Affine mapping and hyperplanes Let $E$ be an affine space attached to a $K$-vector space $T$ and let $u:E\rightarrow K$ be a non-constant affine mapping. Take $\lambda\in K$ and write $S:=\{x\in E\ |\ u(x)=\lambda\}$. We want to show that $S$ is a hyperplane of $E$: i.e. that $S\ne\emptyset$, $S$ is an affine subset and $\text{codim}_K(D)=1$, where $D$ is the direction $S$.
First of all, why should $S$ be nonempty? Supposing $S$ is indeed nonempty, we can show that $S$ is affine.
Now, how about $\text{codim}_K(D)=1$? Take $a\in S$. Then $D=\{(x-a)+(y-a)\ |\ x,y\in S\}$. How should I proceed? Any suggestions?
 A: Let's let $T$ be a $K$-vector space, and $E$ be the associated affine space. If $u: E \to K$ is a non-constant affine mapping, then there exist ${\bf x}, {\bf y} \in E$ with $u({\bf x}) \neq u({\bf y})$, let us put
$$u({\bf x}) - u({\bf y}) = \alpha \neq 0.$$
By the definition of an affine mapping, for any $c\in K$,
$$u(c {\bf x} + (1-c) {\bf y}) = c u({\bf x}) + (1-c) u({\bf y}) = u({\bf y}) + c(u({\bf x}) - u({\bf y})) = u({\bf y}) + c\alpha$$
so putting $c = \frac{\lambda - u({\bf y})}{\alpha}$ gives us a vector $\bf v$ with $u({\bf v}) = \lambda$, and so $S:=\{x\in E\ \mid u(x)=\lambda\}$ is nonempty.
Showing we have a hyperplane is straightforward; there is a unique linear mapping $v: T \to K$ such that $u({\bf x} + {\bf y}) = u({\bf x}) + v({\bf y})$ for all ${\bf x}, {\bf y} \in T$ (this is slightly different from Bourbaki's wording, but is equivalent; notice that $u$ and $v$ can both be considered as maps $E \to K$ or $T \to K$, but $u$ is affine and $v$ is linear). Since $u$ is nonconstant, $v$ is a nonzero linear functional, and so $H = \{ {\bf x} \mid v({\bf x}) = 0 \}$ is a hyperplane of $T$.
