Number of solutions a polynomial can have as a function of the field?

Question

Is there any limitation (upper bound) for number of solutions of polynomial equations? Having a background in engineering, my knowledge of higher algebra is rather limited, but I do know of

1. Real numbers having At Most N roots for an N-deg polynomial
2. Complex numbers Always having N roots for an N-deg polynomial
3. Multivariable real valued polynomials which can have infinite number of roots.

Famous example $x^2+y^2-1=0$ having solutions all along the unit circle.

4. Quaternions having ?? roots for an N-deg polynomial(?)

Are different types of numbers divided into different types or classes depending on how many solutions polynomials in them can have? Given definition of a type of number, is there some way to easily decide which of these classes or types they will belong to?

Answer 1

For fields we have

A polynomial equation of degree $n$ has at most $n$ solutions.

For an algebraically closed field, such as $\mathbb C$, we have

A polynomial equation of degree $n$ has exactly $n$ solutions, if you count them with multiplicity.

Perhaps a bit surprisingly, the commutativity of fields is essential for this:

The equation $x^2+1=0$ has infinitely many solutions.

Indeed, if $a, b, c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$, then $ai+bj+ck$ is a solution of $x^2+1=0$ in the quaternions.

For multivariable real polynomials $f(x_1,\ldots, x_d)$, one natural generalization is

How many connected components does $f(x_1,\ldots, x_d)=0$ have?

Harnack's curve theorem gives an answer in the bivariate case:

An algebraic curve of degree $n$ in the real projective plane has at most $\dfrac{(n-1)(n-2)}{2}+1$ connected components. This bound is tight.

Answer 2

I do not know if I' m answering your question i think this will help you. Suppose you have ab irreducible polynomial $f(x)$ with coefficients over a field $F$ then you can consider the extension of the field $F$, $F\leq K$, where all the roots of the polynomial $f(x)$ exists there. It is easy to prove that such a field exists for every irreducible polynomial $f(x)\in F[x]$, where $F$ is a field, just observe that if $p(x)\in F[x]$ is an irreducible factor of $f(x)$ then $F[x]/I$, where $I=<p(x)>$, is an extension of the field $F$(this can be proved by the first theorem of isomorphisms) where $f(x)$ has a root. Especially the root in the extension is $x+I$. I hope I helped.