Đa thức và mở rộng trường
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| irreducible/ɪrɪˈdjuːsɪbl/ | Not expressible as a product of polynomials of lower positive degree over the given field. |
| tower law/ˈtaʊə lɔː/ | The rule that degrees of field extensions multiply along a chain. |
| minimal polynomial/ˈmɪnɪml ˌpɒlɪˈnəʊmɪəl/ | The monic polynomial of least degree over the base field with the given root. |
Four elements, not mod 4
- $\mathbb Z/4$ has $2 \cdot 2 = 0$, so it is not a field. The field with four elements is built differently: adjoin a root of $x^2 + x + 1$ to $\mathbb F_2$.
- GRE field theory tests irreducibility, quotient fields $F[x]/(p)$, degrees of extensions and roots of unity.
Irreducibility depends on the field
- Degree 2 or 3: irreducible 不可约的 $\iff$ no root in the field. Degree 4+: no roots is not enough — $(x^2+1)(x^2+2)$ has no real root but factors over $\mathbb R$.
- $x^2 - 2$ is irreducible over $\mathbb Q$, splits over $\mathbb R$. Always name the base field.
Irreducible over which field?
Decide irreducibility relative to the named field.
Which polynomial is irreducible over $\mathbb F_2$?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The others have a root: 1, 0 or 0.
A degree-4 polynomial over $\mathbb Q$ with no rational root can still be reducible over $\mathbb Q$.
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True. It may factor into two quadratics with no rational roots, such as $(x^2 + 1)(x^2 + 2)$.
Building a field: $F[x]/(p)$
- If $p$ is irreducible of degree $d$ over $F$ with $q$ elements, $F[x]/(p)$ is a field with $q^d$ elements. Reduce with $p(\alpha) = 0$.
Over $\mathbb F_2$, $x^2 + x + 1$ has no root (values 1 at 0 and 1). With $\alpha^2 = \alpha + 1$, the elements are $0, 1, \alpha, \alpha+1$, and $\alpha(\alpha+1) = \alpha^2 + \alpha = 2\alpha + 1 = 1$. So $\alpha^{-1} = \alpha + 1$, and $\alpha^3 = 1$.
$p$ is irreducible of degree 3 over $\mathbb F_5$. How many elements does $\mathbb F_5[x]/(p)$ have?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
An irreducible p of degree 3 makes 𝔽₅[x]/(p) a field whose elements are the remainders a + bx + cx² with a, b, c ∈ 𝔽₅: 5³ = 125 elements.
In $\mathbb F_2[\alpha]$ with $\alpha^2 = \alpha + 1$, the inverse of $\alpha$ is $\alpha +$____.
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α(α + 1) = 1.
Degrees and the tower law 塔律
- $[M:K] = [M:L][L:K]$. The degree of $\alpha$ is the degree of its minimal polynomial 极小多项式.
- $[\mathbb Q(\sqrt2, \sqrt3):\mathbb Q] = 4$. A degree-2 extension cannot contain an element of degree 3, since 3 does not divide 2.
What is $[\mathbb Q(\sqrt[3]{2}, \sqrt 2) : \mathbb Q]$?
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Divisible by 2 and 3, at most 6.
Roots of unity and Vieta
- Primitive $n$th roots are the roots of $\Phi_n$, of degree $\varphi(n)$. $\Phi_{10} = x^4 - x^3 + x^2 - x + 1$: its roots sum to 1 and multiply to 1.
- For monic degree $d$, the product of the roots is $(-1)^d$ times the constant term.
Recap
- Root test decides irreducibility only for degree 2 and 3.
- $F[x]/(p)$ with $p$ irreducible of degree $d$: a field of $q^d$ elements.
- Degrees multiply up a tower; check which roots of unity are primitive.
- irreducible: Not expressible as a product of polynomials of lower positive degree over the given field.
- tower law: The rule that degrees of field extensions multiply along a chain.
- minimal polynomial: The monic polynomial of least degree over the base field with the given root.