For a real Laurent polynomial f in u/s of width k: if k ≥ 3, or k = 2 and f(−λ̄) ≠ 0, the real pencil space W(λ, f) contains none of u², ū², 1/(u − b)

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and let $s \in \overline{\mathbb{Q}}$ be real with $s^2 = u\bar u$ (so $\bar u = s^2/u$; with $v = u/s$, conjugation acts as $v \mapsto 1/v$ on coefficients). Let $\lambda \in \overline{\mathbb{Q}}$ with $|\lambda| = 1$, and let $f = \sum_{|j| \le k} c_j (u/s)^j$ with $c_j \in \overline{\mathbb{Q}}$, $c_{-j} = \bar c_j$ and $c_k \neq 0$ (so $f$ is real). Let $W(\lambda, f)$ be the \overline{\mathbb{Q}}-span of $1, u, \bar u, f, f/(s + \lambda u)$, and write $f(-\bar\lambda)$ for $\sum_j c_j(-\bar\lambda)^j$, the value at $v = -\bar\lambda$ (the pole of $f/(s + \lambda u)$). Suppose the five numbers are free, and that $k \geq 3$, or $k = 2$ and $f(-\bar\lambda) \neq 0$. Then $W(\lambda, f)$ contains none of $u^2$, $\bar u^2$ and $1/(u - b)$ for non-zero $b \in \overline{\mathbb{Q}}$.

Proof. Put $v = u/s$ (transcendental) and $P(X) = \sum_j c_jX^{j+k}$, so $v^kf = P(v)$, $\deg P = 2k$, $P(0) = c_{-k} \neq 0$. A membership $t = a_1 + a_2u + a_3\bar u + a_4f + a_5g$, multiplied by $N(v)\,s\,v^k(1 + \lambda v)$ ($N = 1$ for $u^2$ and $\bar u^2$, $N = sX - b$ for $1/(u - b)$), is an identity $T$ in \overline{\mathbb{Q}}$[X]$. Its degree forces $a_4 = 0$, except for $u^2$ at $k = 2$. At $X = 0$ it gives $N(0)(a_4s + a_5)c_{-k} = T(0)$, and $T(0) = 0$ except for $\bar u^2$ at $k = 2$. For $k = 2$, at $X = -\bar\lambda$ the factor $1 + \lambda X$ vanishes and $a_5P(-\bar\lambda) = 0$, with $P(-\bar\lambda) = \bar\lambda^2f(-\bar\lambda) \neq 0$. In every case $a_4 = a_5 = 0$, and $t \in \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ is refuted by a cubic in $u$.

Novelty. Not found in the sources read; it is short.