)=\sum _{k=1}^{n}u\circ \pi (R_k)=0$$

Therefore, $R$$R$ is $K$$K$-beautiful. $◻$$\square$

Readers should compare it with the proof of Hilbert’s Third Problem