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According to various places, we define an algebraic group as a group that is also an algebraic variety (along with some compatibility conditions). Many places also list some examples, one of which is the multiplicative group $\mathbb{G}_m$ of a field $k$.

But $\mathbb{G}_m = \mathbb{A}^1\setminus\{0\}$, which is not Zariski-closed in $\mathbb{A}^1$ (i.e. is not a variety). So how exactly is it an algebraic group? I'm sure there's something really obvious that I've missed, but I'm new to this area and have yet to really get to grasps with such basics.

Tim
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2 Answers2

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A different way to view it is as the variety corresponding to the algebra $k[x,y]/(xy-1)$, so it is a closed subvariety of $\mathbb{A}^2$ (you should check that this all works out).

Tobias Kildetoft
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    Sorry, what does it mean to have a variety corresponding to an algebra? I'm really such a beginner at all this! So far my only concept of (affine) algebraic varieties is that of a set of common zeros of some polynomials. Does your answer mean that I should look at $\mathbb{V}(xy-1)\subset\mathbb{A}^2$? – Tim Jun 26 '15 at 14:03
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    @Tim Yes, precisely. – Tobias Kildetoft Jun 26 '15 at 14:04
  • So $\mathbb{V}(xy-1)={(t,t^{-1})\mid t\neq0}$, and this gives us the union of two copies of $\mathbb{A}^1\setminus{0}$. But if we intersect $\mathbb{V}(xy-1)$ and $\mathbb{V}(y)$ (to get just the one copy) we end up with the empty set, since there are no solutions to $xy-1$ with $y=0$... Where am I going wrong? – Tim Jun 26 '15 at 14:16
  • @Tim How do you get that to be two copies? – Tobias Kildetoft Jun 26 '15 at 14:17
  • From the $x$ and $y$ projections? – Tim Jun 26 '15 at 14:19
  • Oh no, obviously ${(t,t^{-1})\mid t\neq0}$ is isomorphic as a group to $\mathbb{G}_m$. So is that enough? – Tim Jun 26 '15 at 14:20
  • @Tim Yes (as long as you mean "isomorphic" to mean both as a group and as a variety). – Tobias Kildetoft Jun 26 '15 at 14:21
  • Is the isomorphism (as varieties) just the one given by $(t,t^{-1})\mapsto (t,0)$? – Tim Jun 26 '15 at 14:25
  • @Tim You want to map it to $\mathbb{A}^1$ (and not hit $0$). – Tobias Kildetoft Jun 26 '15 at 14:26
  • But doesn't that map have $\mathbb{A}^1\setminus{0}$ as its image? Since it maps to ${(t,0)\mid t\neq0}$, which is a copy of $\mathbb{A}^1\setminus{0}$ inside $\mathbb{A}^2$? (Sorry for all the questions!) – Tim Jun 26 '15 at 14:33
  • @Tim Right, but why not just map $(t,t^{-1})$ to $t$? – Tobias Kildetoft Jun 26 '15 at 14:34
  • Right, of course! Lots of thanks. – Tim Jun 26 '15 at 14:38
  • sorry to keep asking... but that's a map into $\mathbb{A}^1$, and so if it's an isomorphism of varieties shouldn't that tell us that the multiplicative group is closed in $\mathbb{A}^1$, which it isn't ? – Tim Jun 26 '15 at 15:28
  • @Tim It is only an isomorphic onto its image. This does not imply that the image is closed. – Tobias Kildetoft Jun 26 '15 at 15:33
  • of course ! thanks again – Tim Jun 26 '15 at 15:43
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To view $G_m$ as an affine variety, one needs to consider it inside $k^2$, namely as the zeros of $yx=0$ and $y=0$ in $k[x,y]$. The affine coordinate ring is $k[x,y]/(xy−1)=(k[x])[x^{−1}]=k[x,x^{−1}]$.

Dietrich Burde
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