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What is the "correct" definition of a free augmented commutative algebra?

At least two definitions come to my mind:

Fix a commutative ring $k$. We need elements $\lambda_1,\dotsc,\lambda_n \in k$. They define an augmentation on the polynomial algebra $k[X_1,\dotsc,X_n]$ via $\varepsilon(X_i) := \lambda_i$. Let us denote this augmented commutative algebra by $k[X_1^{[\lambda_1]},\dotsc,X_n^{[\lambda_1]}]$. This satisfies the universal property (for every augmented commutative algebra $A$) $$\mathrm{Hom}(k[X_1^{[\lambda_1]},\dotsc,X_n^{[\lambda_1]}],A) \cong \{a \in A^n : \varepsilon(a_1)=\lambda_1,\dotsc,\varepsilon(a_n)=\lambda_n\}.$$ So (in contrast to commutative algebras) there is no free augmented commutative algebra with $n$ generators: we need to know their values under the augmentation, and for each list of values there is a different universal solution. This is somewhat similar to the definition of free graded algebras, where for each generator we have to know its degree.

On the other hand, the category of augmented commutative algebras is equivalent to the category of non-unital commutative algebras: We map $A \mapsto \ker(\varepsilon)$, and $B \mapsto B^{+}$ (unitalization) in the other direction. The category of non-unital commutative algebras is finitary algebraic and hence has free objects in the usual way. Specifically, they are algebras of polynomials without a constant term, let's denote them by $k[X_1,\dotsc,X_n]_+$. The corresponding augmented commutative algebra is just $k[X_1,\dotsc,X_n]$ with $\varepsilon(X_i)=0$, so it is $k[X_1^{[0]},\dotsc,X_n^{[0]}]$ with the above notation. It is kind of strange that we only get this special case. Right?

Anyway, my motivation for asking is basically that I need a small-as-possible dense subcategory of the category of augmented commutative algebras. What is a good choice here? By the second approach above, the $k[X_1^{[0]},\dotsc,X_n^{[0]}]$ should be sufficient, but it obviously leaves out elements with non-zero augmentation. How can you explain this?

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For any choice of $\lambda_1,\dots,\lambda_n$ there is an isomorphism:

$$ k[X_1^{[\lambda_1]},\dots,X_n^{[\lambda_n]} ] \simeq k[Y_1^{[0]},\dots,Y_n^{[0]} ] $$

Which is given by $X_i \leftrightarrow Y_i+e\lambda_i$ where $e$ is the unit.

So the two constructions actually give you the same objects.

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  • $\begingroup$ This is what I was looking for :). It is quite easy, but for some reason I didn't see it. $\endgroup$ – Martin Brandenburg Feb 14 at 21:05
  • $\begingroup$ By the way, the link in your mathoverflow profile doesn't work, but normalesup.org/~henry does. :) $\endgroup$ – Martin Brandenburg Feb 14 at 21:24
  • $\begingroup$ Thanks ! I fixed it $\endgroup$ – Simon Henry Feb 14 at 21:29
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The free functor is left adjoint to the forgetful functor. There are two forgetful functors from augmented algebras to vector spaces. One views the algebra as a vector space, the other removes the identity summand first. The second one is probably more natural (in particular, it commutes with products, which you would like to have if you are to define a left adjoint), and its adjoint is the ordinary free commutative algebra with standard augmentation.

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  • $\begingroup$ +1 for the hint about products. So the "naive" forgetful functor here has actually no left adjoint. This is why I agree that the other forgetful functor, which maps $A \mapsto \ker(\varepsilon)$, is more natural. But still it is kind of strange that the "elements" of my augmented algebra should therefore be only those elements of augmentation $0$. In some sense, we are not allowed to treat the unit as an element... $\endgroup$ – Martin Brandenburg Feb 15 at 7:11
  • $\begingroup$ Well there is an alternative point of view you might like better, which is to consider the forgetful functor from augmented algebras to augmented vector spaces (vector spaces with a map to k). This functor also commutes with products and its adjoint is the ordinary free commutative algebra, $V\mapsto k[V]$ (with natural augmentation). I suspect this may answer your original question better, since this augmentation is precisely your list of $\lambda_i.$ $\endgroup$ – Dmitry Vaintrob Feb 16 at 0:31

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