What's the difference between polynomials and polynomial functions?

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What's the difference between polynomials (as elements of a ring of polynomials) and polynomial functions??
 

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  • #2
disregardthat
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There is just a technical difference, a polynomial function has a domain and co-domain associated to it, whereas a polynomial in a polynomial ring does not. A polynomial in [tex]\mathbb{Q}[x][/tex] may be viewed as a function from the integers, rationals, reals, complex numbers, real nxn matrices, function spaces, sequence spaces or anything with a ring structure. The domain need to be specified before you technically can say it is a function (the co-domain is usually implicit).

There are also differences in some formal operations. For example in the field of formal rational functions over, say R, you can say that x*1/x = 1, but as functions from the reals they are not equal; the former is undefined at 0. Also, in the ring of power series we formally have the equality (1-x)(1+x+x^2+...)=1, but as functions from the reals you will need to treat such series as limits (you can't have an infinite sum of numbers), and thus specify where it actually makes sense (determining the range of convergence). Algebraic polynomials are as such more well-behaved than polynomial functions.

The main difference is that the variables in polynomials are indeterminates, not numbers (or matrices etc.). Thus "solving" for 3x+2=5 as formal polynomials is meaningless because 3x+2 and 5 are different polynomials (compare with "solving" for 3=10). x is not considered a number you can solve for, but rather an indeterminate with certain formal rules for manipulation.
 
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  • #3
mathwonk
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this is not correct over finite fields, i.e. they are only the same over infinite fields. I.e. over a finite field, the map from polynomials to polynomial functions has a huge kernel.

e.g. over Z/pZ, the non zero polynomial (x-1)(x-2).....(x-p) corresponds to the zero function on Z/pZ.
 
  • #4
Hurkyl
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you can say that x*1/x = 1, but as functions from the reals they are not equal; the former is undefined at 0.
A nitpick; if x is a real-valued indeterminate, then those aren't functions from the reals. At least, 1/x is not. Depending on the specific language you are using, 1/x is either a partial function or a grammatical error.
 
  • #5
Landau
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Let R,S be rings.

A ring homomorphism

[tex]\phi:R\to S[/tex]

induces a ring homomorphism

[tex]\Phi:R[X]\to S[X][/tex]
[tex]\sum a_iX^i\mapsto \sum \phi(a_i)X^i.[/tex]

For every s in the center of S (i.e. s commutes with all elements of S), there is a ring homomorphism

[tex]\psi_s:S[X]\to S[/tex]
[tex]p=\sum b_iX^i\mapsto \sum b_is^i,[/tex]

usually we write [itex]\psi_s(p)=p(s)[/itex].

These two combine to give the evaluation homomorphism:

For every s in the center of [itex]\Phi(R)[/itex] it is the composition:

[tex]\Phi_s:=\psi_s\circ \Phi:R[X]\to S[/tex]
[tex]p=\sum a_iX^i\mapsto \sum \phi(a_i)s^i=\psi_s(\Phi(p))=[\Phi(p)](s)[/tex]
 
  • #6
disregardthat
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A nitpick; if x is a real-valued indeterminate, then those aren't functions from the reals. At least, 1/x is not. Depending on the specific language you are using, 1/x is either a partial function or a grammatical error.

Yes, you are right, but my point was that polynomials as such does not depend on any domain on which they are well defined (right hand side is well-defined as a function on the reals, the left hand side not, while both being algebraically equal)
 
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  • #7
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this is not correct over finite fields, i.e. they are only the same over infinite fields. I.e. over a finite field, the map from polynomials to polynomial functions has a huge kernel.

e.g. over Z/pZ, the non zero polynomial (x-1)(x-2).....(x-p) corresponds to the zero function on Z/pZ.

"this is not correct over finite fields, i.e. they are only the same over infinite fields."i do not understand........can someone explain any further????
 
  • #8
Landau
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Well, read the next line you quoted. Mathwonk gave an illustration of that statement!
 

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