New Series: Ring Theory

摘抄一下定理，性质，笔记。

Lec 8. Properties of Ideals (Suppose $1\not=0$)

$A\subseteq R$.

Def: ideal generated by $A$ : Smallest ideal of $R$ containing $A$ denoted by $(A)$.

Def: $RA=\{\sum r_ia_i\}$, similar define the $AR$ and $RAR$.

Def: The ideal generated by a single element $(a)$ is called a principal ideal.

Def: The ideal generated by a finite set is called a finite generated ideal.

我们知道，可数多个 ideals 的交也是 ideal，所以我们有 $(A)=\bigcap_{A\subseteq I} I$, $I$ is an ideal.

$RA$ 是 left ideal generated by $A$。因为 $RA$ 对左乘 $r$ 元素，加法封闭（依定义），同时如果包含 $A$ 并且是 ideal，必须满足 $ra$ 都是 $I$ 的元素。故 $RA$ is minimal.

Similar $AR$ is right ideal generated by $A$, $RAR$ is ideal generated by $A$.

In particular, $R$ is commutative $\implies$ $AR=RA=RAR$.

$\implies $If $R$ is commutative, $(a)$ is only the set $S=\{ra\}$. But if $R$ is not commutative, $\{ras\}$ is not necessary to be a ideal since $ras$ is not close under addition.

我们能看到交换环的美好性质，如果 $b\in (a)$，那么就说明 $b=ra$ 也就是 $b$ 是 $a$ 的divisor。

Prop. 9

$I$ is an ideal of $R$.

$I=R$ $\iff$ $I$ contains a unit.

Proof: Assume $u\in I$ is a unit, $uv=1$. Then $r(uv)=(ru)v$, that is $ru\in R,v\in I$. $R=I$.
Assume $R$ is commutative. $R$ is a field $\iff$ only ideals are $0$ and $R$.

Proof: $R$ is a field $\iff$ every non-zero element is a unit(definition). Thus, every ideal contains a unit, $0$ and $R$.

Conversely, $\forall a\in R$, $1\in (a)$, Since $R$ is commutative, $1=ba$, thus $a$ is a unit.$\square$

Corollary. 10

$R$ is a field. Then any homo. $\varphi:R\to S$, $\varphi=0$ ($\ker \varphi=R$) or $\varphi$ ($\ker\varphi=0$) is injective.

Lec 9. Ring Theory Continued

定义 $M$ 是一个 maximal ideal，当且仅当 $S$ 的 ideal 包含 $M$ 的只有 $M$ 与 $S$。(not necessary commutative)

Prop. 11

有单位元的环的 Proper ideal 都有一个 maximal ideal.

证明可能用到选择公理状物。

Prop. 12

Assume $R$ is commutative. $M\le R$ is maximal $\iff$ $R/M$ is a field.

$M$ is maximal $\iff$ no ideal contains $M$ is not $R$. By lattice iso. thm., any ideals containing $M$ correspond bijective with the ideals of $R/M$. That is a ring has no non-simple ideal $\iff$ field.

Def: prime ideal: Assume $R$ is commutative. An ideal $P$ is called a prime ideal $\iff$ $P\not=R$ and whenever $ab\in P$ $\implies$ $a\in P$ or $b\in P$.

Prop. 13

Assume $R$ is commutative. prime ideal $\iff$ $R/P$ is an integral domain.

$(a+P)(b+P)=0+P$ $\iff ab+P=0+P$ $\iff a\in P$ or $B\in P$ i.e. $a+P=0$ or $b+p=0$. $R/P$ is an integral domain.

Corollary. 14

Let $R$ be commutative. Every maximal ideal is a prime ideal.

$R/M$ is field then $R/M$ is an integral domain. Then $M$ is prime.

Rings of Fractions (R is commutative)

Thm. 15

We can construct a new ring that contain $R$. $D$ be a subset that not contains zero divisor.

$Q=\{rd^{-1}\},r\in R,d\in D$.

......

Euclidean Domains (R is commutative)

Define a norm function $N:R\to \mathbb{Z}^+\cup\{0\}$, $N(0)=0$ on an integral domain $R$.

If $\forall a\not=0$, $N(a)>0$, then $N$ is called a positive norm.

The integral domain $R$ is said to be a Euclidean Domain if $\exist N,\forall a,b\in R,b\not=0$,$\exist q,r\in R$. s.t. $a=qb+r$, $N(b)>N(r)$.

$q$ : quotient, $r$ : remainder.

Prop.1

Every ideal in a Euclidean Domain is principal. More precisely, if $I$ is any non-zero ideal in $R$, then $I=(d)$ where d is any non-zero element with minimum norm in $I$.

Proof : $d\in I$, $(d)\subseteq I$. $\forall a\in I$, $a=qd+r$, $r=a-qd\in I$. Thus $r$ must equal to $0$.(minimum norm).

$qd\in (d)$. $I=(d)$. $\square$

In a Euclidean Domain

Def: $a|b$ $\iff$ $b=qa$. (commutative)

Prop.2

If $a$ and $b$ are non-zero elements in $R$(commutative), $(a,b)=(d)$ ,then $d$ is the gcd of $a,b$.

The condition is not necessary. Consider in $\mathbb{Z}[x]$, $\gcd(x,2)$ is $1$,however, $(x,2)\not=(1)=R$.

Def: An Integral domain where $\forall a,b\in R$, $(a,b)$ is principal is called Bezout Domain.

Prop.3

If $(a)=(b)$, then $\exist u\in R$ is a unit, $a=ub$.

moreover, if $d$ and $d'$ are both $\gcd(a,b)$, then $d=ud'$.

Proof: $d'\in(d)$ then $d'=ud$.

Thm.4

Assume $R$ be a Euclidean Domain, we can use Euclid algorithm to obtain the $\gcd$.

Assume the last nonzero remainder in the algorithm for $a,b$. Then:

$\gcd(a,b)=r_n$, $(a,b)=(r_n)$.

Proof:

First of all, the $\gcd$ always exists since $R$ is Euclidean Domain and Prop.1.

$r_{n-1}=q_{n}r_n$, then $r_n|r_{n-1}$. We can prove $r_n|a,r_n|b$ by induction. We can find $(a,b)\subseteq (r_n)$

Then We can see $r_n$ must have a form of $ax+by$,then $r_n\in(a,b)$, that is $(r_n)\subseteq (a,b)$

P.I.D. (Principal Ideal Domains)

Def: an integral domain in which every ideal is principal is called P.I.D. (necessary)

Euclidean Domain is P.I.D.. But P.I.D is not necessary to be Euclidean Domain.

P.I.D is must be a Bezout domain, however Bezout domain may have non-finite generated ideal.

Prop. 6

We can define the $\gcd$ similar to the Euclidean domain.

Prop. 7

In a P.I.D, every non-zero prime ideal must be a maximal ideal.

Proof: $I=(p)$, suppose $I\subseteq (m)\subseteq R$.

$p=rm$ , since $(p)$ is prime, the either $r\in(p)$ or $m\in(p)$.

$m\in (p)\iff (m)\subseteq(p)$, $r\in(p)\iff p=r\cdot m\in (p)$.

Corollary. 8

If $R$ is any commutative ring s.t. $R[x]$ is P.I.D then $R$ is necessary a field.

$R$ is a subring of $R[x]$ and $R[x]$ is an integral domain ,then $R$ is also an integral domain.

ideal $(x)$ is a nonzero prime ideal in $R[x]$ since $R[x]/(x)$ is isomorphic to the integral domain $R$. Since $(x)$ is maximal $R[x]/(x)$ is a field. $\square$

Unique Factorization Domain U.F.D

Def: Let $R$ be a integral domain.

Suppose $r\in R$ is nonzero and not a unit. $r$ is called irreducible if whenever $r=ab$, $a$ or $b$ is a unit.
The nonzero element $p$ is called prime iff $(p)$ is prime ideal.
Def: $a,b\in R$ is said to be associate iff $a=ub$ where $u$ is a unit.

Prop.10

In a Integral domain a prime element is always irreducible. $\forall p=ab,a\in(p)$, $a=up$, $p=upb$ then $ub=1$. $b$ is unit.

For the simple Integral domain a irreducible element is not necessary prime.

However, in P.I.D a irreducible element is always prime.

Prop.11

In a P.I.D a nonzero element is a prime iff it is irreducible.

$(p)\subseteq (m)=M$, $p\in(m),p=mr$,$p$ is irreducible, thus, $m$ is unit $(p)=(1)$ or $r$ is unit $pr^{-1}=m$, $(p)=(m)$.

这也正说明，如果 $p=ab$, $p\in(a)$, 用上述结论即可。

Def: U.F.D

A U.F.D is an integral domain in which every nonzero element which is not a unit obeys:

$r=\prod p_i$ where $p_i$ all is irreducible.
The decomposition is unique up to associate ($a=ub$,$u$ is a unit, then we say $a,b$ is associate)

Field is a trivially U.F.D since all elements are unit.

Exp: $\mathbb{Z}[\sqrt{-5}]$ is not a U.F.D, $6=2\times 3=(1+\sqrt{-5})(1-\sqrt{-5})$.

Prop.12

In a U.F.D a nonzero element is a prime iff it is irreducible.

Only need to prove is irreducible $\implies$ prime.

$ab\in(p)$, $a\in(p)$ or $b\in(p)$. $ab=pc$, write the decomposition of $ab$, $p$ must be associate to one of the factor of $a$ or $b$.

that is $p|a$ or $p|b$.

Prop.13

let $a=u\prod p_i^{a_i},b=v\prod p_i^{b_i}$, $\gcd(a,b)=\prod p_i^{\min(a_i,b_i)}$.

Thm.14

Every P.I.D is a U.F.D. In particular, every Euclidean Domain is a Unique Factorization Domain.

Field $\subset$ Euclidean Domain $\subset$ P.I.D. $\subset$ U.F.D $\subset$ integral domain

Gaussian Integers

Polynomial ring

$R$ is an integral domain, then, $\deg(p(x)q(x))=\deg(p(x))+\deg(q(x))$.

the unit of $R[x]$ are just the unit of $R$.

$R[x]$ is an integral domain.

Prop.2

Let $(I)=I[x]$ denote the ideal of $R[x]$ generate by $I$(the polynomials with coefficients in $I$)

\[R[x]/(I)\cong (R/I)[x] \]

In particular, if $I$ is a prime ideal of $R$, then $(I)$ is also a prime ideal of $R[x]$

We can construct a nature map: $\varphi:R[x]\to (R/I)[x]$. $r\to \overline{r}$. it's obviously a ring homo.

$\ker\varphi=(I)$

Polynomial Rings over Field

Let F be a field, $F[x]$ is a Euclidean domain, we can use the $\deg F$ be the norm.

Prop.5 (Gaussian lemma)

Let $R$ be a U.F.D with Field of fractions $F$ and $P(x)\in R[x]$, if $P(x)$ can be factorized in $F[x]$, $P(x)=A(x)B(x)$. Then $P(x)$ can also factorized in $R[x]$, $P(x)=a(x)b(x)$.

Proof: $dP(x)=A'(x)B'(x)$ (multiplication the denominator), then assume $d=\prod p_i$, $p_i$ is all prime element. RHS,LHS mod $p_i$, then we get $0=\overline{A'(x)}\overline{B'(x)}$, since $R/pR$ is an integral domain, $(R/pR)[x]$ is also an integral domain, that is one of the $\overline{A'(x)},\overline{B'(x)}$ must be zero.

We can divide $p_i$ both LHS and RHS to prove the conclusion by induction.

Corollary. 6

Let $R$ be a U.F.D with field of fraction $F$ and let $p(x)\in R[x]$.

Suppose $\gcd$ of all coefficient is $1$. Then $p(x)$ is irreducible in $R[x]$ iff it is irreducible in $F[x]$.

正面：$F[x]\to R[x]$: Gaussian lemma. 反面 $R[x]\to F[x]$ 显然。

Thm. 7

$R$ is U.F.D iff $R[x]$ is U.F.D.

$R\subseteq R[x]$, so $R[x]$ is U.F.D $\implies$ $R$ is .

证明思路 $R$ is U.F.D, $F$ is Field. $F[x]$ is Euclidean domain(U.F.D).

把 $p(x)$ 在 $F[x]$ 分解，然后根据Gaussian lemma, 得到一组在 $R[x]$ 的解。 $p(x)=q_1(x)\cdots q_k(x)$.（假设 $p(x)$ 的系数 $\gcd=1$,那么每一个 $q_i(x)$ 系数 $\gcd=1$）

我们接下来证明唯一性。假设 $p(x)=\prod q_i(x)=\prod q_i'(x)$，那么 $q_i(x),q'_j(x)$ 都是在 $F[x]$ irreducible 的，不妨假设 $q_i(x)=\frac{a}{b}q'_i(x)$，$bq_i(x)=aq'_i(x)$, 考察两边的 $\gcd$，又由于 $R$ 是 U.F.D 的，故 $b=ua$, $u$ is a unit. $q_i(x),q_i'(x)$ associate 得证。

Coro. 8

多元 $R[x_1,x_2,\cdots ,x_k]$ 是 U.F.D $\iff$ $R$ 是 U.F.D

归纳。

Irreducibility criteria

Prop.9

$R$ is a field, $R[x]$ is a Euclidean Domain, then $p(x)\in R[x]$ has a factor of $\deg=1$ iff $p(a)=0$.

$p(x)=q(x)(x-a)+r$, $r=0\iff (x-a)|p(x)$.

Prop.10

$p(x)\in F[x]$, where $F$ is a Field $p(x)$'s degree is $2$ or $3$, then if $p(x)$ is reducible iff $p(x)$ has a root.

Prop.11

$\mathbb{Z}[x]$ 上判别 $\mathbb{Q}$ 根的存在性。 $p(x)=\frac{s}{r}$ （$s$ and $r$ 互质 relatively prime），那么需要 $r|a_n$, $s|a_0$。证明考虑：$s^np(\frac{s}{r})=0$ 也就是：

$s^na_0+s^{n-1}ra_1+\cdots+r^na_n=0$, mod $r$ and $s$ ，即可得到结论。

Prop. 12

$I$ is proper ideal in the integral domain $R$ and let $p(x)\in R[x]$ and $p(x)$ is monic.

$p(x)\to \overline{p(x)}$ ($R[x]\to(R/I)[x]$ ) ， if $\overline{p(x)}$ is irreducible, then $p(x)$ is irreducible.

显然的， $p(x)=a(x)b(x)$,那么 $\overline{p(x)}=\overline{a(x)}\overline{b(x)}$，又由于 monic，$\overline{a(x)},\overline{b(x)}$ 不会退化。

很遗憾的，就算对于每一个 ideal 都满足 reducible, 但 $p(x)$ 还是有可能 irreducible的。

Eisenstein's Criterion

posted @ 2024-05-22 16:42 Cage123 阅读(6) 评论(0) 编辑收藏举报

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Protons

New Series: Ring Theory

New Series: Ring Theory

Lec 8. Properties of Ideals (Suppose \(1\not=0\))

Prop. 9

Corollary. 10

Lec 9. Ring Theory Continued

Prop. 11

Prop. 12

Prop. 13

Corollary. 14

Rings of Fractions (R is commutative)

Thm. 15

Euclidean Domains (R is commutative)

Prop.1

Prop.2

Prop.3

Thm.4

P.I.D. (Principal Ideal Domains)

Prop. 6

Prop. 7

Corollary. 8

Unique Factorization Domain U.F.D

Prop.10

Prop.11

Def: U.F.D

Prop.12

Prop.13

Thm.14

Gaussian Integers

Polynomial ring

Prop.2

Polynomial Rings over Field

Prop.5 (Gaussian lemma)

Corollary. 6

Thm. 7

Coro. 8

Irreducibility criteria

Prop.9

Prop.10

Prop.11

Prop. 12

Eisenstein's Criterion

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