Lemma 109.48.1. There exists a group scheme $G$ over a base $S$ whose identity component is flat over $S$ but which is not flat over $S$.

## 109.48 A non-flat group scheme with flat identity component

Let $X \to S$ be a monomorphism of schemes. Let $G = S \amalg X$. Let $m : G \times _ S G \to G$ be the $S$-morphism

which maps the summands $X \times _ S X$ and $S$ into $S$ and maps the summands $X$ into $X$ by the identity morphism. This defines a group law. To see this we have to show that $m \circ (m \times \text{id}_ G) = m \circ (\text{id}_ G \times m)$ as maps $G \times _ S G \times _ S G \to G$. Decomposing $G \times _ S G \times _ S G$ into components as above, we see that we need to verify this for the restriction to each of the $8$-pieces. Each piece is isomorphic to either $S$, $X$, $X \times _ S X$, or $X \times _ S X \times _ S X$. Moreover, both maps map these pieces to $S$, $X$, $S$, $X$ respectively. Having said this, the fact that $X \to S$ is a monomorphism implies that $X \times _ S X \cong X$ and $X \times _ S X \times _ S X \cong X$ and that there is in each case exactly one $S$-morphism $S \to S$ or $X \to X$. Thus we see that $m \circ (m \times \text{id}_ G) = m \circ (\text{id}_ G \times m)$. Thus taking $X \to S$ to be any nonflat monomorphism of schemes (e.g., a closed immersion) we get an example of a group scheme over a base $S$ whose identity component is $S$ (hence flat) but which is not flat.

**Proof.**
See discussion above.
$\square$

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