Polynomials and field extensions
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| irreducible polynomial/ɪrɪˈdjuːsɪbl ˌpɒlɪˈnəʊmɪəl/ | 不可约多项式 | bù kě yuē duō xiàng shì |
| extension degree/ekˈstenʃn dɪˈɡriː/ | 扩张次数 | kuò zhāng cì shù |
A decision before an answer
- A field with four elements does not use arithmetic modulo 4. It needs a new element satisfying a polynomial relation.
- Your goal: Test polynomial irreducibility over the specified field.
Read the relationship
- A polynomial is irreducible over a field if it has positive degree and no factorisation into polynomials of smaller positive degrees there. A quadratic or cubic is irreducible exactly when it has no root in that field. This root test alone fails for degree four or higher; for instance (x²+1)(x²+2) has no real root but is reducible over R. Always name the base field.
- Construct small finite fields from irreducible polynomials.
Which polynomial is irreducible over F2?
The quadratic x²+x+1 has no root among 0 and 1. The others factor as x·x, (x+1)², x(x+1) and x²(x+1), respectively, over F2.
Use the defining rule
- The quotient F[x]/(p) is a field when p is irreducible. Reduce powers using p(alpha)=0, where alpha is the residue class of x. If F has q elements and p has degree d, the quotient has q^d elements represented by polynomials of degree below d. Z/4Z has four elements but has zero divisors, so it is not the field with four elements.
- Use extension degrees and the tower law.
If [L:K]=3 and [M:L]=2, what is [M:K]?
The tower law multiplies finite degrees, giving 2 times 3=6.
Check the conditions
- Over F2, p(x)=x²+x+1 has values 1 at both 0 and 1 and is irreducible. In its quotient alpha²=alpha+1 because subtraction equals addition in characteristic two. The four elements are 0,1,alpha,alpha+1. All three nonzero elements must be units; compute their products rather than treating alpha as an ordinary real number.
- Use cyclotomic roots and coefficient relations to compute sums and products.
In F2[alpha] with alpha²+alpha+1=0, multiply alpha(alpha+1) = alpha² + alpha = (alpha+1) + alpha = 1. Thus alpha^−1=alpha+1. Also alpha³=1 and alpha is not 1, so its multiplicative order is 3. This produces a field of four elements. By contrast 2·2=0 in Z/4Z, proving that quotient is not a field.
In the field F2[alpha] above, alpha³ equals ____.
alpha²=alpha+1 gives alpha³=alpha²+alpha=1.
Apply the task format
- For nested finite-degree fields K inside L inside M, the tower law gives [M:K]=[M:L][L:K]. The degree of an algebraic element is the degree of its minimal polynomial. A finite extension of degree two does not contain an element of degree three over the base field. Over Q, sqrt(2) has degree two; adjoining sqrt(3) as well produces a degree-four extension, since sqrt(3) is not in Q(sqrt(2)). Over C, primitive nth roots of unity have exact order n and are the roots of the cyclotomic polynomial Φ_n. For n=10, divide x⁵+1 by x+1 to exclude the order-two root −1: Φ_10=x⁴−x³+x²−x+1. Vieta’s formulas give sum 1 and product 1 of its four primitive roots. Do not sum all tenth roots, or assume every nontrivial tenth root is primitive; a root’s order must be checked. For a monic degree d polynomial, product of its roots is (−1)^d times the constant coefficient.
- Use cyclotomic roots and coefficient relations to compute sums and products.
Polynomial reducibility changes with the coefficient field. The polynomial x²−2 is irreducible over Q but splits over R.
Which answer fits this case?
Test polynomial irreducibility over the specified field
A degree-four polynomial without real roots is necessarily irreducible over R.
(x²+1)(x²+2) has no real roots and is explicitly reducible.
Keep the distinctions
- irreducible polynomial 不可约多项式 — A positive-degree polynomial with no factorisation into smaller positive degrees over the stated field.
- extension degree 扩张次数 — The dimension of an extension field as a vector space over its base field.
- Test polynomial irreducibility over the specified field.
- Construct small finite fields from irreducible polynomials.
- Use extension degrees and the tower law.
- Use cyclotomic roots and coefficient relations to compute sums and products.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.