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Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(\mathbb{A}^1_k,\mathbb{A}^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(\mathbb{A}^1,\mathbb{A}^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $\mathbb{A}^1_S\to \mathbb{A}^1_S$ of $S$-schemes.)

Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.

Let X be a (positive-dimensional) variety and let $f:\mathbb{A}^1_k\to X$ be a finite morphism. How does one show that the hom-functor $\mathrm{Hom}_k(\mathbb{A}^1_k,X)$ is not representable by an algebraic space?

What have I tried? Well, there is a natural morphism Isom$(\mathbb{A}^1,\mathbb{A}^1) \to \mathrm{Hom}(\mathbb{A}^1_k,X)$ which sends $g$ to $f\circ g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?

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  • $\begingroup$ Probably stupid question: Isn't $\text{Isom}(\mathbb{A}^1_k,\mathbb{A}^1_k)$ just $\mathbb{A}^1_k\rtimes\mathbb{G}_{m,k}$? That looks representable (by a scheme!) to me...? $\endgroup$ – Will Chen Sep 16 '18 at 20:18
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    $\begingroup$ @WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf $\endgroup$ – Anne F. Sep 16 '18 at 20:21
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Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $\mathcal{O}_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $\text{Spec}(k)$. The Hom functor $\text{Hom}_{k-\text{Sch}}(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.

If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $\Omega_{X/k,x}\otimes_{\mathcal{O}_{X,x}} k = \mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2$ has positive dimension as a $k$-vector space.

Denote by $$c_x:Y\to X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$\text{Hom}_{\mathcal{O}_Y}(c_x^*\Omega_{X/k},\mathcal{O}_Y) = \mathcal{O}_Y \otimes_k \left(\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.

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    $\begingroup$ Thank you for this clear answer. I did not realise that the argument which proves Isom($\mathbb{A}^1,\mathbb{A}^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".) $\endgroup$ – Anne F. Sep 16 '18 at 23:36
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    $\begingroup$ @AnneF. Thank you for catching the typo. $\endgroup$ – Jason Starr Sep 17 '18 at 1:19

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