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As I've started studying Polynomial Ring on my own, I would like to verify/ask the concept/questions occurred to me. I've noticed over some ring the polynomials are of little/no interests as a function and all we're concerned about is the components obtained using $x^n~(n\ge0)$ as 'separator'. It's the reason why over $\mathbb Z_3$ even though $x^3$ and $x^5$ are identical as a function, differ as polynomial. Consequently, the associated set in a ring of polynomials is the family of all sequences over the ring whose all but finitely many terms are zero.

I wonder why I always thought it as a function earlier when I studied the basic polynomials (over $\mathbb R$ or $\mathbb C$) and why its specialty didn't occur to me then. I guess it's because of the behavior of the polynomials over those fields. As I can see that no two distinct polynomials can occur in $\mathbb R$ or $\mathbb C$ which are functionally identical, since any polynomials over $\mathbb R$ or $\mathbb C$ can at most finite number of roots.

Am I right?

Sriti Mallick
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2 Answers2

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Yes, you are correct. It is worth noting that for many applications it is helpful to think of polynomials as both functions and formal objects. For example:

Factor Theorem: Let $f(x)$ be a polynomial over a field $F$. Then $(x-c)\mid f(x)$ (which is a statement about polynomials as formal objects) iff $f(c)=0$ (which is a statement about polynomials as functions).

Additionally, the (I believe) most common proof of the Fundamental Theorem of Algebra, which states that any nonconstant polynomial over $\mathbb C$ has a root in $\mathbb C$ uses in a crucial way the fact that real polynomials are continuous functions, in the form of this lemma:

Lemma: Any polynomial $f(x)$ over $\mathbb R$ with odd degree has a real root.

Proof: Since $f(x)$ has odd degree, the signs of $f(x)$ disagree for very negative $x$ and very positive $x$. Since polynomials are continuous, by the Intermediate Value Theorem we have some $c\in\mathbb R$ such that $f(c)=0$.

Alex Becker
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  • Though all this began for a verification of my realization your answer definitely added a different dimension to the whole discussion saying that polynomial isn't something different from function rather something more than a function. – Sriti Mallick Jun 04 '13 at 08:03
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Your question, as I understand it, is essentially this:

For which fields $F$ can any polynomial $g \in F[x]$ be uniquely determined by its values on $F$?

Suppose that we have polynomials $f, g \in F[x]$ such that $f(c) = g(c)$ for each $c \in F$. If we put $h = f - g$, this means that $h(c) = 0$ for each $c \in F$, since $h(c) = f(c) - g(c) = 0$. So, equivalently, we may ask which fields $F$ have a nonzero polynomial $h \in F[x]$ such that $h(c) = 0$ for every $c \in F$.

If $F$ is an infinite field, then this is impossible, since $h$ would then have infinitely many roots, and consequently have infinite degree. So this can only happen over a finite field. Over a finite field, of course, you can just multiply together a term of the form $(x - c)$ for each element $c \in F$ and get a polynomial that vanishes at each element of $F$. Over $\mathbb{F}_3$ this polynomial is

$$x(x-1)(x - 2) = x^3 - 3 x^2 + 2 x = x^3 - x.$$

As you've pointed out above, over $\mathbb{F}_3$ the polynomials $x^5$ and $x^3$ take on the same values everywhere, which we can see as their difference $x^5 - x^3$ is a multiple of $x^3 - x$, namely $x^2(x^3 - x)$.

Of course as you say $\mathbb{R}$ and $\mathbb{C}$ are examples of infinite fields. So are certain positive-characteristic fields like $\mathbb{F}_q(x)$ and $\overline{\mathbb{F}_q}$ (i.e. the algebraic closure of a finite field).

To get more to the heart of things, yes, polynomials are sets of coefficients, rather than the functions that can be defined by those coefficients. This definition makes everything work out better, especially since polynomials don't have to be defined over fields and can be in more than one variable, in which things get much messier than the situation I described above.

Daniel McLaury
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  • For every $\mathbb{F}_p$, the polynomial $x^p - x$ is $0$ everywhere. IIRC, this is also the same as $x(x-1)(x-2)\dots(x-(p-1))$. Are all everywhere-$0$ polynomials multiples of this one? – ShreevatsaR Jun 04 '13 at 05:55
  • Well, remember, if $q = p^n$ then $\mathbb{F}_q$ is not $\mathbb{Z}/q\mathbb{Z}$ in the same way that $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$. The ring $\mathbb{Z}/q\mathbb{Z}$ is not a field for $n > 1$. – Daniel McLaury Jun 04 '13 at 06:00
  • However a statement of the form you've described is true. – Daniel McLaury Jun 04 '13 at 06:12
  • Yes I know that... I should have been clearer; by the notation $p$ I meant $\mathbb{F}p$ for prime $p$. I asked for prime $p$ first instead of all $p^n$, because I thought verifying the statement (that all everywhere-0 polynomials are multiples of $x^p - p$) for these finite fields would be a first step to verifying it in general for all $\mathbb{F}{p^n}$. – ShreevatsaR Jun 04 '13 at 06:13