I am reading Springer's Linear Algebraic Groups and have a question about how ordinary group multiplication ( in the ordinary group theory sense) translates to that in term so linear algebraic groups.

Consider $\Bbb{C}$ as an affine algebraic variety (we may view $\Bbb{C}$ as the vanishing locus of the zero ideal) and also as a group under addition. Now if I am right, addition and taking inverses should now translate as a morphism (of varieties) from $\Bbb{C}$ to itself. Let us call these $\mu$ and $i$ respectively.

Now $\mu$ and $i$ are then supposed to define maps from the coordinate ring of $\Bbb{C}$ (namely $\Bbb{C}[T]$, $T$ an indeterminate) to $\Bbb{C}[T \times T]$ and $\Bbb{C}[T]$ respectively. I call these maps $\Delta : \Bbb{C}[T] \longrightarrow \Bbb{C}[T] \otimes_\Bbb{C} \Bbb{C}[T]$ and $\iota : \Bbb{C}[T] \longrightarrow \Bbb{C}[T]$.

My question: This may perhaps be elementary, but how do we know that the ordinary group laws of addition and taking inverses descend to these $\Bbb{C}$ - algebras as maps $\Delta(T) = T + T$ and $\iota(T) = -T$? Also, the way I have defined $\Delta$ now looks like I can only add an element to itself. What's happening here?


  • $\begingroup$ That $\mathbb{C}[T \times T]$ looks a bit strange. $\endgroup$
    – t.b.
    Sep 10, 2012 at 0:24
  • $\begingroup$ @t.b. What should I change it to? $\endgroup$
    – user38268
    Sep 10, 2012 at 0:26
  • $\begingroup$ I don't know, but I'd write something like $\mathbb{C}[T,U] \cong \mathbb{C}[T] \mathbin{\otimes_{\mathbb{C}}} \mathbb{C}[T]$. $\endgroup$
    – t.b.
    Sep 10, 2012 at 0:30
  • $\begingroup$ @t.b. I agree it would be better to introduce another indeterminate. Then in Qiaochu's answer below I would have $T \otimes 1 + 1 \otimes T \mapsto T + U$ yes? $\endgroup$
    – user38268
    Sep 10, 2012 at 0:34
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$
    – t.b.
    Sep 10, 2012 at 0:42

2 Answers 2


The short answer is that addition and inverses are both given by polynomials (namely $(x, y) \mapsto x + y$ and $x \mapsto -x$ respectively). Everything else follows from the fact that a polynomial map between affine varieties is the same thing as an algebra homomorphism between rings of regular functions in the other direction. (If you haven't thoroughly worked through the details of how this works I strongly recommend that you do so.)

Your definition of $\Delta$ is wrong; you actually want $\Delta(T) = T \otimes 1 + 1 \otimes T$. It is crucial that the two $T$s on the RHS are different; in particular, under a homomorphism out of $\mathbb{C}[T] \otimes\mathbb{C}[T]$ they will in general take two different values, which are the two different elements of the group you want to add. Remember that when you pass between algebra and geometry all of the morphisms change direction.

  • $\begingroup$ I will try to explain how the addition map $\phi : \Bbb{C} \times \Bbb{C} \to \Bbb{C}$ that sends $(x,y) \to x+ y$ corresponds to a map between the $\Bbb{C}$ - algebras above. I believe the key point comes in understanding why such a $\phi$ gives rise to a map $$\psi : \mathcal{O}_{\Bbb{C}}(\Bbb{C})\to \mathcal{O}_{\Bbb{C}\times \Bbb{C}}(\Bbb{C}\times \Bbb{C}).$$ Now the function polynomial $f(T) = T \in \Bbb{C}[T]$ can be viewed as a $\Bbb{C}$ - valued function. If I "precompose" this with $\phi$, $\endgroup$
    – user38268
    Sep 10, 2012 at 0:47
  • $\begingroup$ @Ben: I'm having trouble following your explanation. It would be much cleaner if you used a different symbol for the additive group $\mathbb{C}$ because you are using $\mathbb{C}$ in two different ways. I would write the addition map as a map $\phi : G \times G \to G$ and then talk about polynomial functions $G \to \mathbb{C}$. $\endgroup$ Sep 10, 2012 at 0:57
  • $\begingroup$ I have reposted an answer below. $\endgroup$
    – user38268
    Sep 10, 2012 at 1:22

I agree with Qiaochu that there is a cause for confusion here in that the group in question and our ground field are now the same. I will try to explain this again. Let us view $G = \Bbb{A}_\Bbb{C}^1$ as a group under addition. Now we have the addition map $\phi : G\times G \to G$ that sends $(x,y) \to x+y$. I now want to understand how this gives rise to a map $\psi : \mathcal{O}_G(G) \to \mathcal{O}_{G\times G}(G \times G)$. Now suppose I have a polynomial function $f: G\to \Bbb{C}$. If I precompose this with $\phi$, I get a polynomial function $f \circ \phi : G\times G \to \Bbb{C}$. If I define

$$\begin{eqnarray*} \psi : &\mathcal{O}_G(G)& \to \mathcal{O}_{G\times G}(G \times G)\\ &f& \mapsto f \circ \phi \end{eqnarray*} $$

this should be enough to explain why there is a map from the coordinate ring of $G$ (which I call $\Bbb{C}[G]$) to the coordinate ring $\Bbb{C}[G\times G]$ of $G \times G$.

Now I realise that I defined my $\Delta$ wrongly before. Now because $G$ is now the vanishing locus of the ideal $(0) \subset \Bbb{C}[T]$, the ideal of all functions that vanish on $G$ is $(T)$. I can then call the coordinate ring of $G$ as $\Bbb{C}[T]$. Similarly $G \times G$ is the vanishing locus of the zero ideal in $\Bbb{C}[T,U]$, and so the ideal of all functions that vanish at $0$ is the ideal $(T,U) \subseteq \Bbb{C}[T,U]$. The coordinate ring of $G \times G $ is isomorphic to $\Bbb{C}[T] \otimes_\Bbb{C}[U] \cong \Bbb{C}[T,U]$.

  • $\begingroup$ $G$ is not the vanishing locus of the ideal $(T) \subset \mathbb{C}[T]$; it's the vanishing locus of the zero ideal. $\endgroup$ Sep 10, 2012 at 1:27
  • $\begingroup$ @QiaochuYuan I have edited the last paragraph above. $\endgroup$
    – user38268
    Sep 10, 2012 at 1:33
  • $\begingroup$ $G \times G$ is also not the vanishing locus of the ideal $(T, U) \subset \mathbb{C}[T, U]$. I am not sure exactly what this answer is supposed to accomplish; are you trying to prove that $\psi$ is in fact $\Delta$? $\endgroup$ Sep 10, 2012 at 1:34
  • $\begingroup$ Same mistake with $(T,U)$: that should be the zero ideal, too. Also, what's this $G \times G \cong \mathbb{C}[T]\mathbin{\otimes_{\mathbb{C}}}\mathbb{C}[U]$ about? $\endgroup$
    – t.b.
    Sep 10, 2012 at 1:35
  • $\begingroup$ @QiaochuYuan Well I originally posted this to check if my understanding was correct. $\endgroup$
    – user38268
    Sep 10, 2012 at 1:36

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