Map Induced by the Multiplication of $\mathbb G_m$ I realize that since $\mathbb G_m$ is equal to $\mathrm{Spec}\, k[x, x^{-1}]$, the multiplication $m : \mathbb G_m \times \mathbb G_m \rightarrow \mathbb G_m$ is induced by some algebra homomorphism $k[x, x^{-1}] \rightarrow k[x, x^{-1}] \otimes k[x, x^{-1}]$.
My question is: how can we describe this homomorphism explicitly?
In particular, in Milne's book on algebraic groups, this map is denoted $\Delta$, and it is claimed that $\Delta : x \mapsto x \otimes x$. How can I see that this statement is true?
 A: $m$ is the restriction of the multiplication map $\mathbb{A}^1 \times \mathbb{A}^1 \to \mathbb{A}^1$. What is such a map? it's exactly a polynomial in two variables. What polynomial must it be? By definition, it's the polynomial $z = xy$. But this says exactly that the function $z : \mathbb{A}^1 \to \mathbb{A}^1$ (given by the identity, but thought of as a coordinate function) pulls back to the function $xy = x \otimes y : \mathbb{A}^1 \times \mathbb{A}^1 \to \mathbb{A}^1$. So $\Delta(z) = x \otimes y$.
This may seem tautological; it's a bit hard to keep track of what's going on here because $\mathbb{A}^1$ is playing four different roles in this argument. You can verify, for example, that the corresponding map for the additive group $\mathbb{G}_a \times \mathbb{G}_a \to \mathbb{G}_a$ is $\Delta(z) = x \otimes 1 + 1 \otimes y$, to get a sense of how it works. Alternatively you can argue via functors of points. The general pattern is that $\Delta$, the comultiplication, literally just outputs the polynomial(s) defining multiplication; this is not supposed to be esoteric. For an extra challenge you can try to work out the comultiplication corresponding to matrix multiplication $GL_n \times GL_n \to GL_n$.
A: Consider the homomorphism $k[x,x^{-1}]\to k[x_1,x_1^{-1}]\otimes k[x_2,x_2^{-1}]\cong k[x_1^{\pm1},x_2^{\pm1}]:x\mapsto x_1\otimes x_2\mapsto x_1x_2$. The prime ideal $(x_1-a,x_2-b)$ for $a,b\in k^\times$ pulls back to $\{x^{-n}p(x)\in k[x^{\pm1}]:p(x_1x_2)\in (x_1-a,x_2-b)\}=\{x^{-n}p(x)\in k[x^{\pm1}]:p(ab)=0\}=(x-ab)$.
