Power-of-two projective spaces do not embed in twice the dimension minus one

We first compute the inverse. Since $m=2^r$, [F2] gives $(1+a{)}^m=1+a^m$, so the telescoping identity in ${\mathbb{F}}_2[a]$ reads $(1+a)(1+a+\cdots +a^{m-1})=1+a^m$. Multiplying by the same factor $(1+a{)}^m=1+a^m$ gives $(1+a{)}^{m+1}{\sum }_{i=0}^{m-1}{a}^i=(1+a^m{)}^2=1+a^{2m}\text{in }{\mathbb{F}}_2[a],$ using $(u+v{)}^2=u^2+v^2$ in characteristic two. Because $2m\ge m+1$, the term $a^{2m}$ vanishes in $R={\mathbb{F}}_2[a]\mathrm{/}(a^{m+1})$, so $(1+a{)}^{m+1}{\sum }_{i=0}^{m-1}{a}^i=1$ in $R$: the polynomial ${\sum }_{i=0}^{m-1}{a}^i$ is the inverse of $(1+a{)}^{m+1}$, and by [F1] $\bar{w}({\mathbb{R}\mathbb{P}}^m)=(1+a{)}^{-(m+1)}={\sum }_{i=0}^{m-1}{a}^i=1+a+\cdots +a^{m-1}\mathrm{.}$

The top coefficient is ${\bar{w}}_{m-1}({\mathbb{R}\mathbb{P}}^m)=a^{m-1}$, which is nonzero in ${\mathbb{F}}_2[a]\mathrm{/}(a^{m+1})$ because $m-1<m+1$ and the powers $1,a,\dots ,a^m$ are linearly independent. Suppose ${\mathbb{R}\mathbb{P}}^m$ embedded smoothly in ${\mathbb{R}}^{2m-1}$; here $m\ge 2$ and $k=m-1\ge 1$, so [F3] with this codimension forces ${\bar{w}}_{m-1}(T{\mathbb{R}\mathbb{P}}^m)=0$, contradicting the computed nonzero class. Hence no such embedding exists.

The cases $r=1$ and $r=2$ give $m=2$ and $m=4$: ${\mathbb{R}\mathbb{P}}^2$ does not embed in ${\mathbb{R}}^3$, and ${\mathbb{R}\mathbb{P}}^4$ does not embed in ${\mathbb{R}}^7$. The argument proves only non-embeddability: no assertion is made about the existence of an immersion of ${\mathbb{R}\mathbb{P}}^m$ in ${\mathbb{R}}^{2m-1}$, nor about embeddability in ${\mathbb{R}}^{2m}$ or in larger codimension. The only choice used is the AC assumed by the class and embedding suppliers.