0
$\begingroup$

Let $k$ be a field, and $A$ an algebra. I keep seeing reference to $k \cdot 1_A$; what does this mean?

For example, if $A$ is an augmented algebra, via $\epsilon : A \to k$, then $A$ is canonically isomorphic to $k\cdot 1_A \oplus \ker \epsilon$.

I would like to prove the above, but I haven't seen the notation $k \cdot 1_A$ defined.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think you have seen the notation before, at least in other contexts. Like $Hg$ when talking about the (right) cosets of a subgroup $H$ in some group $G\ni g$. Or when we write $Ra$ for the (left) ideal generated by an element $a$ in a ring $R$.

The notation $k\cdot 1_A$ means $$ \{x\cdot 1_A\mid x\in k\} $$ Intuitively, it's the image of the inclusion map $k\to A$.

$\endgroup$
3
  • $\begingroup$ Thank you. So just to be sure, $x \cdot 1_A$ is the function that assigns $a \mapsto xa$. And this is generally different from $k$ because $x$ and $y$ can be different while $x\cdot 1$ and $y\cdot 1$ the same? $\endgroup$ Mar 8, 2018 at 23:42
  • $\begingroup$ @NotablyYours No, $x\cdot 1_A$ isn't a function. It's an element of $A$. It's the element you get if you multiply $x$ (which is an arbitrary element of $k$) with $1_A$ (which is the unit element of $A$), using the $k$-algebra multiplication. And a priori, $x\cdot1_A$ and $y\cdot1_A$ could be the same element, but because $k$ is a field it's impossible (consider what element $(x-y)^{-1}(x-y)\cdot 1_A$ is). $\endgroup$
    – Arthur
    Mar 8, 2018 at 23:46
  • 1
    $\begingroup$ It's probably worth noting that it is a very common abuse of notation to identify $x\cdot 1_A$ with $x$ for $x\in k$ and thus $k$ with a subset of $A$. So you're very likely to see things like $k\oplus\ker\epsilon$. $\endgroup$ Mar 9, 2018 at 1:06

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.