Let $A$ be a smooth Hopf Algebra over a field $k$ with comultiplication map $\Delta$ and Augmentation Ideal $I$.

Then I can regard the composition of $\Delta$ with the natural projection of $A \otimes A$ to the square of $I$, i.e. the map

$A \rightarrow A \otimes_k A \rightarrow A/I^2 \otimes_k A/I^2$.

Now my questions:

(1) Can one say what this map does with an element $x$ of $A$ explicitly? In particular, what is it's kernel? I'm not quite sure if one can really say something in this general setting...

(2) Is $A/I^2 \otimes_k A/I^2$ naturally isomorphic as $k-$algebra to something more familiar? Can it be that it is $(A\otimes_kA)/I\otimes_kI$? Or what else? And how does the above map then look like?

Thanks a lot for your effort!

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2) I think the formula you're looking for is $A/I\otimes B/J = (A\otimes B)/(I\otimes B + A \otimes J)$. I would advise to look at a simple example like $A=k[x]$. –  AFK Jan 12 '12 at 10:44
In that example you get as kernel $I^3$, but I don't know if one can generalize this... –  Cyril Jan 14 '12 at 11:57
Elementary questions are not necessarily easy, I guess. –  HbCwiRoJDp Jan 14 '12 at 13:19
But easily formulated...;) –  Cyril Jan 14 '12 at 19:33

(1) The kernel is $I^3$ if $\mathrm{char}(k)\ne 2$, and is $I^2$ otherwise.

Denote by $\delta: A\to A\otimes A \to A/I^2\otimes A/I^2$. The first observation is $A/I^2\otimes_k A/I^2=(A\otimes_k A)/(I^2\otimes A+ A\otimes I^2)$. See YBL's comments. So the kernel of $\delta$ is $\Delta^{-1}(I^2\otimes A+ A\otimes I^2)$.

Let's describe $\Delta(x)$ for $x\in A$. Note that $A=k+I$. So $$A\otimes_k A=k+k\otimes_k I+ I\otimes_k k+ I\otimes_k I.$$
Decompose $\Delta(x)=\lambda+1\otimes a+ b\otimes 1+ \sum_i u_i\otimes v_i$ as above. By the definition of $I$, the composition $A\to A\otimes A\to A/I \otimes A =A$, $x\mapsto \Delta(x)\mapsto \lambda+ a$ is identity. Hence $a=x-\lambda=b$ (by symmetry). From now on, suppose $x\in I$. Then $\lambda=0$ and $$\Delta(x)=1\otimes x + x\otimes 1 + \sum_i u_i\otimes v_i.$$ This implies that $$\Delta(I^2)\subseteq I\otimes I + I^2\otimes A + A\otimes I^2$$ and $$\Delta(I^3)\subseteq (k\otimes I + I\otimes k + I\otimes I)^3 \subseteq A\otimes I^2 + I^2\otimes A.$$ On the other hand, again considering $A\to A\otimes A \to A/I \otimes A$, we see that if $x\in\ker \delta$, then $x\in I^2$. So $$I^3\subseteq \ker \delta \subseteq I^2.$$

The above inclusions imply that $\Delta$ induces a map $$\tilde{\Delta}: I^2/I^3 \to I/I^2\otimes I/I^2$$ and we have $\ker\delta/I^3=\ker\tilde{\Delta}$. Now consider $\phi : I/I^2\otimes I/I^2 \to I^2/I^3$ induced by $A\otimes A\to A$, $u\otimes v\mapsto uv$. Let's show that $\phi\circ \tilde{\Delta}=4.\mathrm{Id}$. Let $y\in I$ and $\Delta(y)=1\otimes y + y\otimes 1 + \sum_jw_j\otimes z_j$ avec $w_j, z_j\in I$. A direct computation shows that $$\Delta(xy) = x\otimes y + y\otimes x + 1\otimes (xy)+ (xy)\otimes 1+ t, \quad t\in I^2\otimes I+ I\otimes I^2.$$ This implies that Then $\phi\circ \tilde{\Delta}(\overline{xy})=4\overline{xy}$ and the same equality for all $z\in I^2$.

End of the proof. If $\mathrm{char}(k)\ne 2$ and $z\in \ker\delta$. Then $4\bar{z}=\phi\circ\tilde{\Delta}(\bar{z})/4=0$ in $I^2/I^3$. So $z\in I^3$ and $\ker\delta =I^3$.

Suppose $\mathrm{char}(k)=2$. Let $x\in I$. Then $$\Delta(x^2)=\Delta(x)^2\in 2(x\otimes x) + I^2\otimes A + A \otimes I^2=I^2\otimes A + A\otimes I^2.$$ As $I^2$ is generated by the $x^2$, $x\in I$ (in characteristic $2$ !), we have $I^2\subseteq \ker\delta\subseteq I^2$ and we are done.

This is a little complicated, there should be a more direct proof using Hopf algebra properties...

(2) I don't have much to say. The tensor product $A/I^2\otimes A/I^2$ is a local Artinian $k$-algebra whose vector dimension is $(1+d)^2$ where $d=\dim I/I^2$ is the dimension of the tangent space of $\mathrm{Spec} A$ at the origin. Except when $A$ is Artinian, $A/I^2\otimes A/I^2$ is not isomorphic to $(A\otimes A)/(I\otimes I)$ because the latter is not Artinian.

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Applause for this! I didn't think one could answer it in such general form. You helped me a lot, thanks! –  Cyril Jan 15 '12 at 18:21