## 109.23 Taking scheme theoretic images

Let $k$ be a field. Let $t$ be a variable. Let $Y = \mathop{\mathrm{Spec}}(k[t])$ and $X = \coprod _{n \geq 1} \mathop{\mathrm{Spec}}(k[t]/(t^ n))$. Denote $f : X \to Y$ the morphism using the closed immersion $\mathop{\mathrm{Spec}}(k[t]/(t^ n)) \to \mathop{\mathrm{Spec}}(k[t])$ for each $n \geq 1$. In this case we have

The scheme theoretic image (Morphisms, Definition 29.6.2) of $f$ is $Y$. On the other hand, the image of $f$ is the closed point $t = 0$ in $Y$. Thus the underlying closed subset of the scheme theoretic image of $f$ is not equal to the closure of the image of $f$.

The formation of the scheme theoretic image does not commute with restriction to the open subscheme $V = \mathop{\mathrm{Spec}}(k[t, 1/t]) \subset Y$. Namely, the preimage of $V$ in $X$ is empty and hence the scheme theoretic image of $f|_{f^{-1}(V)} : f^{-1}(V) \to V$ is the empty scheme. This is not equal to $Y \cap V$.

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## Comments (2)

Comment #6729 by Wei Chen on

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