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I try to understand what is meant by the expression that $\mathbb{A}^1 \setminus \{0\}$ is the underlying variety of the multiplicative group $\mathbb{G}_m$ over the field $K$ .

I know that

  • $\mathbb{G}_m$ is the algebraic group over $K$ whose group action is given by multiplication $$\mu(x,y)=x \cdot y ~~~ \text{ and } ~~~ \iota(x)=x^{-1}$$

  • we can define $\mathbb{G}_m =$ Spec$(K[T, T^{-1}])$, i.e. it is an affine variety

  • $\mathbb{G}_m$ is the functor sending a $K$-scheme $X$ to the multiplicative group of invertible global sections of the structure sheaf, i.e. $\mathbb{G}_m(X)= \mathcal{O}_X(X)^*=K^*$

    But what exactly is meant by the above expression? Thank you for your explanations!

(EDIT: In this post How is the multiplicative group an algebraic variety? they even use it as an identity.)

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SallyOwens
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An algebraic group is a quadrauple $(G, m, i, 1)$ where $G$ is an algebraic variety over a field $k$, and $$m : G \times_k G \to G, i : G \to G, 1 : \mathrm{Spec}(k) \to G$$ are morphisms of varieties, satisfying some commutative diagrams as here. So the algebraic variety underlying an algebraic group $(G, m, i, 1)$ is simply the algebraic variety $G$.

By the way, it is not quite exact to say that an algebraic group is « a group that is also an algebraic variety ». What is true is that for any $k$-algebra $R$, the set $G(R)$ has a group structure (and actually the functor of points $R \mapsto G(R)$ factors through a functor $k$-$\sf Alg$ $\to \sf Grp$). The problem is that the set underlying $G \times_k G$ is not the set $|G| \times |G|$ (where $|G|$ is the set underlying the variety $G$), and it is difficult to compute in general. Therefore, a group structure on the set $|G|$, given in particular by a multiplication map $ \mu : |G| \times |G| \to |G|$, will not give you a morphism $m : G \times_k G \to G$ required in the definition of algebraic group (see also here).

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