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Let $A$ be a ring then, a homomorphism $A\rightarrow A[x_1,\cdots,x_n]$ induces a finite type morphism between spectrums. I want to find the map which is a locally finite type but not finite type....

Sang Cheol Lee
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    Try $k[x]\rightarrow k(x)$. – Marci Jul 06 '13 at 05:15
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    Dear Marci, $k[x]→k(x)$ is the inclusion?? Then, $\operatorname{spec}k(x)={0}$... so is $\operatorname{spec}k(x) \rightarrow \operatorname{spec}k[x]$ is finite type?? – Sang Cheol Lee Jul 07 '13 at 15:22
  • The map $k[x]\rightarrow k(x)$ is not of finite type, $k(x)$ is not finitely generated $k[x]$-algebra. – Marci Jul 07 '13 at 18:38
  • sorry... but then is this map locally finte type??? – Sang Cheol Lee Jul 08 '13 at 04:04
  • Assume k is algebraically closed. Then every maximal ideal is given by $x=a$ for some $a\in k$. Then locally (for example around 0) $k[x]$ becomes $k[x]{(x)}$. Moreover $k(x)$ is finitely generated over $k[x]{(x)}$, you only need to add $1/x$. – Marci Jul 08 '13 at 04:12
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    The comments by Marci are wrong. For morphisms of affine schemes, locally of finite type = finite type. Affine morphisms are quasi-compact. – Martin Brandenburg May 20 '15 at 14:25

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A morphism is called of finite type if it is quasi-compact and locally of finite type. If $f_i : X_i \to Y$ are morphisms of finite type for $i \in I$, then clearly the induced morphism $f : \coprod_{i \in I} X_i \to Y$ is locally of finite type, but it is usually not of finite type when $I$ is infinite. So, for example, if $A$ is any commutative ring $ \neq 0$, then the canonical morphism $\coprod_{n \geq 0} \mathrm{Spec}(A) \to \mathrm{Spec}(A)$ locally of finite type, but it is not of finite type.

Martin Brandenburg
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