Polynomials and Polynomial Functions in I_m = Z/mZ

[Math Amateur]

I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 3. Polynomials

I need help with the a statement of Rotman's concerning the polynomial functions of a finite ring such as \(\displaystyle \mathbb{I}_m = \mathbb{Z}/ m \mathbb{Z} \)

The relevant section of Rotman's text reads as follows:https://www.physicsforums.com/attachments/4530In the above text we read the following:

" ... ... The reader should realize that polynomials and polynomial functions are distinct objects. For example, if \(\displaystyle R\) is a finite ring (e.g. \(\displaystyle \mathbb{I}_m\)), then there are only finitely many functions from \(\displaystyle R\) to itself; a fortiori, there are only finitely many polynomial functions. ... ... "

My question is as follows:

How, exactly (indeed, rigorously and formally) can we demonstrate that only finitely many functions from \(\displaystyle R\) to itself implies that there are only finitely many polynomial functions ... ... indeed, how would we prove this statement ...Hope someone can help ... ...

Peter

 

[caffeinemachine]

I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 3. Polynomials

I need help with the a statement of Rotman's concerning the polynomial functions of a finite ring such as \(\displaystyle \mathbb{I}_m = \mathbb{Z}/ m \mathbb{Z} \)

The relevant section of Rotman's text reads as follows:In the above text we read the following:

" ... ... The reader should realize that polynomials and polynomial functions are distinct objects. For example, if \(\displaystyle R\) is a finite ring (e.g. \(\displaystyle \mathbb{I}_m\)), then there are only finitely many functions from \(\displaystyle R\) to itself; a fortiori, there are only finitely many polynomial functions. ... ... "

My question is as follows:

How, exactly (indeed, rigorously and formally) can we demonstrate that only finitely many functions from \(\displaystyle R\) to itself implies that there are only finitely many polynomial functions ... ... indeed, how would we prove this statement ...Hope someone can help ... ...

Peter

If a set i finite then each of its subsets is also finite. Here we have the set of all the functions mapping $R$ into $R$ finite if $R$ is finite. A polynomial function is, in particular, a function.