In particular, how is addition defined on it? For finite number of unknowns, we can define it using multi index, but I couldn't generalize it for uncountably many unknowns as multi index is defined only for a finite number of tuples.

Also, is degree defined in such polynomial rings?


  • $\begingroup$ What's your rigorous definition of a single variable polynomial ring? I think a good answer should build on that, otherwise the connection between the two will probably be missing. But as a teaser, polynomial rings are a special case of a monoid ring, where the monoid is $\mathbb N^n$ for an $n$-variable polynomial ring. This can be used to define it for infinitely many variables. en.wikipedia.org/wiki/Monoid_ring $\endgroup$ Sep 14 '20 at 8:00
  • $\begingroup$ It's a special case of semigroup ring or monoid ring construction, e.g. see this answer for some interesting results on such. $\endgroup$ Sep 14 '20 at 8:03

It doesn't matter how many unknowns there are. The polynomial ring is still spanned by monomials consisting of a finite number of the variables at a time, so multi-indices work fine and so does the degree.

Formally, as hinted at in the comments, the polynomial ring $k[S]$ on a set $S$ over a commutative ring $k$ is the monoid algebra over $k$ on the direct sum $\bigoplus_S \mathbb{Z}_{\ge 0}$ of $S$ many copies of $\mathbb{Z}_{\ge 0}$, or in other words, the free commutative monoid on $S$. (Not the direct product $\mathbb{Z}_{\ge 0}^S$, that's much larger.)

Abstractly, if we define the polynomial $k$-algebra functor $k[S]$ as the left adjoint to the forgetful functor from commutative $k$-algebras to sets, the forgetful functor factors as a composite

$$\text{CAlg}(k) \xrightarrow{U(-)} \text{CMon} \xrightarrow{U'(-)} \text{Set}$$

through commutative monoids, where $U$ is given by forgetting everything except the multiplication, and so its left adjoint factors as a composite going the other way ("adjoints compose")

$$\text{Set} \xrightarrow{F'(-)} \text{CMon} \xrightarrow{F(-)} \text{CAlg}(k)$$

where $F'(-)$ takes the free commutative monoid on a set and $F(-)$ takes the monoid $k$-algebra of a commutative monoid.


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