I have some (hopefully super) basic questions about the exterior algebra functor

$$ \wedge:R\text{-Mod}\rightarrow R\text{-Alg}. $$

As I (think I) understand it, if one considers it as a functor into graded $R$-algebras (denoted here by $(R\text{-Alg})_{\text{gr}}$), this is a free construction, so this functor admits a right adjoint given by the forgetful functor $F:(R\text{-Alg})_{\text{gr}}\rightarrow R\text{-Mod}$ sending a graded $R$-algebra to its degree 1 component. As such, $\wedge$ should be right exact.


  • I'm not crazy, and that forgetful functor is a right adjoint to $\wedge$, right?
  • If considered as first stated, i.e. as a functor into the category of $R$-algebras rather than graded $R$-algebras, does this still hold? Does $\wedge$ still admit a right adjoint? My thinking is that it no longer does (thinking in terms of a simple, concrete example: let $R$ be a field, say, and take $W$ to be some 3-dimensional subspace of a 5-dimensional $R$-vector space. Then $\wedge^2$, say, being right exact would imply that it satisfies $\wedge^2(V/W)\cong\wedge^2(V)/\wedge^2(W)$. This can't be true though, since

$$ \text{dim}(\wedge^2(V/W))= {2\choose 2}=1\neq 7={5\choose 2}-{3\choose 2}=\text{dim}(\wedge^2(V)/\wedge^2(W)) $$

$\quad \ \ $ Is my reasoning actually correct here? And can this non-right-exactness be extended to the $\quad \ \ $ functor $\wedge$?)

  • What about a left adjoint? I cannot think of one, but is there some easy way to see that one doesn't exist? Or does it admit a left adjoint as well?

Any help with these would be greatly appreciated! Thanks!

  • 4
    $\begingroup$ No, the forgetful functor is not right-adjoint to $\wedge$, unless your "graded" means "commutative superalgebras" (i.e., superalgebras which obey the axioms $ab = \left(-1\right)^{pq} ba$ for any homogeneous elements $a$ and $b$ of degrees $p$ and $q$, and $aa = 0$ for any odd $a$). It is all the less right-adjoint to $\wedge$ without the "graded". $\endgroup$ Commented Apr 8, 2015 at 3:21
  • $\begingroup$ Ah, you are completely right! Thank you, this cleared up quite a bit for me -- I'll go to work out the details a bit. $\endgroup$ Commented Apr 11, 2015 at 8:04

1 Answer 1


As darij grinberg mentions, the exterior algebra is the free graded-commutative algebra on a vector space. It's not the free algebra or the free graded algebra. This doesn't rule out the possibility of "unexpected" right adjoints.

Your dimension-counting argument doesn't work, because in order for $\operatorname{dim}(A/B) = \operatorname{dim} A - \operatorname{dim} B$ to hold in an algebra-like category, you need to verify that $B$ is an ideal in $A$. In fact, it isn't in your case where $A = \Lambda^2(V)$ and $B = \Lambda^2(W)$. Categories of algebras are generally not abelian, so watch out! But I think it should be easy to show that $\Lambda$ does not preserve coproducts when it takes values in the wrong category.

To show that $\Lambda$ doesn't have a left adjoint, you can show that it doesn't preserve products. After all, finite products and coproducts coincide in module categories but they don't coincide in algebra categories (where coproducts are given by tensor products and products are taken pointwise). So a functor from modules to algebras can preserve one or the other, but not both!

  • $\begingroup$ Yes, I see that now. I'll see if I can figure out any "unexpected" right adjoints (I kind of doubt they exist, but I suppose they wouldn't be unexpected if I didn't doubt them!). And yes, I see what you mean. I figured something was a bit fishy... alright, I will take a stab at showing $\wedge$ doesn't preserve coproducts (I also expect this will be straight forward). Lastly, that makes perfect sense about the left adjoint. That's a nice fact, I'll have to keep in mind. Thanks! $\endgroup$ Commented Apr 11, 2015 at 8:05

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