There are two complete fields with respect to archimedean valuations – i.e. valuations for which |x + y| <= max{|x|, |y|} does not always hold – are R and C. This is because Ostrowski’s theorem says that every complete archimedean field containing Q must also contain R, i.e. be R or C, and if a field doesn’t contain Q then it must contain F(p), which admits no archimedean valuations.
Now, C is algebraically closed, so a polynomial over C is irreducible iff it is linear. R is a bit more complicated, but still easy because its algebraic closure is C, and [C:R] = 2. Suppose f is in R[x]. Then f splits into linear factors over C, and if z in C is a root of f, then so is z, the complex conjugate of z; this is because complex conjugation is a field isomorphism from C to itself that fixes every element of R, so that it preserves polynomial equations with coefficients in R.
Now, (x – z)(x – z) = x^2 – (z + z) + zz, where z + z and zz are always real, so if f has a proper complex root, then it has a factor of degree 2. If it has no proper complex roots, then it of course splits into linear factors. In other words, every irreducible polynomial over R has degree 1 or 2. Further, a linear polynomial is always irreducible, while a quadratic ax^2 + bx + c is irreducible iff its two complex roots are proper complex, i.e. iff (-b (+/-) SQRT(b^2 – 4ac))/2a is not real, i.e. iff b^2 < 4ac.
Local fields are a lot more involved, because their algebraic closures have infinite degree over them. The same of course applies to global fields – recall that A has infinite degree over Q, for instance. Hensel’s lemma provides one local field analog of polynomial factorization in R, which more or less corresponds to the statement that every real polynomial of odd degree has a real root.
There’s another lemma that vaguely resembles the b^2 < 4ac condition, which is also useful in constructing extensions of local fields (more on this later): if K is a local field and f is an irreducible polynomial over K, a(n)x^n + … + a0, then |a(i)| <= max{|a0|, |a(n)|} for all i. Equivalently, the coefficient of f whose valuation is the highest is either a0 or a(n).
To see why, first multiply by a suitable power of a prime element p in the discrete valuation ring R of K, so that f becomes a polynomial over R. If all coefficients of f are in P, then divide by some power of p to make f have at least one coefficient not in P. In this configuration, the coefficients of f whose valuation is the highest must have valuation exactly 1, so that they’re the only ones outside P.
Now, let j be the highest index for which |a(j)| = 1. It suffices to show that j = 0 or n. But if 1 <= j <= n-1, then we can reduce f mod P; as a(i) is in P for i > j, the lowest term in f with nonzero coefficient mod P is x^j. But then f = x^j(a(n)x^(n–j) + … + a(j)) mod P, and x^j and a(n)x^(n–j) + … + a(j) are coprime since a(j) != 0 mod P. Also, j and n–j are not zero, since by assumption 1 <= n <= n-1. Therefore, by Hensel’s lemma, f is reducible as the product of polynomials equivalent to x^j and a(n)x^(n–j) + … + a(j) mod P, contradicting the assumption that f is irreducible.
Note that this lemma says |a(i)| <= max{|a0|, |a(n)|}, not |a(i)| < max{|a0|, |a(n)|}. That is, it’s entirely possible for a coefficient other than a0 and a(n) to attain the maximum valuation, as long as it shares it with a0 or a(n).
Abstract Nonsense 