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Source. Published pp. 351–354, Theorem 16 (published scan).

Statement. Let $F$ be a field, $c_1,\ldots,c_k\in F$, and $b\in F$ with $b\ne0$. There is a finite coloring $\chi$ of $F$ for which

$$\sum_{i=1}^k c_i(x_i-x_i')=b$$

has no solution satisfying $\chi(x_i)=\chi(x_i')$ for every $i$. Different pairs may have different colors. Coefficients equal to zero can be deleted; if none remain, the assertion is immediate.

Complete proof. We prove the result first for prime fields, then show that it survives adjoining one transcendental and a finite algebraic extension, and finally reduce the arbitrary field to those cases.

For a finite prime field, give every element its own color. Equal-colored pairs have zero difference. For $F=\mathbb Q$, multiply the equation by a common denominator to make all $c_i,b$ integers. Choose a prime $p$ not dividing the nonzero integer $b$ and an integer $M\ge\max(1,\sum_i|c_i|)$. Give a rational $x$ the color

$$\left(\lfloor x\rfloor\bmod p,\ \lfloor M\{x\}\rfloor\right), \qquad \{x\}=x-\lfloor x\rfloor\in[0,1).$$

If every pair has the same color, then $\sum_i c_i(\lfloor x_i\rfloor-\lfloor x_i'\rfloor)$ is divisible by $p$, whereas

$$\left|\sum_i c_i(\{x_i\}-\{x_i'\})\right| <\frac{\sum_i|c_i|}{M}\le1.$$

The first sum cannot be the integer $b$ modulo $p$, and its distance from $b$ is at least one. This contradicts their sum being $b$. The strict inequality comes from both fractional parts lying in the same half-open interval of length $1/M$.

Now assume the theorem for a field $F$ and consider $F(t)$ with $t$ transcendental. Clear denominators, so $c_i(t)$ and $b(t)\ne0$ are polynomials. If $F$ is infinite, replace $t$ by $u+a$ for an $a\in F$ with $b(a)\ne0$. Such an $a$ exists because a nonzero polynomial has at most its degree many roots. If $F$ is finite, first pass to a finite extension $E$ with more than $\deg b$ elements, on which the theorem is trivial by injective coloring, choose such an $a\in E$, and work in $E(u)$. Restricting a resulting coloring back to $F(t)$ suffices. Arbitrarily large finite extensions exist elementarily: over a field of size $q$, the polynomial $X^q-X-1$ has no root, so an irreducible factor gives a proper finite extension; repeating produces unbounded sizes. We may therefore assume that $b(0)\ne0$ over a base field where the theorem is known.

Let $m=\max_i\deg c_i$ and write $c_i(t)=\sum_{j=0}^m c_{ij}t^j$. Each rational function has a unique finite principal part at zero,

$$x=tA(t)+\sum_{j\ge0}a_jt^{-j},$$

where $A$ is regular at zero and all but finitely many $a_j$ vanish. To see this, factor a power of $t$ from numerator and denominator; the remaining denominator has nonzero constant term and a uniquely determined formal power-series inverse. The finitely many terms of exponent at most zero give the displayed principal part. Formal coefficient extraction preserves addition and multiplication of rational functions.

The constant coefficient of the proposed equation is

$$\sum_{i=1}^k\sum_{j=0}^m c_{ij}(a_{ij}-a_{ij}')=b(0)\ne0.$$

The base-field theorem supplies a finite coloring $\psi$ ruling out this equation whenever each displayed pair has equal $\psi$-color. Color $x$ by $(\psi(a_0),\ldots,\psi(a_m))$. Equal colors of each original pair imply exactly those equalities, a contradiction. This proves the transcendental extension step.

For a finite extension $L/F$, fix an $F$-basis $\omega_1,\ldots,\omega_d$ and reorder it so that the coefficient $b_1$ of $b$ at $\omega_1$ is nonzero. Write

$$c_i=\sum_\alpha c_{i\alpha}\omega_\alpha,\quad x_i=\sum_\beta a_{i\beta}\omega_\beta,\quad \omega_\alpha\omega_\beta=\sum_\gamma \lambda_{\alpha\beta\gamma}\omega_\gamma.$$

The coefficient at $\omega_1$ gives

$$\sum_{i,\alpha,\beta}c_{i\alpha}\lambda_{\alpha\beta1} (a_{i\beta}-a_{i\beta}')=b_1\ne0.$$

Apply the theorem over $F$ to this finite list of coefficients and color $x=\sum_\beta a_\beta\omega_\beta$ by all $d$ colors of its coordinates. Equal original colors imply equality for every required pair of coordinates, which is impossible by the chosen base-field coloring. Repeated coordinate pairs in the displayed sum cause no difficulty.

Finally let $F$ be arbitrary and let $F_0=\Pi(c_1,\ldots,c_k)$, where $\Pi$ is its prime field. This finitely generated field is finite algebraic over a purely transcendental extension of $\Pi$: take a maximal algebraically independent subfamily of the finitely many generators; each remaining generator is algebraic, and finitely many finite algebraic extensions have finite total degree. The steps already proved give the theorem over $F_0$ with right side $1$.

Choose an $F_0$-basis of $F$ containing the nonzero element $b$. Let $\pi:F\to F_0$ be the coefficient of $b$, so that $\pi(b)=1$ and $\pi$ is $F_0$-linear. Color $x\in F$ by the established color of $\pi(x)$. Applying $\pi$ to a forbidden equation would give $\sum_i c_i(\pi(x_i)-\pi(x_i'))=1$ with every pair equally colored, contradiction. This completes every field case. $\square$

The basis step uses the usual vector-space basis principle (choice). The source's Laurent and finite-extension arguments are expanded above; the field theorem itself is not treated as an unexplained external input.

Bears on. #174.