The ring extension produced by adjoining x to the ring R, written R[x], is the smallest ring that contains R and x. Here x is the adjoined element. If x is an indeterminant, a variable, R[x] is the set of polynomials in x with coefficients in R. And of course, R could be a field. Thus F[x] is the polynomials in x over F.
The field extension produced by adjoining x to the field F, written F(x), is the smallest field that contains F and x. If x is an indeterminant, a variable, F(x) is the set of rational functions in x with coefficients in F. Recall that a rational function is a reduced quotient of polynomials.
Adjoining the set S means finding the smallest ring or field that conttains the base ring or field, and S. If the elements of S are all indeterminants, we have all polynomials (ring), or quotients of polynomials (field), with variables taken from S and coefficients taken from the base ring or field.
Since extensions are defined in terms of minimum rings or fields containing certain elements, the order of inclusion is not significant. Thus F(x)(y) = F(y)(x) = F(x,y).
If S, the set of adjoined elements, is finite, the extension is of finite type, or finitely generated. When a single element is adjoined the extension is cyclic.
Assume x is not an indeterminant, hence some power of x is set equal to a linear combination of lower powers of x. Said another way, the powers of x are not linearly independent, not a basis for the extension. Perform polynomial math in F[x] as usual, but replace each nth power of x with its lower degree polynomial as you go. This is similar to performing arithmetic mod n.
If p(x) is xn minus its equivalent lower degree polynomial, then p(x) = 0, and x is a root of p. The extension is sometimes written F[x]/p(x), the ring of polynomials mod p(x). In fact, we sometimes shorten the notation to F/p(x), or even F/p. Mathematicians love to overload their operators.
Let F be the reals and let p(x) = (x2+1)2. Let E be the extension F[x]/p(x). every time we run into x4, replace it with -2x2-1. This is a perfectly good ring, but it isn't an integral domain, much less a field. The element x2+1, when squared, produces 0. We can think of x as the imaginary number i, since x2+1 is the square root of 0, but the ring is more than the complex numbers. It is more like two copies of the complex numbers, defined by 1 and x, and x2 and x3.
In general, F/p admits zero divisors whenever p = q×r. The polynomials q and r are the zero divisors, their product being p. Conversely, if F/p contains zero divisors, say q and r, then q×r is a multiple of p. If p is an irreducible polynomial it divides q or r, yet the degree of p exceeds that of q and r, hence p factors into smaller polynomials and cannot be irreducible. Therefore the extension F/p is an integral domain iff p is irreducible.
Note, this argument assumes unique factorization in the polynomials over F. In other words, certain polynomials are "prime", and every polynomial is a unique product of these primes. We proved this using a gcd argument.
In an earlier section we proved that a finite extension is an integral domain iff it is a field. We can now make a stronger statement involving irreducible polynomials. The extension F/p is a field iff p is irreducible.
Let s be an element not in F, and let E be F(s), that is, F adjoin s. Assume E/F is finite, thus s is the root of some irreducible polynomial p(x). If s is the root of some other polynomial q(x) then use the gcd algorithm to show s is the root of a common polynomial r(x). Since r(x) is a factor of p(x), and p is irreducible, r = p, or at least r is an associate of p. There is one irreducible polynomial associated with F(s).
Had we chosen t instead of s, we might get a different polynomial, even though the extension is the same. We can extend the reals into the complex numbers by adjoining i, x2+1 = 0, or 2i, x2+4 = 0. These are different polynomials, with different roots, yet they generate the same field extension.
If E/F is a finite extension, take any element s in E and consider successive powers of s, until one of them is spanned by the earlier powers. This defines a polynomial p(x), with root s. The extension F(s) is a subfield of E. Now E is a finite extension of F(s) with a lower dimension. Repeat the above procedure until the subfield becomes E. Thus E is F adjoin a finite set of elements. Every finite extension is of finite type.
If x is an indeterminant, F[x] is an extension with a countably infinite dimension. Use the powers of x as a basis for F[x]. If the powers of x are dependent, we have a polynomial in x that is equal to 0, and we've already dealt with that.
The field extension F(x) contains more basis elements, to handle the denominators. Let's see what happens when F is the real numbers. The field F(x) includes 1/(x+c) for every real number c. Suppose a linear combination of these fractions sums to 0. Equate the polynomials in F(x) with their corresponding functions in the xy plane. Thus 1 over x+c becomes a real valued function; add c to x and take the reciprocal. We can graph this; it's a hyperbola. It is defined everywhere, except at -c. If a linear combination of polynomials drops to 0, that same combination of functions must be identically zero across the x axis. Yet this is impossible near -c. The other functions are bounded near -c, while 1 over x+c approaches infinity. For some ε, the linear combination can't possibly be 0 near -c. Therefore the fractions 1/(x+c) form an independent set. The field extension F(x) has an uncountable basis. Of course, if F is countable, such as the rationals, then the set F(x) is also countable, and the field F(x) has countable dimension over F.
As mentioned earlier, F[x] is finitely generated as a ring, with only one generator, namely x. The field F(x) is quite a different story. Suppose F(x) is finitely generated as a ring. Each generator is a reduced quotient of polynomials. Since F[x] is a ufd, there are finitely many primes in each denominator. Together the generators bring in finitely many primes, and since there are infinitely many primes, select one, and its reciprocal is not spanned. Therefore F(x) is not a finitely generated F algebra. Of course this assumes there are infinitely many primes - a prime being an irreducible polynomial. If F is infinite, each monomial x+c is prime. If F is a finite field there are plenty of irreducible polynomials over F. Thus F(x) is an infinitely generated F algebra.
The same proof shows finitely many rational numbers cannot conspire to build all of Q.
As you can see, the term "finitely generated" is rather ambiguous. The generators could be building a group, a vector space, a module, a ring, or a field. We are simply looking for the smallest structure containing those generators.