# Affine $n$-space over a scheme

In an exercise of Eisenbud-Harris The Geometry of Schemes, they ask to prove the following:

Let $$S$$ be any scheme. Let $$\mathbb{A}_{\mathbb{Z}}^n = \mathrm{Spec}\mathbb{Z}[x_1, \dots , x_n]$$ be the affine space over $$\mathrm{Spec} \mathbb{Z}$$. show that the affine space $$\mathbb{A}_S^n$$ over $$S$$ may be described as a product: $$\mathbb{A}_S^n = \mathbb{A}_{\mathbb{Z}}^n \times_{\mathrm{Spec} \mathbb{Z}} S$$.

The problem is that the definition for the fibered product of schemes $$X \times_S Y$$ they give in the book works when $$S$$ is not affine and we have a covering of $$S$$ by affine spectra $$\mathrm{Spec} R_i$$. Then we cover $$Y$$ and $$X$$ with the preimages of $$\mathrm{Spec} R_i$$ under the maps $$X \to S$$ and $$Y \to S$$ and glue together the resulting fibered product of affine schemes.

However, in this case, both $$\mathbb{A}_{\mathbb{Z}}^n$$ and $$\mathrm{Spec} \mathbb{Z}$$ are affine, and thus I don't know how apply their construction to this case. I thought of the following.

If $$S = \mathrm{Spec} R$$, then $$\mathrm{A}_{\mathbb{Z}}^n \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} R = \mathrm{Spec}( \mathbb{Z}[x_1, \dots , x_n] \otimes_{\mathbb{Z}} R) = \mathrm{Spec}(R[x_1, \dots , x_n]) = \mathbb{A}_R^n \, .$$

Then, if we cover $$S$$ by affine schemes $$U_i = \mathrm{Spec}R_i$$, we can consider the affine schemes $$\mathbb{A}_{\mathbb{Z}}^n \times_{\mathrm{Spec} R} \mathrm{Spec} R_i = \mathbb{A}_{R_i}^n$$ and then glue them by the maps induced by the identity maps on $$U_i \cap U_j$$.

If this indeed gives us a fibered product, then the construction agrees with the one they do when defining $$\mathbb{A}_S^n$$, but I'm not entirely sure.

• I don't understand -- if $S$ is affine then it has a singleton cover by affines (itself) and then you can directly apply the Eisenbud-Harris definition. – hunter Feb 10 '19 at 19:26
• @hunter But then I can't see how I get a construction that is analogous to $\mathbb{A}_S^n$. – user313212 Feb 10 '19 at 19:28

I believe you would like to see $$\mathbf{A}_S^n$$ as spectrum of some "ringed" thing even if $$S$$ is not affine.

This is not only reasonable, but actually true! However, the tool you need is not the algebraic spectrum of a ring but the relative Spec construction. If you are not familiar with it, see for example Stacks project.

Let $$S$$ be any scheme and let $$\mathscr{O}_S$$ be the structure sheaf. For every $$n>0$$ one can construct the quasi-coherent sheaf of $$\mathscr{O}_S$$-algebras $$\mathscr{O}_S[t_1,\ldots,t_n]:= \bigoplus_{i=1}^n \mathscr{O}_S$$.

Then we defined the affine space of dimension $$n$$ over $$S$$ as $$\mathbf{A}_S^n :=\mathbf{Spec} (\mathscr{O}_S[T_1 ,\ldots ,T_n])$$

This definition leads to the one you were given by considering the properties of the base change for the relative spectrum, that is to say: for every morphism of schemes $$f:U\to S$$ and for every sheaf of $$\mathscr{O}_S$$-algebras $$\mathscr{A}$$, one has

$$U\times _S \mathbf{Spec}(\mathscr{A}) \simeq \mathbf{Spec}(f^* \mathscr{A})$$

Now, in our case it is immediate to see that $$f^* \mathscr{O}_S[T_1,\ldots, T_n] \simeq \mathscr{O}_U [T_1,\ldots, T_]$$ so

$$U\times_S \mathbf{A}^n_S \simeq \mathbf{A}^n_U$$

So taking $$S = \mathrm{Spec}\,\mathbb {Z}$$ and $$f$$ as the structure morphism of $$U$$ we obtain your definition.