1
$\begingroup$

An augmented algebra is equipped with a morphism of algebras $ \epsilon: A \to \mathbb{K}$. In this case $ A \equiv \mathbb{K} \oplus \ker(\epsilon)$. May you please clarify what the meaning of $ A \equiv \mathbb{K} \oplus \ker(\epsilon)$? What is the role of augmented algebras in the case of relations between non-unital algebra and unital ones?

$\endgroup$
2
  • 1
    $\begingroup$ Should probably be $\ker(\epsilon)$. The author may be using informal/different notation, just as people sometimes talk about the "kernel of a matrix". $\endgroup$
    – David P
    Dec 24 '19 at 16:40
  • $\begingroup$ Right, it must be $ker(\epsilon)$. I correct that. Now what will be the meaning of augmentation ideal? $\endgroup$
    – user40491
    Dec 24 '19 at 16:44
3
$\begingroup$

An augmentation for a unital algebra $A$ is a morphism of unital algebras $\varepsilon \colon A \rightarrow \mathbb{K}$. In particular, it satisfies $\varepsilon(1_A) = 1_{\mathbb{K}}$ and so it is non-zero and onto. Let's denote the kernel by $\overline{A} = \ker \varepsilon$. The kernel $\overline{A}$ is a two-sided ideal of $A$ calle the augmentation ideal and can be thought of as a not-necessarily unital algebra. We have a vector space (internal) direct-sum decomposition

$$ A = \mathbb{K}\cdot 1_A \oplus \overline{A}. $$

This means that an element $x \in A$ can be written uniquely as $x = c \cdot 1_A + y$ where $c\in \mathbb{K}$ and $y \in \overline{A}$. Explicitly, we have

$$ x = \varepsilon(x) \cdot 1_A + (x - \varepsilon(x) \cdot 1_A). $$


The relation between not-necessarily unital and augmented algebra is as follows. Given a not-necessarily unital algebra $I$, one can adjoin to $I$ and define $$ I_{+} = \mathbb{K} \oplus I$$ (here, on the right side we have an external direct sum of vector spaces). The multiplication on $I_{+}$ is defined by

$$ (\lambda + a) \cdot (\mu + b) = \lambda \mu + (\lambda b + \mu a + ab) $$ where $\lambda,\mu \in \mathbb{K}$ and $a,b\in I$. This gives $I_{+}$ the structure of a unital algebra with unit $1_{\mathbb{K}} + 0_{I}$ and in addition, it has a "canonical" copy of $I$ sitting inside it. Stated differently, $I_{+}$ is not only a unital algebra but it has a natural augmentation $\varepsilon \colon I_{+} \rightarrow \mathbb{K}$ given by the projection onto the first factor and we have $\overline{I_{+}} = I$.

Hence, we have two operations:

  1. Given an augmented unital algebra $(A,\varepsilon)$ over $\mathbb{K}$, we can get a not-necessarily unital algebra $\overline{A}$.
  2. Given a not-necessarily unital algebra $I$, we get an augmented unital algebra $\overline{I}_{+}$.

One readily sees that the two operations are functorial and are inverses of each other (up to natural identifications) so we get an equivalence of categories between the category of unital augmented algebras and the category of not-necessarily unital algebras.

$\endgroup$
2
  • $\begingroup$ Thank you for your detailed answer. May I ask you to add a brief description on augmentation ideal to the above answer, because I would like to add your answer to my library. (I want to consider that as a quick reference for this concept). $\endgroup$
    – user40491
    Dec 25 '19 at 2:45
  • 1
    $\begingroup$ @user40491: The kernel $\ker \varepsilon = \overline{A}$ is called the augmentation ideal, I don't think I have much to add other than the name. $\endgroup$
    – levap
    Dec 25 '19 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.