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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.

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  • $\begingroup$ I wrote a computer program that confirms this for $n\le 4$. It is brute force and runs over all $2^{2^n}$ elements in the exterior algebra so without any optimization bigger $n$ seem to be hopeless. $\endgroup$
    – HenrikRüping
    Commented Mar 3, 2016 at 1:45
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    $\begingroup$ Nice question. Maximal rank of $\cdot r$ seems to be $2^{n-1}-2^{\lceil\frac{n-1}2\rceil}$, attainable at $x_1x_2+x_3x_4+...$ (while the equality would give rank $2^{n-1}$). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 3, 2016 at 8:30

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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.

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