0
$\begingroup$

Let $ X $ be a monoid and $ R $ be a (two-sided) congruence relation on $ X $ which is generated by some relations $ u_i \equiv_R v_i $ for any $ i $ in some index set $ J $. Let $ K $ be a field, $ K[ X ] $ (resp. $ K[ X / R ] $ ) be the monoid algebra of $ X $ (resp. the quotient monoid $ X / R $ ) over $ K $. Let $ I $ be the the ideal in $ K[ X ] $ which is generated by the elements $ u_i - v_i $ for any $ i \in J $.

If I am not mistaken, it immediately follows that the canonical $ K $-algebra morphism $$ K[ X ] \to K[ X / R ] \text{ via } \sum_\omega c_\omega \omega \to \sum_\omega c_\omega [ \omega ]_R $$ has kernel $ I $. But, I cannot find any source on that. Am I wrong or is it too trivial? Is there a good reference for this kind of introductory questions on monoid algebras (or monoid ring)? All my google searches immediately run into more advanced topics.

Thanks

Improve this question
$\endgroup$
6
  • 5
    $\begingroup$ This is trivial. Just use the universal property. $K[X/R]$ and $KX/I$ are universal for K-algebras A with a monoid homomorphism $X\to A$ where $u_i=v_i$ $\endgroup$
    – Benjamin Steinberg
    Commented Mar 11, 2021 at 12:30
  • 1
    $\begingroup$ Or you can use the monoid algebra construction is a left adjoint and the quotient can be described as a colimit $\endgroup$
    – Benjamin Steinberg
    Commented Mar 11, 2021 at 12:31
  • 2
    $\begingroup$ If you don't like categorical arguments observe that $I$ is contained in the kernel J of the projection KX\to K[X/R] and it is easy to check the kernel J is spanned by all differences x-y with x,y in the same congruence class of R. Now if you follow a derivation that x is equivalent to y from the generators of R you can prove x-y is in I $\endgroup$
    – Benjamin Steinberg
    Commented Mar 11, 2021 at 12:36
  • 1
    $\begingroup$ @BenjaminSteinberg, ok, so I am not mistaken, the last reasoning was also the one I had in mind. Still. Is there a source where this kind of simple statements are at least stated? $\endgroup$
    – diddy
    Commented Mar 11, 2021 at 12:44
  • $\begingroup$ I don't know a concrete reference. It is one of those abstract nonsense things so people I think just use it without comment $\endgroup$
    – Benjamin Steinberg
    Commented Mar 11, 2021 at 13:12

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .