Ring of Polynomials and Ring of Polynomial Functions

[Mandelbroth]

Recently, I've developed a habit of trying to separate the idea of a function from the idea of the image of the function. This has mostly just confused me, but I am adamant about sticking to it.

I think the two terms, "ring of polynomials" and "ring of polynomial functions," are not equivalent. Here's my reasoning:

If we talk about a polynomial, we talk about something of the form $\sum a_nx^n$. When we talk about a polynomial function, we talk about a function that takes every $x$ in a domain to a member of a codomain of the form $\sum a_nx^n$. If we were to try to establish a bijection between them, we'd have to forget about the domain and codomain.

Is there any reason to think of them differently, or are they really just two terms for the same thing?

 

[lavinia]

Algebraically, as rings, the two are the same. The bijection is an isomorphism of rings.

Evaluation at some element of the ring of scalars is a ring homomorphism.

 

[jgens]

Algebraically, as rings, the two are the same. The bijection is an isomorphism of rings.

This is not true. The polynomial x²-x is non-zero in the polynomial ring Z/2[x] but is zero in the ring of polynomial functions over Z/2.

 

[lavinia]

This is not true. The polynomial x²-x is non-zero in the polynomial ring Z/2[x] but is zero in the ring of polynomial functions over Z/2.

Right. I was thinking of a base field of characteristic zero. I should have said so.

 

[jgens]

Right. I was thinking of a base field of characteristic zero. I should have said so.

In that case you're covered

 

[economicsnerd]

The general picture:
- The polynomial ring (1), the ring of polynomial functions (2), and the base ring (3) are all rings.
- Taking a polynomial to the function (on the base ring) it defines is a ring homomorphism from (1) to (2).
- Fixing any element of the base ring, the map taking a polynomial function to its evaluation (at the element we fixed) is a ring homomorphism from (2) to (3).