多項式と体拡大
導入済み| English | 意味 |
|---|---|
| irreducible polynomial/ɪrɪˈdjuːsɪbl ˌpɒlɪˈnəʊmɪəl/ | 不可約多項式 |
| extension degree/ekˈstenʃn dɪˈɡriː/ | 拡大次数 |
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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]?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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 ____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
(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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.