I am looking into when quotients of finite-dimensional exterior algebras over $\mathbb{F}_2$ can be embedded into the original exterior algebra.
It is clear that all such quotients can be embedded into their exterior algebra when there are less than three generators. For example, $E(x,y)/xy$ can be identified with the subalgebra $\lbrace 1,x,xy\rbrace$ of $E(x,y)$ with $x(xy)=0$.
I suspect that in general such embeddings do not exist. Right now I'm trying to prove that there is no monomorphism of unital $\mathbb{F}_2$-algebras $E(x,y,z)/xz\to E(x,y,z)$. Here is what I have:
Suppose that such a monomorphism exists; call it $\phi$. Then $$\lbrace 1,\phi(x),\phi(y),\phi(z),\phi(x)\phi(y),\phi(y)\phi(z)\rbrace$$ is linearly independent in $E(x,y,z)$ with $\phi(x)\phi(z)=0$. There exist $w,\alpha\in E(x,y,z)$ with $w^2=\alpha^2=0$ completing it to a basis. I've tried expressing $\phi(x)$, $\phi(z)$ as linear combinations and seeing what comes out the other end, which has just become a horrible mess and not got me anywhere.
How might I go about finding a contradiction here? Also any insights on the more general issue of embedding quotients into the exterior algebra would be much appreciated.
$ \newcommand\Cl{\mathrm{Cl}} \newcommand\End{\mathrm{End}} \newcommand\lintr{\mathbin\lrcorner} \newcommand\act{\mathbin.} \newcommand\lcontr{\mathbin\rfloor} \newcommand\proj[1]{\langle#1\rangle} $This is not an answer but is too long for a comment. Let $V$ be our vector space. I will use $\wedge$ for the exterior product.
You're looking for injective homomorphisms $E(V)/I \to E(V)$ for some ideal $I \subset E(V)$. This is the same as an endomorphism $E(V) \to E(V)$ with kernel exactly $I$. The algebra of endomorphisms $\End(E(V))$ is isomorphic to the Clifford algebra $\Cl(V^* \oplus V)$ with quadratic form $v^* + v \mapsto v^*(v)$ where the action of $v^* + v$ on $X \in E(V)$ is $$(v^* + v)\mathbin\act X = v^*\lintr X + v\wedge X$$ where $v^*\lintr X$ is the interior product. $E(V)$ is naturally a subalgebra of $\Cl(V^* + V)$, and in fact we can realize $\Cl(V^*\oplus V) \cong E(V^*\oplus V)$ as vector spaces by using the nonsymmetric bilinear form $\beta(v^* + v, w^* + w) = v^*(w)$ and defining a Clifford product on $E(V^*\oplus V)$ via $$ (v^*+v)X = (v^*+v)\lcontr X + (v^*+v)\wedge X = v^*\lintr X + v\wedge X + v^*\wedge X $$ where $(v^*+v)\lcontr X = \beta(v^* + v,\cdot)\lintr X = v^*\lintr X$. The projection $\proj\cdot_{V}$ of $E(V^*\oplus V)$ onto the subalgebra $E(V) \subseteq E(V^*\oplus V)$ is well-defined; careful consideration will then show that when $X \in E(V)$ and $Y \in E(V^*\oplus V)\cong \Cl(V^*\oplus V)$ $$ Y\act X = \proj{YX}_V. $$ So ther kernel of $Y$ as an element of $\End(E(V))$ is the set of all $X \in E(V)$ such that the Clifford product $YX$ has no $E(V)$ component.
I don't have anything past that, but it may be that this line of reasoning allows you to turn the question into a geometric question about subspaces of $V$.