# Polynomial in Z[X]

• May 17th 2008, 03:11 AM
Banach
Polynomial in Z[X]
Hi! I just have a short question: I want to proof the following claim: Let K be a field with characteristic p>0 and f be a polynomial in Z[X] (with Z being the set of integers). Then f(y^p)=f(y)^p for any y in K.

My consideration is this: f=a_0 + a_1 x + a_2 x^2 + ... + a_n x^n Using the binomial theorem it easily follows that f(y)^p=(a_0)^p + (a_1 y)^p + (a_2 y^2)^p + ... + (a_n y^n)^p = a_0 + a_1 y^p + ... + a_n (y^p)^n = f(y^p), as (a_i)^p=a_i. But this last implication only holds if the coefficients are in Z/pZ which is not demanded.

So my questions: Is there a mistake in the description of the exercise? Or are the coefficients to be treated modulo p; if yes, why? Thankful for any tips, Banach!
• May 17th 2008, 04:31 AM
KGene
Quote:

Originally Posted by Banach
Hi! I just have a short question: I want to proof the following claim: Let K be a field with characteristic p>0 and f be a polynomial in Z[X] (with Z being the set of integers). Then f(y^p)=f(y)^p for any y in K.

My consideration is this: f=a_0 + a_1 x + a_2 x^2 + ... + a_n x^n Using the binomial theorem it easily follows that f(y)^p=(a_0)^p + (a_1 y)^p + (a_2 y^2)^p + ... + (a_n y^n)^p = a_0 + a_1 y^p + ... + a_n (y^p)^n = f(y^p), as (a_i)^p=a_i. But this last implication only holds if the coefficients are in Z/pZ which is not demanded.

So my questions: Is there a mistake in the description of the exercise? Or are the coefficients to be treated modulo p; if yes, why? Thankful for any tips, Banach!

Every field of characteristic p>0 contains the base field $\mathbb{Z}/p\mathbb{Z}$, so the equation ${a_i}^p\equiv a_i$ is perfectly all right.
• May 17th 2008, 05:35 PM
ThePerfectHacker
The binomial theorem still holds in over a field.
Thus, $(a+b)^p = a^p + b^p + \sum_{k=1}^{p-1}{p\choose k}a^pb^{p-k}$.
But the coefficients in the summand is each divisible by $p$ so each one dies off and gives you zero.
• May 18th 2008, 12:14 AM
Isomorphism
Quote:

Originally Posted by Moo
Hello,
Assuming that p is prime...
Or did I miss a point in the original text ?

Yes you did. Characteristic of a finite field is always prime.
• May 18th 2008, 12:17 AM
Moo
Quote:

Originally Posted by Isomorphism
Yes you did. Characteristic of a finite field is always prime.

Ok, I have to wake up, I missed this sentence in the wiki..
• May 18th 2008, 12:32 AM
KGene
Quote:

Originally Posted by Moo
Ok, I have to wake up, I missed this sentence in the wiki..

you can see this by writing out $\underbrace{1+1+1+\ldots+1}_\text{m times}=0$ and use the fact that a field has no zero divisor.
• May 18th 2008, 01:23 AM
Banach
Hi! Thanks everyone for your answers. My only problem was the structure that is given between elements of the field and integers, but as i remembered the definition everything cleared up. I got it now. Have a nice day! Banach