Let $E$ be an affine space over a field $k$ and let $V$ its vector space of translations. Denote by $X=\operatorname{Aff}(E,k)$ the vector space of all affine-linear transformations $f:E\to k$, that is, functions such that there is a $k$-linear form $Df:V\to k$ satisfying
$$
f(x+v) = f(x) + Df(v) \quad \text{for all } x\in E, v\in V,
$$
and let $X^*=\operatorname{Aff}(E,k)^*$ its dual space. Consider the following $k$-linear map:
$$
\varepsilon:V\to X^*, \quad v\mapsto \varepsilon(v),
$$
where $\varepsilon(v):X\to k$ is given by evaluation of $Df$ at $v$, that is
$$
\varepsilon(v)(f) = Df(v) \quad \text{for all } f\in X.
$$
Clearly $\varepsilon$ is injective. The claim is that $\varepsilon$ embeds $V$ as a (linear) hyperplane of $X^*$. This is obvious if $E$ is of finite dimension. For arbitrary dimension I defined
$$
\alpha:X^* \to k, \quad \varphi \mapsto \alpha(\varphi)=\varphi(1),
$$
where $1:E\to k$ is the constant function taking the constant value $1\in k$. Then it is clear that
$$
\varepsilon(V)\subseteq \ker(\alpha).
$$
I believe that equality holds but I don't know how to prove it. This will prove the claim.
Is this the right approach? Is there a better one?
Context: I know this has to be a basic fact, so to be honest I feel a bit ashamed for asking this question. I'm reading Macdonald's Affine Hecke Algebras and Orthogonal Polynomials; this claim is stated without comments on page 1.
Best Answer
The claim is false if $E$ is not assumed to be finite dimensional: Take $E$ to be of infinite dimension and fix $x_0\in E$. Consider the linear map $$ \theta:F\to k, \quad f\mapsto f(x_0). $$ Then the map $D:\ker(\theta)\to V^*$ given by $f\mapsto Df$ is easily seen to be a $k$-linear isomorphism. Thus as $\ker(\theta)$ is of codimension one in $F$, the first isomorphism theorem implies that $\dim F = \dim V^* +1$. But as $V$ is of infinite dimension, then $\dim F = \dim V^*$. But, because the algebraic dual space is strictly larger than the original space, we have the following inequality of infinite cardinals: $$ \dim V < \dim V^* = \dim F < \dim F^*. $$ This shows that $\varepsilon(V)$ cannot be of codimension one in $F^*$, but $\ker(\alpha)$ is, so the inclusion $\varepsilon(V)\subseteq \ker(\alpha)$ is strict in this case.