1
$\begingroup$

I have a very basic question in number theory. How could we define a polynomial over a finite field which has "prime power" elements? As far as I know, if a field has prime power elements, it will include polynomials as field elements. But if we try to define a polynomial over this field, we will get a polynomial for which the coefficients are also polynomials (field elements).

I'm confused. Could someone give me a simple example?

$\endgroup$
5
  • $\begingroup$ What is "a prime power element"? And why do you think elements in a field are polynomials?? $\endgroup$
    – DonAntonio
    Feb 24 '17 at 0:07
  • $\begingroup$ prime power element means that the number of elements inside the field is a prime power like 8 (2 to the power of 3) $\endgroup$ Feb 24 '17 at 0:29
  • $\begingroup$ Oooh! Well, I couldn't have ever guessed that. Anyway, if you want a standard notation for that you can use $\;\Bbb F_{p^n}=\;$ the field (the unique one up to isomorphism) with $\;p^n\;$ elements, $\;p\;$ a prime. $\endgroup$
    – DonAntonio
    Feb 24 '17 at 0:31
  • 1
    $\begingroup$ A better phrasing is "a prime power number of elements". The way you've written it sounds like you are asserting that each element of a finite field is a prime power. $\endgroup$
    – user14972
    Feb 24 '17 at 0:39
  • 1
    $\begingroup$ Incidentally, there's nothing wrong with polynomials having coefficients named by polynomials. Or even coefficients that are actually drawn from a polynomial ring. In fact, it is fairly useful to view the bivariate polynomial ring $R[x,y]$ as being the ring of univariate polynomials $S[y]$ over the ring $S$, where $S = R[x]$ is the ring of univariate polynomials over $R$. $\endgroup$
    – user14972
    Feb 24 '17 at 0:40
1
$\begingroup$

It may help to understand that there is more than one way of thinking about a polynomial. One way that you may be more familiar with is as a function that transforms the elements of a field. Consider for example the polynomial $p(x) = x^2 + 1$ over the field $\mathbb{F}_2$ (we'll consider prime powers later). As a function, $p(0) = 1$ and $p(1) = 0$. We could also consider the polynomial $q(x) = x + 1$ over $\mathbb{F}_2$. As a function, $q$ corresponds to the same mapping as $p$. However, as polynomials, we cannot say that $p(x) = q(x)$ because they have different coefficients.

In a similar fashion, a polynomial over a field of prime power order is defined by its coefficients. To see this, it may help to use a different indeterminate to write out the coefficients if you must write them explicitly. For an arbitrary example, you might write the polynomial where the coefficient of $x^2$ is $y + 2$ and the constant term is $3$ over $\mathbb{F}_{25}$ as $(y + 2) x^2 + 3$. In this context, this would be a polynomial of a single variable because $y + 2$ is a coefficient and not a second indeterminate. You cannot "simplify" a polynomial by mixing coefficients with indeterminates. Conceptually, we could also define the polynomial as the vector $(y + 2, 0, 3)$ to make the distinction even clearer.

$\endgroup$
0
$\begingroup$

Not sure of what you mean by prime power elements, but I'm afraid you're confusing polynomials and polynomial functions.

For instance, let's consider the prime field $\mathbf F_p=\mathbf Z/p\mathbf Z$, and the polynomial ring $\mathbf F_p[X]$. In this ring, the non-zero polynomial $X^p-X$ induces the polynomial function \begin{align}f\colon\mathbf F_p&\longrightarrow \mathbf F_p\\x&\longmapsto x^p-x, \end{align} which is $0$ by Little Fermat.

$\endgroup$
3
  • $\begingroup$ In short: with polynomials it's all about coefficients, with polynomial functions it's all about values. $\endgroup$ Feb 25 '17 at 20:09
  • $\begingroup$ Absolutely. Let me add there's an isomorphism between them in characteristic $0$. $\endgroup$
    – Bernard
    Feb 25 '17 at 21:21
  • $\begingroup$ @Morgan Rodgers: I'm not sure either, as the O.P. question was rather confusing. I answered what seemed most likely to me, waiting for some reaction to adjust what I could explain $\endgroup$
    – Bernard
    Feb 27 '17 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.