Field norm

In mathematics, the (field) norm is a mapping defined in field theory, to map elements of a larger field into a smaller one.

Formal definitions

1. Let K be a field and L a finite extension (and hence an algebraic extension) of K. Multiplication by α, an element of L, is a K-linear transformation

$m_\alpha:L\to L\;.$

That is, L is viewed as a vector space over K, and m_α is a linear transformation of this vector space into itself. The norm N_L/K(α) is defined as the determinant of this linear transformation. Properties of the determinant imply that the norm belongs to K and

N_L/K(αβ) = N_L/K(α)N_L/K(β)

so that the norm, when considered on non-zero elements, is a group homomorphism from the multiplicative group of L to that of K.

2. If L/K is a Galois extension, the norm N_L/K of an element α of L is the product of all the conjugates

g(α)

of α, for g in the Galois group G of L/K.

Example

The field norm from the complex numbers to the real numbers sends

x + iy

to

x² + y². This corresponds to the Galois group having two elements, the identity element which sends x + iy to itself, and complex conjugation, which sends x + iy to x − iy.

Notice that in this case the norm is the square of the "usual" norm, in the sense that we would normally consider the above number x+iy as having norm equal to the Euclidean distance from the origin in the complex plane, which is $\sqrt{(x^2+y^2)}$. In some sense therefore we need to speak of this field norm as an `arithmetic norm' as opposed to the usual Euclidean (distance) norm.

This is particularly important in algebraic number theory where we may have a number field K (a finite extension of the rationals $\Q$ which for simplicity we assume to be Galois), which is naturally itself a vector space of finite dimension d say over $\Q$. The field norm from K down to $\Q$ of any element will be as defined above, namely the product of all of its d Galois conjugates. However we may also embed K in a standard way into the real or complex numbers $\mathbb{R}$ or $\mathbb{C}$ (according to whether K contains any imaginary numbers), and we then consider the Norm (mathematics) or absolute value of the element within the larger field in the usual way. This in general will be very different from the field norm. We give an example below.

Finally, if (as is common in algebraic number theory) we consider K as embedded inside a real vector space $K \otimes_\Q \mathbb{R} \cong \mathbb{R}^d$ then both of these norms will in general be very different from the Euclidean norm inside $K \otimes_\Q \mathbb{R}$ (which is basis-dependent - see below).

As an example of this, consider the number field $K=\Q(\sqrt{2})$ obtained from $\Q$ by adjoining $\sqrt{2}$, which is the splitting field for the polynomial $X^2-2$. The Galois group of this extension is of order d=2, generated by the element which sends $\sqrt{2}$ to $-\sqrt{2}$, and so the norm of the number $1+\sqrt{2}$ is:

$(1+\sqrt{2})*(1-\sqrt{2}) = -1.$

We may also see this using the terminology of "formal definition 1" above if we fix a basis for the extension $\Q(\sqrt{2}):\Q$, say $\{1,\sqrt{2}\}$: then multiplication by the number $1+\sqrt{2}$ sends 1 to $1+\sqrt{2}$ while $\sqrt{2}$ is mapped to $2+\sqrt{2}$. Hence the determinant of the mapping is that of the matrix which sends the vector $(1,0)^T$ (corresponding to the first basis element, ie 1) to $(1,1)^T$ and the vector $(0,1)^T$ (which represents the second basis element $\sqrt{2}$) to $(2,1)^T$, viz.:

$\begin{bmatrix}1 & 2 \\1 & 1 \end{bmatrix}.$

The determinant of this matrix is again -1, showing that the two "formal" definitions of norm agree. However, if we view K as a subfield of the complex numbers $\mathbb{C}$ in the normal way, then the norm (ie the absolute value) is of course $1+\sqrt{2}$.

Finally if we embed K inside $K \otimes_\Q \mathbb{R} \cong \mathbb{R}^2$ then taking $\{1,\sqrt{2}\}$ as a basis for K over $\Q$ we see that $1+\sqrt{2}$ is represented by the vector $(1,1)$ and so its Euclidean norm inside $\mathbb{R}^2$ is $\sqrt{2}$.

Further properties

The norm of an algebraic integer is again an integer, because it is equal (up to sign) to the constant term of the minimal polynomial.

In algebraic number theory one defines also norms for ideals. This is done in such a way that if I is an ideal of O_K, the ring of integers of the number field K, N(I) is the number of residue classes in O_K/I – i.e. the cardinality of this finite ring. Hence this norm of an ideal is always a positive integer. When I is a principal ideal αO_K then N(I) is equal to the absolute value of the norm to Q of α, for α an algebraic integer.

See also

• Field trace
• Ideal norm
• Norm field

References

• Roman, Steven (2006), Field theory, Graduate Texts in Mathematics 158 (Second ed.), Springer, Chapter 8, ISBN 978-0-387-27677-9