What happens if the coefficients of polynomials are not taken from a field of real numbers?

Question

I saw in my abstract algebra textbook that defines the gcd of a polynomial over a field (i.e. the coefficients of the polynomial is taken from a field). My question is that what happens if the field is not a real numbers? I mean intuitively, I always think of it as real numbers. Is there anything wrong in taking of it this way. Can someone give me an example of a polynomial taken from other fields such that thinking of it as real numbers will result in a misconception. Thank you.

Answer 1

Over the real numbers, every polynomial can be factored into quadratic polynomials. (This is a consequence of the mean value theorem from calculus.) But in other fields, some polynomials of high degree cannot be factored at all. For example, the polynomial $x^p - x + a$ cannot be broken into factors of smaller degree when the coefficients are taken to lie in the finite field $\mathbb F_p$ ($p$ a prime).

There are also some fields which are much larger than $\mathbb R$. Thinkings of elements of these fields as real numbers would simplify them too much. For example, the set of rational functions from $\mathbb R$ to $\mathbb R$ forms a field, called $\mathbb R(t)$. Here's an example of a polynomial with coefficients in that field: $$ \frac{t^4 + t + 1}{2t - 6} x^2 + \frac{t^8-t}{t^3+t^2-t-1}x + \frac{t+1}{t-1}. $$

Answer 2

In $\mathbb Z_7$ - the field with the elements $0,1,2,3,4,5,6$ and addition and multiplication defined modulo $7$ we have $$x^7-x=x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$$which is a factorisation you don't get in the real numbers, $[x(x+1)(x-1)(x^2+x+1)(x^2-x+1)]$.

You get a similar factorisation for any prime instead of $7$.

One of the reasons for exploring different fields is that they can give extra information about the structures we are studying.

Answer 3

Say the answers to this question have already convinced you, and you're thinking, "Okay, real numbers aren't enough for me". Your first instinct will probably be to look at the following fields: $\mathbb{Q}$, $\mathbb{C}$, and $\mathbb{F}_{p^n}$. But even that won't tell you the full story. Here's a phenomenon that can't occur in these fields:

For any of these fields (call it $F$), "$p$ irreducible" implies "all roots of $p$ are distinct". (We can talk about the roots that lie outside $F$, because we can just look at its algebraic closure.) It turns out this is a really nice property (separability).

Unfortunately, this isn't always true. Let $F = \mathbb{F}_{p^n}(t)$. The polynomial $x^p - t$ is irreducible (what would $t$ factor as?). But all of its roots are the same! Let $u$ be a root of $p$. Since $u^p = t$: $$ (x - u)^p = x^p - u^p = x^p - t $$

Fields where every polynomial is separable are called "perfect", and as the name implies, are much nicer than those that are not.

Answer 4

The first misconception you get if you always think of real numbers is that there is always infinitely many polynomial of degree $k$ (for some $k\in \mathbb N$). If the coefficients are from a finite field (like $\mathbb Z/2\mathbb Z$), you have finitely many polynomials of degree $k$. Such polynomials on finite fields are used in many areas, especially related to computer science, like in error-correcting codes.

Another crucial problem is that on $\mathbb R$, some non-constant polynomials have no root, while non-constant polynomials on $\mathbb C$ always have roots.