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Structure of a Finite Field

Recall that a field F has characteristic p if p is the smallest positive integer such that for every nonzero element \alpha in F, we have p \alpha = 0. If no such integer exists, then F has characteristic 0.

Structure of a Finite Field

Recall that a field \(F\) has characteristic \(p\) if \(p\) is the smallest positive integer such that for every nonzero element \(\alpha\) in \(F\), we have \(p \alpha = 0\). If no such integer exists, then \(F\) has characteristic \(0\). From we know that \(p\) must be prime. Suppose that \(F\) is a finite field with \(n\) elements. Then \(n \alpha = 0\) for all \(\alpha\) in \(F\). Consequently, the characteristic of \(F\) must be \(p\), where \(p\) is a prime dividing \(n\). This discussion is summarized in the following proposition.

Throughout this chapter we will assume that \(p\) is a prime number unless otherwise stated.

Proposition

If \(F\) is a finite field of characteristic \(p\), then the order of \(F\) is \(p^n\) for some \(n \in {\mathbb N}\).

Proof

Let \(\phi : {\mathbb Z} \rightarrow F\) be the ring homomorphism defined by \(\phi(n) = n \cdot 1\). Since the characteristic of \(F\) is \(p\), the kernel of \(\phi\) must be \(p {\mathbb Z}\) and the image of \(\phi\) must be a subfield of \(F\) isomorphic to \({\mathbb Z}_p\). We will denote this subfield by \(K\). Since \(F\) is a finite field, it must be a finite extension of \(K\) and, therefore, an algebraic extension of \(K\). Suppose that \([F : K] = n\) is the dimension of \(F\), where \(F\) is a \(K\) vector space. There must exist elements \(\alpha_1, \ldots, \alpha_n \in F\) such that any element \(\alpha\) in \(F\) can be written uniquely in the form \[\begin{aligned}\end{aligned}\], where the \(a_i\)'s are in \(K\). Since there are \(p\) elements in \(K\), there are \(p^n\) possible linear combinations of the \(\alpha_i\)'s. Therefore, the order of \(F\) must be \(p^n\).

Let \(F\) be a field. A polynomial \(f(x) \in F[x]\) of degree \(n\) is separable if it has \(n\) distinct roots in the splitting field of \(f(x)\); that is, \(f(x)\) is separable when it factors into distinct linear factors over the splitting field of \(f\). An extension \(E\) of \(F\) is a separable extension of \(F\) if every element in \(E\) is the root of a separable polynomial in \(F[x]\).

The unique finite field with \(p^n\) elements is called the Galois field of order \(p^n\). We will denote this field by \(\gf(p^n)\). \(\gf(p^n)\) Galois field of order \(p^n\)

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Енді сен Бұл есепті шешетін калькулятор жоқ, бірақ оның бөлшектері есептеледі. Төмендегілерді қолданып көріңіз немесе өзіңіздіңіңізді келтіріңіз.

Өзіңіздің жұмысыңызды жалғастыру

Тіркелгіңізде әрбір сабақта жазбалар, орындалған тапсырмалар, шешілген тапсырмалар бір жерде сақталады, және мұғалімге осы бет туралы сұрақ қоюға болады. Математикалық сабақтар жүйеге кірген немесе кірмеген кез келген адамға ашық.

Жазылу Кіру

Мұнда қолданылатын символдар

Тұтас анықтамасын, суретін және әрбір әріптің мағынасын көру үшін кез келген символды түртіңіз.

Адамдар қоятын сұрақтар

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Бұл парақтағы кейбір бөлшектер мынадан алынған Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Бұл жерде қайталап түсіндірілді; қателер біздің кінәміз.

Келесіде Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula