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.



Ideals in exterior algebras over the field with two elements
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.


