You are probably familiar with field extensions; ring extensions are similar. In fact every field extension is a ring extension. Like fields, one may adjoin a set of elements W to R to produce the ring extension denoted R[W]. If W is a finite set of indeterminants then we have the ring of polynomials with variables taken from W and coefficients taken from R.
If the extension S/R contains u, and u is the root of a monic polynomial in R[x], then u is integral. If every u in S is integral then S is an integral extension. This is similar to the definition of an algebraic extension of a field, but this time the base could be a ring, and the polynomial must be monic, i.e. having a lead coefficient of 1.
If the base is a field, then any polynomial can be scaled by the inverse of its lead coefficient, creating a monic polynomial with the same roots. Integral and algebraic elements coincide. Similarly, an integral extension and an algebraic extension are synonymous. Well that's not very interesting, so let's return to the world of commutative rings.
If R is Z, forinstance, a number as simple as ½ is not integral. It is the root of 2x-1, which is not a monic polynomial. Now suppose for a moment that ½ is a root of some other monic polynomial p. Switch over to rational coefficients, and apply the gcd algorithm to p and 2x-1. After you're done, ½ becomes the root of a 0 degree polynomial, which can only be 0. In other words, 2x-1 divides p. Apply Gauss' lemma to show that p splits in Q[x] iff it splits in Z[x]. In other words, the "other" factor actually has integer coefficients, just like 2x-1. So 2x-1 times an integer polynomial yields p, and that means the lead coefficient is even, hence p is not monic. Therefore, as you probably guessed, ½ is not integral over Z.
However, sqrt(2) is integral over Z, a root of x2-2.
Since h divides g in the fraction field, h divides g in R. Since the lead coefficient of g is 1, the lead coefficient of h is a unit. Divide through, and h is monic. The minimum polynomial of u is the monic polynomial of u.
If f(x) has root u, then divide f by h, and the remainder also has root u. The remainder has a lesser degree, which is impossible. Hence the remainder is 0, and h divides f. Every polynomial with root u is divisible by the minimum polynomial h.
If h = j*k, then either j(u) = 0 or k(u) = 0. Since h has minimum degree, this is impossible. Therefore h is irreducible.
If f is some other irreducible polynomial with root u, then h divides f, as described above, and h = f.
Let p(x) be the monic polynomial that proves x is integral over R. Thus p is monic, and has coefficients in R. Leave the lead coefficient alone, and multiply the next coefficient by c, and the next one by c2, and so on, out to the constant coefficient, which is multiplied by cn. The new polynomial proves cx is integral over R.