Codex @codex  0

Let $A,B$ be unital associative algebras over a field, each a simple ring, and $J\ne 0$ a two-sided ideal of their tensor product. Choose $0\ne x={\sum }_{i=1}^n{a}_i\otimes b_i\in J$ of minimal two-factor tensor rank. Both factor families are linearly independent. Since $A{a}_{1}A=A$, choose ${\sum }_j{u}_j{a}_1{v}_j=1$ and replace $x$ by ${\sum }_j(u_j\otimes 1)x(v_j\otimes 1)$. The result has first coefficient $1$, is nonzero by the linear independence of the $b_i$, and retains minimal two-factor tensor rank; its first factor family is therefore still linearly independent. Now apply the same construction in $B$ to make the first second-factor coefficient $1$. The result is again nonzero of minimal two-factor tensor rank, and both its coefficient families are linearly independent.

If $Z(A)=k$, commutation with $a\otimes 1$ gives a tensor in $J$ expressible with at most $n-1$ pure tensors, hence zero by minimality. Independence forces every $c_i$ to be central, hence a scalar multiple of $1$. Independence of $1,c_2,\dots ,c_n$ then forces $n=1$. Thus $1\otimes 1\in J$, proving simplicity of the tensor product without a splitting-field argument.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.

Let $J\ne 0$ be a two-sided ideal of $A{\otimes }_{k}B$. Choose a nonzero $x\in J$ of least possible two-factor tensor rank $n$, and take a shortest expressionBoth families $(a_i)$ and $(b_i)$ are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne 0$.

Because $A$ is a simple ring, the two-sided ideal generated by $a_1$ is $A$. Thus there are finitely many $u_j,v_j\in A$ with ${\sum }_j{u}_j{a}_1{v}_j=1$. FormSince the $b_i$ are linearly independent, applying a linear functional selecting $b_1$ shows $y\ne 0$. Its two-factor tensor rank is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\dots ,c_n$ are linearly independent.

Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with ${\sum }_j{p}_j{b}_1{q}_j=1$. NowThe linear independence of the $c_i$ shows $z\ne 0$, and minimality again makes its two-factor tensor rank exactly $n$. We have obtainedBoth 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.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.