Lemma 59.95.4. Let $L/K$ be a field extension. Then we have $\text{cd}(L) \leq \text{cd}(K) + \text{trdeg}_ K(L)$.

**Proof.**
If $\text{trdeg}_ K(L) = \infty $, then this is clear. If not then we can find a sequence of extensions $L= L_ r/L_{r - 1}/ \ldots /L_1/L_0 = K$ such that $\text{trdeg}_{L_ i}(L_{i + 1}) = 1$ and $r = \text{trdeg}_ K(L)$. Hence it suffices to prove the lemma in the case that $r = 1$. In this case we can write $L = \mathop{\mathrm{colim}}\nolimits A_ i$ as a filtered colimit of its finite type $K$-subalgebras. By Lemma 59.95.2 it suffices to prove that $\text{cd}(A_ i) \leq 1 + \text{cd}(K)$. This follows from Lemma 59.95.3.
$\square$

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