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Let $A$ be a ring, and let $S$ be the scheme $\textrm{Spec }A$. Then the $S$-scheme $G = \textrm{Spec }A[T]$ has the structure of a group scheme over $S$, via the morphisms of $S$-schemes $$m: G \times_S G \rightarrow G, \iota: G \rightarrow G, e: S \rightarrow G$$

whose corresponding $A$-algebra homomorphisms are given respectively by

$$m^{\ast}: A[T] \rightarrow A[X,Y], \space \space f(T) \mapsto f(X+Y)$$ $$\iota^{\ast}: A[T] \rightarrow A[T], \space \space f(T) \mapsto f(-T)$$

$$e^{\ast}: A[T] \rightarrow A, \space \space f(T) \mapsto f(0)$$

If $X$ is any scheme over $S$, then we can identify $$G(X) = \textrm{Hom}_{S-\textrm{sch}}(X,G) = \textrm{Hom}_{A-\textrm{alg}}(A[T], \mathcal O_X(X)) = \mathcal O_X(X)$$

and the morphisms endow $G(X) = \mathcal O_X(X)$ with a group structure identical to the existing additive group structure on $\mathcal O_X(X)$.

Now, what if we replace the affine scheme $S$ with an arbitrary scheme? Does there exist an $S$-group scheme $G$ such that $G(X) = (\mathcal O_X(X),+)$ for all $S$-schemes $X$? The first thing I thought to try was $\textrm{Spec } \mathcal O_S(S)[T]$, but this probably isn't the correct object.

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    $\begingroup$ Don't know why this was downvoted, +1 $\endgroup$ – user223391 Jun 5 '17 at 22:17
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As I typed my question, I realized the correct object to look at was

$$G = S \times \textrm{Spec } \mathbb{Z}[T]$$

(product in the category of schemes). For an $S$-scheme $X$, to give a morphism of $S$-schemes $X \rightarrow G$ is the same as giving a morphism of schemes $X \rightarrow \textrm{Spec } \mathbb{Z}[T]$, so we can identify

$$G(X) = \textrm{Hom}_{\textrm{Sch}}(X, \textrm{Spec } \mathbb{Z}[T]) = \textrm{Hom}_{\textrm{Ring}}(\mathbb{Z}[T], \mathcal O_X(X)) = \mathcal O_X(X)$$

The group operations on $G$ are obtained by taking the obvious group operations on $\textrm{Spec } \mathbb{Z}[T]$, and tensoring with $1_S$.

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  • $\begingroup$ Another way to think about this is to use the general fact that if $H$ is a group scheme over $S_0$, and $S \rightarrow S_0$ is a morphism of schemes, then $H \times_{S_0} S$ is a group scheme over $S$. Now take $H = \textrm{Spec } \mathbb{Z}[T], S_0 = \textrm{Spec } \mathbb{Z}$. $\endgroup$ – D_S Jun 5 '17 at 20:44

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