Đại số trừu tượng và lý thuyết số
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| integral domain/ˈɪntɪɡrəl dəˈmeɪn/ | A commutative ring with 1 and no zero divisors. |
| kernel/ˈkɜːnl/ | The set of elements a homomorphism sends to the identity. |
| zero divisor/ˈzɪərəʊ dɪˈvaɪzə/ | A nonzero element whose product with some nonzero element is zero. |
When division fails
- Modulo 8, $3 \cdot 3 = 9 \equiv 1$, so 3 has an inverse. But $2x \equiv 1 \pmod 8$ has no solution: every $2x$ is even.
- GRE algebra and number theory test which structure you are in — group, ring, integral domain 整环, field — and what that structure permits.
Groups
- A group needs closure, associativity, an identity and inverses. Commutativity is extra (abelian).
- $(\mathbb Z, +)$ is a group; $(\mathbb Z, \times)$ is not (2 has no inverse); $(\mathbb Q \setminus \{0\}, \times)$ is.
Group or not?
Check the group axioms on familiar sets.
Which is a group?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The others lack inverses (or closure).
Homomorphisms and kernels 核
- $\varphi(ab) = \varphi(a)\varphi(b)$. The kernel is a normal subgroup, and $G/\ker\varphi \cong \operatorname{im}\varphi$.
- Reduction $\mathbb Z \to \mathbb Z_n$ has kernel $n\mathbb Z$, so $\mathbb Z / n\mathbb Z \cong \mathbb Z_n$. $\varphi$ is injective exactly when the kernel is trivial.
A homomorphism is injective exactly when its ____ is trivial.
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Only the identity maps to the identity.
Rings, domains, fields
- Integral domain: commutative ring with $1$ and no zero divisors 零因子. Field: every nonzero element invertible.
- $\mathbb Z$ is a domain, not a field. $\mathbb Z_6$ has $2 \cdot 3 = 0$, so it is not a domain. $\mathbb Z_p$ ($p$ prime) is a field. Every finite domain is a field.
$\mathbb Z_7$ is an integral domain.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. $\mathbb Z_n$ has no zero divisors exactly when n is prime, and 7 is prime; in fact $\mathbb Z_7$ is a field.
Congruences and inverses
- $a$ is invertible mod $n$ exactly when $\gcd(a, n) = 1$. Cancel a factor only if it is invertible.
Solve $3x \equiv 2 \pmod 8$. The inverse of 3 is 3, so $x \equiv 6$. Check: $18 = 16 + 2$. In contrast, $2x \equiv 4 \pmod 8$ has two solutions, $x \equiv 2$ and $x \equiv 6$: cancelling 2 would lose one.
Which residue has a multiplicative inverse modulo 9?
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$\gcd(4, 9) = 1$; $4 \cdot 7 = 28 \equiv 1$.
Find the least non-negative $x$ with $5x \equiv 3 \pmod 7$.
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$5 \cdot 2 = 10 \equiv 3$.
Recap
- Check all group axioms; inverses are where most candidates fail.
- Kernel trivial $\iff$ injective; $G/\ker \cong$ image.
- Inverse mod $n$ exists iff $\gcd(a,n) = 1$; $\mathbb Z_n$ is a field iff $n$ is prime.
- integral domain: A commutative ring with 1 and no zero divisors.
- kernel: The set of elements a homomorphism sends to the identity.
- zero divisor: A nonzero element whose product with some nonzero element is zero.