How does Hartshorne's definition of group schemes encode the law for the neutral element?

Question

Hartshorne's Algebraic Geometry says

A scheme $X$ with a morphism to another scheme $S$ is a group scheme over $S$ if there is a section $e\colon\;S\to X$ (the identity) and a morphism $\rho\colon\;X\to X$ over $S$ (the inverse) and a morphism $\mu\colon\;X\times X\to X$ over $S$ (the group operation) such that

(1) the composition $\mu\circ(\operatorname{id}\times\rho)\colon\;X\to X$ is equal to the projection $X\to S$ followed by $e$, and

(2) the two morphisms $\mu\circ(\mu\times\operatorname{id})$ and $\mu\circ(\operatorname{id}\times\mu)$ from $X\times X\times X\to X$ are the same.

Clearly those two demands formalize that $\rho$ is a right-inverse and $\mu$ is associative. However, I miss some statement concerning the (right-)neutrality of $e$: I would expect something like

The morphism $\mu\circ(\operatorname{id}\times e)\circ(X\overset\sim\to X\times_S S)$ is the identity.

Does this somehow already follow from the cited definition?

Answer 1

As stated it doesn't appear to work. For simplicity set $S = \mathrm{Spec} (k)$.

Here's a counterexample: let $X$ be any $k$-variety, $e : S \to X$ any point of $X$, and $\rho : X \to X$ any morphism.

Set $\mu : X \times X \to X$ to be the constant morphism $\mu(x,y) = e$. This clearly satisfies properties (1) and (2) that you listed (since $\mu(\mu(x,y),z) = \mu(x,\mu(y,z))\ ( = e)$ and $\mu(x,\rho(x)) = e$), but does not make $X$ a group scheme over $\mathrm{Spec}(k)$.

In particular, it fails the last property you pointed out, $\mu(e,x) \ne x$.