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My abstract algebra textbook* has taken great care in the last few sections to stress that, "polynomial are NOT functions to just 'plug' values of x into". To fully 'get' this, and to 'get' what polynomials really are then, I read through the proofs and explanations in the appendix that build up the idea of what a polynomial rings really is, and how we know a polynomial ring does exist for all rings.

In the current section of the book however, the book simply stated that "every polynomial induces a function from R to R", and then started to use polynomials as functions again, and the way it's doing so has me confused- what does it mean that it "induces" a function? The book will use the polynomial as function in a proof, and then make a conclusion about the polynomial itself- why, really, is that step valid?

The easiest example of this is it's proof for the polynomial remainder theorem.

The remainder theorem for polynomials.

The proof.

I get that, by the division algorithm for polynomials and the fact that (x - a) is degree 1, the following is a true statement: $\exists q(x),r(x)\in R[x] \; s.t \; f(x) = (x - a)q(x) + r(x)$ and $r=0 _r$ or $degree(r) = 0$

Isn't the next statement, namely that $f(a) = (a - a)q(a) + r(x) $, just a statement about some elements of in the ring R? Why can we say that the r(x) from the first statement is the same r(x) in the second statement? In other words, why is this second statement true?

/***********************/

I've worked ahead on some of the more complicated and involved stuff, but I have this nagging frustration in some part of my brain, as I don't "get" how these polynomials are really related with their evaluation maps, and when/why we can/cannot use a polynomial's evaluation map to make a conclusion about the polynomial itself.

*Here are some related threads I looked at that I thought about for a bit:

*I feel like there is something very simple I'm missing.. I've spent too much time thinking about this and I think I've muddled my brain. Also, my bad if I've formatted this poorly, this is my first post. Thanks a lot for the help.

Edit: *My textbook: Abstract Algebra, An Introduction - Third Edition, by Thomas W Hungerford and David Leep

Syed M. Akbari
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  • You can always treat a polynomial as a function by evaluating it, but you shouldn't think of them as ring elements this way, because this can lead to the wrong ring structure w/r/t addition and especially multiplication. – Randall Mar 01 '21 at 02:41
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    what textbook do you mean? – Will Jagy Mar 01 '21 at 02:42
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    note that $x^p$ and $x$ are different polynomials, but in $\mathbb Z/p\mathbb Z$ their evaluations are the same – J. W. Tanner Mar 01 '21 at 03:20
  • "abstract algebra, an introduction" - third edition, by Thomas W Hunderford and David Leep – Syed M. Akbari Mar 01 '21 at 03:31

2 Answers2

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Perhaps what the author wanted to say is that albeit polynomials are usual treated as functions through its evaluation map in elementary algebra courses, the more appropriate view of each polynomial $p(x_1,\cdots,x_n)$ is that it is an element of the ring $R[x_1,\cdots,x_n]$.

Note that, the evaluation homomorphism is not always guaranteed to exists if $R$ is noncommutative, as in the thread below.

Evaluation homomorphism for noncommutative rings

Minh Khôi
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If $p(x)$ is irreducible over the field $\Bbb F$, then the evaluation homomorphism at any root $\alpha$ of $p$ is an isomorphism between $\Bbb F[x]/(p)$ and $\Bbb F(\alpha)$, the field you get by adjoining the root to $\Bbb F$.