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$\operatorname{Spec} \mathbb Z[x]$ is a group scheme. Then there is a morphism of schemes: $\operatorname{Spec} \mathbb Z[x]\times\operatorname{Spec} \mathbb Z[x]\rightarrow\operatorname{Spec} \mathbb Z[x]$ induced by the coaddition $a:\mathbb Z[x]\rightarrow \mathbb Z[x] \otimes\mathbb Z[x]$ mapping $x$ to $1\otimes x+x\otimes 1$. Now, the coaddition is not a morphism of rings, since it does not respect multiplication: $a(x^2)=(1\otimes x^2+x^2\otimes 1)\neq(1\otimes x + x\otimes 1)^2=a(x)\cdot a(x)$.

But $a$ should be a morphism of rings, since the addition on $\operatorname{Spec} \mathbb Z[x]$ is a morphism of affine schemes.

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I think you are being confused by the notation. There is a canonical isomorphism $\mathbb{Z}[t] \otimes_{\mathbb{Z}} \mathbb{Z}[t] \cong \mathbb{Z}[x, y]$ such that $t \otimes 1 \mapsto x$ and $1 \otimes t \mapsto y$. In that light, coaddition is the unique ring homomorphism $\alpha : \mathbb{Z}[t] \to \mathbb{Z}[x, y]$ such that $t \mapsto x + y$, since $\mathbb{Z}[t]$ is freely generated by $t$. In particular, $\alpha(t^2) = x^2 + 2 x y + y^2 = \alpha(t)^2$.

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Thank you very much! – Felix Wellen Dec 1 '11 at 15:04

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