Let $k$ be a commutative ring with unity, and let $A$ be the $k$-algebra of an affine group scheme over $k,$ endowed with its usual structure as a Hopf algebra over $k.$ In Waterhouse's textbook, An Introduction to Affine Group Schemes, he discusses the criteria that an ideal $I$ of $A$ must satisfy in order for $A/I$ to represent a closed subgroup scheme. These conditions are
1) If $\Delta$ is the comultiplication, then $\Delta(I) \subseteq A \otimes_k I + I \otimes_k A$.
2) If $S$ is the antipode, then $S(I) \subseteq I$.
3) If $\epsilon$ is the augmentation, then $\epsilon(I) = 0.$
I understand that 3) is both necessary and sufficient for the unit element(s) to belong to the subgroup scheme. I also understand that 1) is sufficient for closure under multiplication for the subgroup scheme and that 2) is sufficient for closure under inversion for the subgroup scheme.
I am guessing that 2) is necessary for closure under inversion because we can just consider the canonical projection $A \to A/I,$ which is a map with kernel equal to $I.$
However, I am confused as to why 1) is necessary for closure under multiplication.
That is, if it is true that for every $k$-algebra $R$ and for every $f, g : A \to R$ with $f(I) = 0 = g(I)$ that we have $\left((f \otimes g) \circ \Delta\right) = 0$, then why must condition 1) follow?
Thank you very much.
As per the comments: we want $\Delta \colon A \to A\otimes A $ to descend to $\Delta \colon A/I \to A/I \otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $\Delta$) if the natural map $$ A\otimes A / (A\otimes I + I \otimes A) \to A/I \otimes A/I,$$ defined and determined by $$ a\otimes b\mapsto \overline a \otimes \overline b,$$ is an isomorphism, where $\overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'\subset M$, $N'\subset N$ are $k$-modules, we want to show that the natural map $$ M\otimes N / (M'\otimes N + M \otimes N') \to M/M' \otimes N/N',$$ defined/determined by $$ m\otimes n + (M'\otimes N + M \otimes N') \to \overline m \otimes \overline n$$ is an isomorphism (and $\overline m = m + M'$, and similarly for $n$). Just to give this last map a name, call it $\alpha$.
To construct an inverse map $\beta$:
Fix $m \in M$. Then there is a $k$-linear map $$\phi_m \colon N \to M\otimes N / (M'\otimes N + M \otimes N') ,$$
defined by $$\phi_m\colon n \to m \otimes n + (M'\otimes N + M \otimes N').$$ Clearly, $\phi_m$ vanishes on $N'$, so $\phi_m$ descends to a $k$-linear map - call it $\phi_m$ again - abuse of notation: $$ \phi_m \colon \bar n \to m\otimes n + (M'\otimes N + M \otimes N').$$
Now, $m \to \phi_m$ is $k$-linear, and depends only on $\bar m$, so we obtain a $k$-bilinear map $$ M/M' \times N/N' \to M \otimes N / (M'\otimes N + M \otimes N'),$$ with $$ (\bar m,\bar n) \to (m\otimes n) + (M'\otimes N + M \otimes N').$$ Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $\beta$ $$ M/M' \otimes N/N' \to M \otimes N / (M'\otimes N + M \otimes N').$$
On generators, the two maps $\alpha$ and $\beta$ are inverses, so $\alpha$ and $\beta$ are $k$-module inverses.