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Let $f:X\rightarrow Y$ a faithfully flat morphism between $k$-schemes. We assume that the fibers are locally of finite type, do we have that $f$ is locally of finite type?

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No. Let $Y$ be $\text{Spec} \mathbb{Z}$. Let $X$ be $\text{Spec} (\mathbb{Z} \times \mathbb{Q})$.

$\textbf{Edit}.$ As pointed out, the OP wants an example over a field. As the commenters explain, the same idea works over a field $k$ with $Y = \text{Spec} k[t]$ and with $X=\text{Spec}(k[t] \times k(t))$.

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it's not over a field. Sorry, I omit to precise but $k$ is a field. –  prochet Jun 23 '13 at 15:10
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Standard dictionary: change $\mathbb{Z}$ to $k[t]$ and $\mathbb{Q}$ to $k(t)$. –  S. Carnahan Jun 23 '13 at 15:59
    
The same trick will work for things over a field. You can let $Y = \text{Spec }k[x]$ and $X = \text{Spec }(k[x] \oplus k(x) )$. $$\text{ }$$ The map $k[x] \to k[x] \oplus k(x)$ is the natural one on each coordinate. All the fibers are finite type, but the map is not finite type. It is also clearly flat. –  Karl Schwede Jun 23 '13 at 16:00

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