# Congruences of abelian monoids which can be extended to (ideal) congruences of polynomials

Some weeks ago I asked the same question at [math.stackexchange][1] but I have not gotten any feedback. The flavour of the question (but see the details later) is about whether to understand congruences in abelian monoids we can always consider the enriched structures obtained by considering polynomial rings.

Let us consider the ring $\langle K[x_1,...,x_n],+,\cdot \rangle$ where $K$ is either a field or the ring of integers. It is well known that congruences of this polynomial ring are characterized by ideals.

On the other hand, we can consider the abelian (multiplicative sub-) monoid $\langle Mon(x_1,...,x_n), \cdot\rangle$ given by considering monic monomials. Let me point out that such monoid is independent of the chosen $K$.

I am interested in the relationship between the congruences of $\langle K[x_1,...,x_n],+,\cdot\rangle$ and the congruences of $\langle Mon(x_1,...,x_n),\cdot\rangle$. It is obvious that all congruences of the polynomial ring $\langle K[x_1,...,x_n],+,\cdot\rangle$ also determine, when restricted, congruences of $\langle Mon(x_1,...,x_n),\cdot\rangle$. My question is about the reciprocal statement (and which has the flavour of the "congruence extension property").

Main Question: Is it true that every congruence $\theta$ of $\langle Mon(x_1,...,x_n), \cdot\rangle$ can be extended to a congruence $\theta'$ of $\langle K[x_1,...,x_n],+,\cdot\rangle$ such that $\theta = \theta' \cap Mon(x1,...,x_n)$?

Indeed, it might be the case that the answer to this question depends on the chosen $K$, but I would be very surprised if this is the case.

Let me add a second question in case that the answer to the main one is negative: is there some characterization (for some field $K$) of which congruences of $\langle Mon(x_1,...,x_n),\cdot\rangle$ can be extended?

By the way, I am also interested in any bibliographic reference where this problem is considered (I have not been able to locate anything).

• If $\sim$ is a congruence on the free commutative monoid, then just take the ideal generated by all differences $m-m'$ with $m\sim m'$. – Benjamin Steinberg Mar 9 '16 at 18:55
• This observation is how Hilbert's basis theorem was used to first show every finitely generated commutative monoid is finitely presented. – Benjamin Steinberg Mar 9 '16 at 18:56
• @Benjamin: Any reference where it is showed that the ideal you propose is not identifying more monomials? – boumol Mar 9 '16 at 22:32
• You don't need one. If M is the quotient monoid then this is the kernel of the natural projection to the monoid algebra KM. – Benjamin Steinberg Mar 10 '16 at 1:44
• The monoid algebra is a functor and the kernel of the induced map does what you want. You can check it is generated as I say but you don't really need it. – Benjamin Steinberg Mar 10 '16 at 1:45

If $K$ is a commutative ring with unit (I guess even this is not needed) an $M$ and $N$ are monoids, then given a homomorphism $\varphi\colon M\to N$, there is an induced homomorphism of monoid algebras $\Phi\colon KM\to KN$ that on the basis agrees with $\varphi$. If $m,m'\in M$ with $m-m'\in \ker \Phi$ if and only if $\Phi(m)=\Phi(m')$ if and only if $\varphi(m)=\varphi(m')$ and so $\ker \Phi$ is the ideal you seek.
It is not difficult to see that in fact $\ker \Phi$ is spanned by all $m-m'$ with $\varphi(m)=\varphi(m')$. I leave that as an exercise.