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  "body": "\n## Proof\n\n1. Let $K \\subseteq L \\subseteq M$ be [[fields|0QA8CHH8MG]]. By [[existence of bases|00MS9VMQ9Z]], choose a [[basis|0VTQTPQGAV]] $(u_\\alpha)_{\\alpha \\in A}$ of $L$ over $K$ and a basis $(w_\\beta)_{\\beta \\in B}$ of $M$ over $L$.\n2. By the [[dimension theorem|0WK0Y8RH1M]], the [[degrees|0GS74FGFEK]] are $[L : K] = \\abs{A}$ and $[M : L] = \\abs{B}$, as [[cardinalities|0JTGRV4WT4]]. The product of cardinals is $\\abs{A} \\cdot \\abs{B} = \\abs{A \\times B}$, with the [[Cartesian product|0C81YBVTWH]].\n3. We show that the family $(u_\\alpha w_\\beta)_{(\\alpha, \\beta) \\in A \\times B}$ is a basis of $M$ over $K$.\n4. Spanning. Let $z \\in M$. Since the $w_\\beta$ [[span|088RYKAWK9]] $M$ over $L$, $z = \\sum_{\\beta \\in B_0} c_\\beta w_\\beta$ for a finite $B_0 \\subseteq B$ and $c_\\beta \\in L$.\n5. Since the $u_\\alpha$ span $L$ over $K$, each $c_\\beta$ is a $K$-combination of finitely many $u_\\alpha$. Let $A_0 \\subseteq A$ be the finite union of the index sets used. Adding zero coefficients, $c_\\beta = \\sum_{\\alpha \\in A_0} b_{\\alpha \\beta} u_\\alpha$ with $b_{\\alpha \\beta} \\in K$.\n6. Then $z = \\sum_{\\beta \\in B_0} \\sum_{\\alpha \\in A_0} b_{\\alpha \\beta} \\, u_\\alpha w_\\beta$, a [[linear combination|0V8F3E48E6]] over $K$ of the family of Step 3.\n7. [[Linear independence|0EVHJFTG9Z]]. Let $A_0 \\subseteq A$ and $B_0 \\subseteq B$ be finite, let $b_{\\alpha \\beta} \\in K$, and assume $\\sum_{\\alpha \\in A_0} \\sum_{\\beta \\in B_0} b_{\\alpha \\beta} u_\\alpha w_\\beta = 0$.\n8. Regroup as $\\sum_{\\beta \\in B_0} c_\\beta w_\\beta = 0$ with $c_\\beta = \\sum_{\\alpha \\in A_0} b_{\\alpha \\beta} u_\\alpha \\in L$.\n9. The $w_\\beta$ are linearly independent over $L$, so $c_\\beta = 0$ for each $\\beta \\in B_0$.\n10. The $u_\\alpha$ are linearly independent over $K$, so $b_{\\alpha \\beta} = 0$ for all $\\alpha \\in A_0$ and $\\beta \\in B_0$.\n11. Every finite subfamily of the family of Step 3 lies in one of the form of Step 7. So the family is linearly independent, and with Step 6 it is a basis.\n12. Two members with different indices are different. Otherwise $1 \\cdot v - 1 \\cdot v = 0$ would be a nontrivial relation, against Step 11. So the basis has $\\abs{A \\times B}$ elements.\n13. By the definition of [[dimension|05G2G0P224]] and Step 2, $[M : K] = \\abs{A \\times B} = \\abs{B} \\cdot \\abs{A} = [M : L] \\cdot [L : K]$.\n\n## Idea\n\nWrite an element of $M$ in the basis of $M$ over $L$, and then write each coefficient in the basis of $L$ over $K$. This shows that the products $u_\\alpha w_\\beta$ span $M$ over $K$. Reading the same computation backwards shows that they are independent.\n",
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