Quadratic Extensions, An Introduction

Introduction

A quadratic field extension is an extension of dimension 2. Let E/F be such an extension and let u be anything in E-F. Now Fu = E, and u satisfies an irreducible polynomial of degree 2. Such a polynomial is called quadratic, hence we have a quadratic extension.

Conversely, adjoin the root of an irreducible quadratic polynomial to a field F to get an extension E/F of dimension 2. The extension is quadratic iff it is F(u), where u is the root of an irreducible quadratic polynomial.

If F = Q, the rationals, the extension is called a quadratic number field.

If the extension E/F lives in the reals, it is a positive quadratic extension. If the extension brings in complex numbers, it is a negative quadratic extension. This terminology comes from the discriminant d of the polynomial p(u), which is positive or negative respectively.

Apply the quadratic formula, and the root u of p(x) is sqrt(d) modified by the coefficients of p, which already live in F. Adjoining u produces the same extension as adjoining sqrt(d). All that matters is the value of d.

If two discriminants d1 and d2 have a ratio that is a square in F, then sqrt(d1) is a scale multiple of sqrt(d2). either spans the other, and they both produce the same extension. The value of d only matters up to the squares in F.

Integrally Closed

Let R be integrally closed, and let F be the fraction field of R. Multiply u by something in R, so that u becomes algebraic over R. This does not change the field extension. since R is integrally closed, the minimum polynomial of u is monic, with coefficients in R. Thus we can look at polynomials x2+bx+c, having discriminant d = b2-4c. If d is a square of something in F then d is integral over R, and since R is integrally closed, d lies in R. We only need ask whether d is a square in R. If not, the extension is quadratic.

PID

Let R be a pid. Now the field extensions and their integral rings correspond. This can help characterize both the quadratic field extensions and the integral rings thereof. Then, if we wish, we may explore smaller subrings, although these subrings are not integrally closed, hence they are not dedekind. They are therefore of limited utility.

Norm and Trace

Let u be an element in E-F. Thus E = F(u). For simplicity, assume u = sqrt(d), for some d in F. Now the conjugate of u is -u, and the norm of a+bu is the product of the conjugates, giving a2-db2. The trace is 2a.

This formula works even when F has characteristic 2, but then the extension is inseparable, and we are usually interested in separable extensions, so this doesn't come up very often.

If a ring extension has the basis 1 and u, with coefficients in R, an endomorphism on R[u] extends to an endomorphism on E, by tensoring with F. The matrix that implements the endomorphism is the same. The trace and norm are the same. Thus trace and norm in a ring extension R[u] can be evaluated in F[u], which is E. In other words, the above formulas apply.