Lemma 110.28.1. There exists a finite type morphism of algebraic spaces $Y \to X$ with $Y$ affine and $X$ quasi-separated, such that there does not exist an immersion $Y \to \mathbf{A}^ n_ X$ over $X$.

## 110.28 Affines over algebraic spaces

Suppose that $f : Y \to X$ is a morphism of schemes with $f$ locally of finite type and $Y$ affine. Then there exists an immersion $Y \to \mathbf{A}^ n_ X$ of $Y$ into affine $n$-space over $X$. See the slightly more general Morphisms, Lemma 29.39.2.

Now suppose that $f : Y \to X$ is a morphism of algebraic spaces with $f$ locally of finite type and $Y$ an affine scheme. Then it is not true in general that we can find an immersion of $Y$ into affine $n$-space over $X$.

A first (nasty) counter example is $Y = \mathop{\mathrm{Spec}}(k)$ and $X = [\mathbf{A}^1_ k/\mathbf{Z}]$ where $k$ is a field of characteristic zero and $\mathbf{Z}$ acts on $\mathbf{A}^1_ k$ by translation $(n, t) \mapsto t + n$. Namely, for any morphism $Y \to \mathbf{A}^ n_ X$ over $X$ we can pullback to the covering $\mathbf{A}^1_ k$ of $X$ and we get an infinite disjoint union of $\mathbf{A}^1_ k$'s mapping into $\mathbf{A}^{n + 1}_ k$ which is not an immersion.

A second counter example is $Y = \mathbf{A}^1_ k \to X = \mathbf{A}^1_ k/R$ with $R = \{ (t, t)\} \amalg \{ (t, -t), t \not= 0\} $. Namely, in this case the morphism $Y \to \mathbf{A}^ n_ X$ would be given by some regular functions $f_1, \ldots , f_ n$ on $Y$ and hence the fibre product of $Y$ with the covering $\mathbf{A}^{n + 1}_ k \to \mathbf{A}^ n_ X$ would be the scheme

with obvious morphism to $\mathbf{A}^{n + 1}_ k$ which is not an immersion. Note that this gives a counter example with $X$ quasi-separated.

**Proof.**
See discussion above.
$\square$

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