What is the "correct" definition of a free [augmented commutative algebra](https://en.wikipedia.org/wiki/Augmentation_%28algebra%29)?


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](https://ncatlab.org/nlab/show/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?



Free augmented algebras
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.


