i \leq m$.

\[lemma:uniform\_bound\_d\_d+1\] Let $K$ be a field and $\ell$ a prime not equal to the characteristic of $K$. If $N$ is an $(n,\ell)$-uniform bound for $K $, then $N$ is also a $(d,
\ell)$-uniform bound for $K $ for all $d \geq n$.

Suppose $N$ is an $(n, \ell)$-uniform bound for $ K$. It is enough to prove the lemma for $d = n+1$. Let $L$ be a finite extension of $K$ and $\alpha_1, \cdots , \alpha_m
\in H^{n+1} (L, \mu_\ell^{\otimes (n+1)})$ be symbols. Then $\alpha_i =
\beta_i \cdot (a_i)$ for some symbols $\beta_i \in  H^{n}(L,
\mu_{\ell}^{\otimes n})$ and $a_i \in L^*$, $1 \leq i \leq m$. Since $N$ is an $(n, \ell)$-uniform bound for $K$, there exists a field extension $E$ of $L$ with $[E : L] \leq N$ and $\beta_ {i_E} = 0$ for $1 \leq i \leq
  m$. Then clearly $\alpha_{i_E} = 0$ for $1 \leq i \leq m$. Thus $N$ is also an $(n+1, \ell)$-uniform bound for $K$.

\[cor:symbol\_length\] Let $K$ be a field and $\ell$ a prime not equal to the characteristic of $K$. Suppose that $N$ is a $(2, \ell)$-uniform bound for $K$. Then for every $n \geq 2$, there exists an 