Ideals in exterior algebras over the field with two elements Suppose we have an exterior algebra over $\mathbb{F_2}$, say $R = \Lambda_{\mathbb{F_2}}V$, where $V$ is an $n$-dimensional $\mathbb{F}_2$ vectorspace. Let $x_1,\ldots,x_n$ be a basis of that vectorspace. 
This ring is a local ring whose maximal ideal $I$ is generated by $x_1,\ldots,x_n$.
The square of any non-unit $r$ in $R$ is zero, so $(r)\subset $Ker$(\cdot r:R\rightarrow R)$. Now I am wondering for which elements $r$ we get equality. 
If $r\in I\setminus I^2$, then the two are equal. (You can find an automorphism of $R$ sending that element to $x_1$ and it is true for $x_1$).
Does the converse also hold?
Also I would like to understand the orbits of the action of the automorphism group of this algebra on itself.
 A: The following work is joint work with Omar Antolin-Camarena.
We want to prove that if $r\in I^2$, then $\dim(r)<2^{n-1}$ (Dimension as an $\mathbb{F}_2$-vectorspace).
We will prove this by induction on the number of variables $n$ and the length of $r$ (number of monomials that are summed up). Rename the variables such that $x_1$ appears in at least one monomial. Let $R_2$ denote the exterior algebra in $x_2,\ldots,x_n$. Write $r=x_1\cdot r'+r''$ with $r',r''\in R_2$.
We have short exact sequences
\[0\rightarrow (x_1)\cap (r)\rightarrow (x_1)\oplus (r)\rightarrow (x_1,r'')\rightarrow 0;\]
\[0\rightarrow (x_1)\cap (r'')\rightarrow(x_1)\oplus (r'')\rightarrow(x_1,r'')\rightarrow 0.\]
Using the additivity of the dimension we get:
\[\dim(r)-\dim(r'')  =\dim (x_1)\cap(r) -\dim(x_1)\cap (r'').\]
We have:
\[(r'') = (r'')_{R_2}\oplus x_1\cdot (r'')_{R_2};\] 
\[(x_1)\cap (r'') = (x_1r'')=x_1\cdot (r'')_{R_2}.\]
Here $(-)_{R_2}$ should denote the ideal in $R_2$ generated by some element.
\[\dim (x_1)\cap (r'') = \dim (x_1r'')=\dim (r'')_{R_2}.\]
Inserting this in our Dimension-formula we get:
\[\dim(r)  =\dim (x_1)\cap(r) +\dim(r'')_{R_2}.\]
It remains to examine $(x_1)\cap(r)$. So we have to find out when a multiple of $r=x_1\cdot r'+r''$ is also a multiple of $x_1$. We have
\[(ax_1+b)(x_1\cdot r'+r'')=(ar''+br')x_1+br''.\]
This is a multiple of $x_1$, iff $b\in Ann_{R_2}(r'')$ and $a\in R_2$ is an arbitrary element. So we get:
\[(x_1)\cap(r) = x_1(r'\cdot Ann_{R_2}(r'') + (r'')_{R_2}).\]
So we are interested in the Dimension of the sum of those two ideals. Note that that sum is contained in $Ann_{R_2}(r'')$, since both ideals are. So we get:
\[\dim(r)\le \dim Ann_{R_2}(r'')+ \dim(r'')_{R_2}=\dim R_2 = 2^{n-1}.\]
To make this inequality strict, we have to prove that 
\[r'\cdot Ann_{R_2}(r'') + (r'')_{R_2}\subsetneq Ann_{R_2}(r'').\]
Let $p:Ann_{R_2}(r'')\rightarrow Ann_{R_2}(r'')/(r'')_{R_2}$ denote the projection and let $f:Ann_{R_2}(r'')/(r'')_{R_2}\rightarrow Ann_{R_2}(r'')/(r'')_{R_2}$ denote the map induced by multiplication with $r'$. By induction assumption we know that  $\dim Ann_{R_2}(r'')/(r'')_{R_2} >0$. The map $f$ cannot be surjective, since $f^2$ is induced by multiplication with $r'^2$ (which is zero. Since $r\in I^2$, $r'$ cannot be a unit). And the sum of ideals that we want to understand is exactly $p^{-1}(Im(f))\subsetneq Ann_{R_2}(r'')$.
This completes the proof.
