Lemma 10.116.6. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Set $X = \mathop{\mathrm{Spec}}(S)$. Let $K/k$ be a field extension. Set $S_ K = K \otimes _ k S$, and $X_ K = \mathop{\mathrm{Spec}}(S_ K)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$ and let $\mathfrak q_ K \subset S_ K$ be a prime corresponding to $x_ K \in X_ K$ lying over $\mathfrak q$. Then $\dim _ x X = \dim _{x_ K} X_ K$.

**Proof.**
Choose a presentation $S = k[x_1, \ldots , x_ n]/I$. This gives a presentation $K \otimes _ k S = K[x_1, \ldots , x_ n]/(K \otimes _ k I)$. Let $\mathfrak q_ K' \subset K[x_1, \ldots , x_ n]$, resp. $\mathfrak q' \subset k[x_1, \ldots , x_ n]$ be the corresponding primes. Consider the following commutative diagram of Noetherian local rings

Both vertical arrows are flat because they are localizations of the flat ring maps $S \to S_ K$ and $k[x_1, \ldots , x_ n] \to K[x_1, \ldots , x_ n]$. Moreover, the vertical arrows have the same fibre rings. Hence, we see from Lemma 10.112.7 that $\text{height}(\mathfrak q') - \text{height}(\mathfrak q) = \text{height}(\mathfrak q_ K') - \text{height}(\mathfrak q_ K)$. Denote $x' \in X' = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$ and $x'_ K \in X'_ K = \mathop{\mathrm{Spec}}(K[x_1, \ldots , x_ n])$ the points corresponding to $\mathfrak q'$ and $\mathfrak q_ K'$. By Lemma 10.116.4 and what we showed above we have

and the lemma follows. $\square$

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