Looking for full reason why polynomials form rings Hello in my class my teacher had been talking about if we have a commutative ring R and form R[x] ie the ring of polynomials, with coefficients in R then this is also a commutative ring. However, it was said that the verifications of axioms are not something there was time for, but I am interested in seeing this. Is there any sort of list that shows why in fact this is true? I understand I would have to verify all the axioms and such, but I am confused on how I would do things like distributivity etc.
Most important to me, would be to see a formal argument that indeed distributive holds for example
Thanks
 A: The deep reason why polynomials form a ring is that by design, they are an abstraction of the ring axioms, in the following sense:
Take the set of all symbolic expressions built using the ring operations from the elements of $R$ and one or more variables. Consider the following equivalence relation:
$$ e_1 \sim e_2 \iff \text{the ring axioms prove that }e_1=e_2 $$
This is an equivalence relation (easy to check), and the equivalence classes are set of expressions. Then it turns out that each equivalence class contains exactly one polynomial in standard form (decreasing order, variables in each term ordered lexicographically, etc). Which is not surprising because the very concepts of "polynomial" of "standard form" have been selected such that this would be the case.
So we can use the polynomials as canonical representatives of the equivalence classes.
And the rules for multiplying and adding polynomials are exactly such that if we take two expressions $e_1$ and $e_2$ and form, for example, the product $e_1\cdot e_2$, then the product of the polynomials that represents the equivalence classes of $e_1$ and $e_2$ is exactly the polynomial that is equivalent to $e_1\cdot e_2$.
(This is not surprising either, because the rules for arithmetic with polynomials have been chosen to make this true -- which is why arithmetic on formal polynomials agree with pointwise arithmetic on polynomial functions).

Above, "the ring axioms" means the axioms of commutative rings, together with arithmetic on known elements of $R$. Polynomials work less well over non-commutative rings; if we try to remove commutativity from the axioms in the definition of $\sim$ we end up with needing many more "polynomials" in order to give all classes a representative. We'd need not only to distinguish between $xy$ and $yx$ when we have multiple variables; each term would need to contain a sequence of coefficients alternating with variables because we can't move the coefficient out in front of the term without commutativity.
A: Here is a common construction of $R[x]$:


*

*$R[x]$ is the set of functions $\mathbb N \to R$ that have finite support, that is, sequences that are eventually zero.

*Define $x^n=\delta_n$, that is, the sequence that is $1$ at position $n$ and zero everywhere else.

*Define $x^n \cdot x^m = x^{n+m}$.

*Identify $R$ with the sequences $(r,0,\dots)$.

*Declare that $R$ commutes with $x$ (and so with all its powers).

*Define multiplication of two polynomials by distributivity, that is, so that distributivity holds.
In this approach, there is no need to check distributivity...
