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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.

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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 $k\to A$.

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  • $\begingroup$ 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? $\endgroup$
    – NotablyYours
    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
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    $\begingroup$ 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 $k\oplus\ker\epsilon$. $\endgroup$
    – Derek Elkins left SE
    Mar 9, 2018 at 1:06

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