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).



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


