Rings¶
Definition¶
A ring satisfies
- \(R\neq \empty\)
- \(R\times R \rightarrow R \quad i.e.(r,s) \rightarrow r+s\)
(R,+) is an Abelian group(associativity, commutativity, inverse element, identity element).
- \(R\times R \rightarrow R \quad i.e.(r,s) \rightarrow r\cdot s\)
(R,\(\cdot\)) satisfying associative law.
- \(a(b+c)=ab+ac \quad (b+c)a=ba+ca\)
e.g.
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\(F\) is a field, which is also a ring.(\(Q,R,C\))
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\(F[x]={\sum\limits_{i=0}^na_ix^i|a_i\in F}\) is a ring.
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\(Z_n=\{\overline0,\overline1,\ldots,\overline{n-1}\}\) where \(\overline k =\{np+k\mid p\in Z\}\)
and it satisfies
- \(\overline k + \overline l =\{n(p+q)+k+l\mid p,q\in Z\}=\overline{k+l}\)
- \(\overline k \cdot \overline l=\{(np+k)(nq+l)\mid p,q \in Z\}=\overline{kl}\)
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\(\overline {k}(\overline l + \overline s)=\overline k \overline l + \overline k \overline s=\overline k\ \overline{l+s}=\overline{kl+ks}\)
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\(M_n(F)=F^{n\times n}=\{(a_{ij}\mid a_{ij}\in F)\}\) and \((M_n(F), +, \cdot)\) is a ring
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\(R_1\times R_2=\{(a,b)\mid a\in R_1,b\in R_2\}\) and we define
\((a_1,b_1)+(a_2,b_2)=(a_1+b_1,a_2+b_2)\)
\((a_1,b_1)(a_2,b_2)=(a_1 b_1,a_2 b_2)\)
Properties¶
Unit, identity(单位元, 恒等元)¶
def: an element \(a\) is called an identity if \(\forall a \in R \quad 1\cdot a = a\cdot 1 =a\)
Not every ring has an identity. e.g. \(2\Z=\{2n\mid n \in \Z\}\)
suppose \(R\) is a ring, \(\Z\times R =\{(n,r)\mid n\in \Z r\in R\}\) and we define \((n,r)+(m,s)=(n+m,r+s)\quad (n,r)\cdot (m,s)=(nm,rs+ns+rm)\)
then \(\Z\times R\) has a unit \((1,0)\)
and we find \((0,r)+(0,s)=(0,r+s)\quad (0,r)\cdot(0,s)=(0,rs)\)
Note:任意环可以通过这样的操作扩充为有单位元的环, 而且还保留了\(R\)上的运算.
Commutative Ring¶
def: A ring is called commutative if \(\forall a,b\in R\quad ab=ba\), otherwise if \(\exist a,b\in R ab\neq ba\) it is called noncommutative.
e.g. \(\mathbb{R},\mathbb{C}[x]\) are commutative, while \(M_n(R)\) is noncommutative.
Domain Ring(整环)¶
def: \(a \in R\) there is a nonzero element \(b\) such that \(ab=0\), \(a\) is called left zero-divisor(左零因子). Similarily we can define right zero-divisor.
- 0 must be zero-divisor.
proof: \(0\cdot a = (0+0)a=0\cdot a + 0\cdot a\)
so \(0=0\cdot a +(- 0\cdot a)=0\cdot a +0\cdot a +(-0\cdot a)=0\cdot a\)
thus we get \(0\cdot a =0\)
def: A ring is called domain ring if
- \(\lvert R \rvert \ge 2\)
- \(R\) has no nonzero zero-divisor.
e.g.
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\(\overline Z_6\) is not a domain ring since \(\overline 2 \cdot \overline 3 = \overline 0\)
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\(H=\{ \left[\begin{matrix} a+bi & c+di\\ -c+di & a-bi \end{matrix} \right] \mid a,b,c,d\in R \ and\ i=\sqrt{-1} \}\)
\(H\) is noncommutative but it is a domain ring.(proof?)
Division Ring(除环)¶
def: \(a\in R\) is invertible if \(\exist b \ s.t.ab=ba=1\)
e.g.
- in \(M_n(R)\), \(A\) is invertible $ \Leftrightarrow\lvert {A} \rvert \neq 0$
- \(0\) is always not invertible since \(0 \cdot a =1 \Rightarrow 0\cdot a = 0 =1\)then \(R=\{0\}\) where \(\forall r \in R,\ r=1\cdot r=0\cdot r =0\)
- \(H\) is a domain ring since \(\left[\begin{matrix} a+bi & c+di \\ -c+di& a-bi \end{matrix} \right]^{-1}=\frac{1}{a^2+b^2+c^2+d^2}\left[\begin{matrix} a-bi & -c+di \\ c+di& a+bi \end{matrix} \right]\)
def: A ring is called division ring(or skew-field斜域) if
- \(\lvert R \rvert \ge 2\)
- nonzero element is invertible.
e.g. Z is not division ring while Q is division ring.
Field(域)¶
def: A commutative division ring is called field.
Note: 域中必有单位元.
Subring¶
def: \(\empty\neq S\subseteq (R,+,\cdot)\) it is called a subring of \(R\) if it satisfies
- \(\forall a,b\in R\quad a-b\in S\)
- \(\forall a,b\in R\quad ab\in R\)
Note: 为什么是\(a-b\in S\)而不是\(a+b\in S\)? 为了确保任意元素的负元也在集合中
而线性空间中因为要对数乘封闭, \((-1)a=-a\) 故负元一定在集合中.
suppose \(S_1,S_2\) are subring of \(R\), then \(S_1+S_2=\{a_1+a_2\mid a_1\in S_1,a_2\in S_2\}\) is not a subring of \(R\), \(S_1\cap S_2\) is a subring, and \(S_1\cup S_2\) is not a subring.
e.g.
- \(R=\mathbb{Q}[\sqrt2,\sqrt3,\sqrt{6}]=\{a+b\sqrt2+c\sqrt3+d\sqrt6\mid a,b,c,d\in \mathbb{Q}\}\) and \(R\leq (\R,+,\cdot)\)
let \(S_1=Q[\sqrt2]=\{a+b\sqrt2\mid a,b\in \mathbb{Q}\}\) \(S_2=Q[\sqrt3]=\{a+b\sqrt3\mid a,b\in \mathbb{Q}\}\) and both are subrings of \(R\)
but \(S_1+S_2=\{a+b\sqrt2+c\sqrt3\mid a,b,c\in \mathbb{Q}\}\) is not a subring since \(\sqrt2\cdot\sqrt 3 =\sqrt 6 \notin S_1+S_2\)
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\(Z\) has an identity, but its subring \(2\Z\) has no identiy.
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\(S=\{\left[ \begin{matrix}a&b&0\\c&d&0\\0&0&0\end{matrix}\right]\mid a,b,c,d\in\R\}\leq M_n(R)\)
their identities are not the same.(\(S:\left[ \begin{matrix}1&0&0\\0&1&0\\0&0&0\end{matrix}\right]\) abd \(M_n(R):\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right]\))
Ideal¶
def: \(\empty\neq S\subseteq (R,+,\cdot)\) it is called a left ideal of \(R\) if it satisfies
- \(\forall a,b\in R\quad a-b\in S\)
- \(\forall a\in S,b\in R \quad ba\in S\)
and we denote it by \(I\lhd R\)
Similarily we define right ideal.
ideal = left ideal + right ideal.
proposition: suppose \(I_1,I_2\) are left ideal, then \(I_1+I_2=\{a+b\mid a\in I_1,b\in I_2\}, I_1\cap I_2=\{a\cdot b\mid a\in I_1,b\in I_2\}\) are also ideals.
proof: suppose \(\forall a_i,b_i\in I_i\) where \(i=1,2\)
- \((a_1+a_2)-(b_1+b_2)=(a_1-b_1)+(a_2-b_2)\in I_1 +I_2\)
- \(\forall r\in R \quad r(a_1+a_2)=ra_1+ra_2\in I_1+I_2\)
0 always in ideal since \(a-a=0\)
we can define \((a\rangle=\cap \{I\) is a left ideal of \(R\) containg \(a\)\(\}\)
proposition: \((a\rangle=\{na+ra\mid n\in Z, r\in R\}\) this ideal containg \(a\) and \(\forall\) ideal containing \(a\) \(\Rightarrow (a\rangle \sube I\)
proof:
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\(a\) in this set \(a = 1*a+0*a\) where \(1\in Z,0\in R\) thus \(a\in (a\rangle\)
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this set is an ideal
- \(\forall n_1a+r_1a, n_2a+r_2a\in(a\rangle\) then \((n_1a+r_1a)-(n_2a+r_2a)=(n_1-n_2)a+(r_1-r_2)a\in (a\rangle\) where \(n_1-n_2\in \Z,r_1-r_2\in R\)
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\(\forall b\in R, na+ra \in (a\rangle\) then \(b(na+ra)=nba+bra\in (a\rangle\) where \(nb\)
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the minimum ideal containg \(a\)
suppose an ideal \(I\) containing \(a\), out target is to prove \((a\rangle\sube I\)
\(\forall\ na+ra\in (a\rangle\) then \(a\in I, r\in R \Rightarrow ra\in I\)
and \(na = \left\{ \begin{matrix}a+a+\ldots+a,n>0 \\ 0,n=0 \\ (-a)+(-a)+\ldots+(-a),n<0\end{matrix}\right.\) so \(na\in I\)
thus \(na+ra\in I\) \(\square\)
suppose \(R\) has an identity, \(na =1_R\cdot a+1_R\cdot a+\dots+1_R\cdot a= (n1_R)\cdot a\) where \(n1_R\in R\)
so \((a\rangle\) can be writed as \(\{ra\mid r\in R \}=Ra\)
Similarily, \((a)=\{\sum\limits_{i=0}^nr_ias_i+ra+as+na\mid r_i,s_i,r,s\in R, n\in\N\}\stackrel{+1_R}=\{\sum\limits_{i=0}^nr_ias_i\mid r_i,s_i\in R, n\in \N\}=RaR\)
suppose \(R\) is commutative, then \(\langle a)=(a\rangle=(a)\)
PID(主理想整环)¶
def: An ideal is called principal if it can be generated by one element.
def: \(R\) is called left(resp. right) principal ring if every left(resp. right)ideal is generated by one element.
def: A commutative principal domain is simply denoted by PID.
e.g.
- \((\Z,+,\dot)\) is a PID.
proof: let \(I\) be an ideal of \(Z\)
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\(I=\{0\}=(0)=Z\cdot0=\{0\}\) \(0\) ideal must be principal ideal.
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\(I\neq\{0\}\) so \(\exist n\in I,n\neq0\)
without loss of generalization, we assume \(n\in I\) and \(n>0\) (since \(-n=0-n\in I\), so both \(n\) and \(-n\) in \(I\))
let \(n\) be the least of the set \(\{n\in I,n>0\}\) our target is to show \(I=(n)=n\Z\)
- \(n\in I \Rightarrow(n)\sube I\)
- \(\forall m\in I\quad m=qn+r(0\le r\leq n-1)\) so \(r=m-qn\in I\) where \(m\in I\) and \(qn\in(n)\sube I\)
- by choice of n, \(r\) must be \(0\). Otherwise \(r\neq0, r<n\) contradicts the assumption that \(n\) is the least element which is gereater than \(0\).
- Thus, \(m=qn\in(n)\) \(\Z\) is a principal ideal.
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Apparently, \(Z\) is commutative and domain.
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By the proof above, we can similarily prove \((F[x],+,\cdot)\) is a PID where \(F\) is a field.
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Every field is a PID.
proof: suppose \(F\) is a field, \(I\) is an ideal of \(F\). We assert that \(I\) can be either \((0)\) or \((1)=F\).
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\(I=\{0\}=(0)\)
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\(I\neq\{0\}\) we get \(a\in I,a\neq0\Rightarrow a^{-1}a=1\in I\) where \(a^{-1}\in F,a\in I\)
\(\forall b\in F,b=b\cdot 1\in I\) where \(b\in F, 1\in I\) so we know \(F\sube I\) and obviously \(I\sube F\)
thus \(I=F=(1)\)
Note: 域(除环)中没有非平凡理想.
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\(M_n(\R)\) is PID.
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\(I=\{0_{n\times n}\}=(0_{n\times n})\)
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\(I\neq\{0_{n\times n}\}\) then \(\forall A=(a_{ij})\in I, A\neq 0\) so \(\exist a_{kl}\neq 0\)
thus \(\forall i,j\in \N,E_{ik}AE_{lj}=a_{kl}E_{ij}\) where \(E_{ij}\) represent a matrix whose \((i,j)\) element is \(1\).
so \(a_{kl}^{-1}E_{ik}AE_{lj}=E_{ij}\in I\) where \(E_{lj}\in M_n(\R), A\in I\) and \(a_{kl}^{-1}E_{ik}=(a_{kl}^{-1}E)E_{ik}\in M_n(\R)\) for the definition of ideal.
$\Rightarrow \forall M=(b_{ij})\in M_n(\R)\quad M=\sum x_{ij}E_{ij}=\sum (x_{ij}E)E_{ij}\in I $
Therefore, \(M_n(\R)\sube I\Rightarrow I=M_n(\R)\)
Note: 非交换, 非整环也可以只有两个理想.
Proposition: \(A\) is similar to a diagonal matrix \(\Leftrightarrow\) there is a splitting polynomical \(f(x)\) which has no mutiplicity roots(重根) \(s.t. f(A)=0\)
proof
Quotient Ring(商环)¶
def: suppose \(I\) is an ideal of \(R\), then \(R/I=\{a+I\mid a\in R\}\) is called quotient ring.
and we define \((a+I)+(b+I)=(a+b)+I,(a+I)(b+I)=ab+I\)
Note:
- 商环中的元素\(a+I=\{a+b\mid b\in I\}\)也是一个集合, 若\(a\in I\)则\(a+I=0+I=0\).
- 若\(R=I\), 则规定\(R/I=0\)
proposition: \(a_1+I=a_2+I\Leftrightarrow a_1-a_2\in I\)
proof:
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\(\Rightarrow\) \(a_1\in a_1+I=a_2+I\) so \(\exist x\in I\ s.t.a_1=a_2+x\) thus \(a_1-a_2=x\in I\)
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\(\Leftarrow\) \(\forall a_1+x\in a_1+I\) where \(x\in I\) so \(a_1+x=a_2+x+(a_1-a_2)\in a_2+I\) since \(x\in I,(a_1-a_2)\in I\)
therefore, \(a_1+I\sube a_2+I\). Similarily, we can prove \(a_2+I\sube a_1+I\) so \(a_1+I=a_2+I\).
Note: 对于商环\(R/I\) 我们需要验证well-defined. 即若\(a+I=a'+I,b+I=b'+I\) 则要满足\(a+b+I=a'+b'+I,ab+I=a'b'+I\) 即不同形式同一本质的元素应该映射后得到的元素应该相同.
Maximal Ideal(极大理想)¶
def: suppose \(I\neq R\) is an ideal of \(R\), \(I\) is called to be maximal ideal if \(\forall J\lhd R,J\supe I \Rightarrow \left\{\begin{matrix}J=I \\ R\end{matrix}\right.\)
Note: 极大理想\(I\)即只被\(I\)和\(R\)包含.(\(R\)是自身的理想.)
Proposition: suppose \(R\) is commutative ring with identity, then \(M\) is a maximal ideal \(\Leftrightarrow\) \(R/M\) is a field.
proof:
Homomorphism(同态)¶
def: \(\phi:R_1\rightarrow R_2\) and it statisfies \(\phi(a+b)=\phi(a)+\phi(b),\phi(ab)=\phi(a)\phi(b),\phi(1)=1\) then it is called a homomorphism.
def: \(\phi\) is called monomorphism(单同态) if \(a\neq b \Rightarrow\phi(a)\neq\phi(b)\)
def: $\phi $ is called epiomorphism(满同态) if \(r\in R_2\ \exist a\in R_1\ s.t.\phi(a)=r\).
isomorphism(同构) = injective + surjective
we define \(\ker\phi=\{a\in R_1\mid \phi(a)=0\}\) \(Im\phi=\{\phi(a)\mid a\in R_1\}\)
e.g. \(\phi:\Z\rightarrow \Z_n\) and \(a\rightarrow \overline{a}=\{a+kn\mid k\in \Z\}\) and it’s easy to verify it is a homomorphism.
\(\ker\phi={a\in\Z\mid \overline a=\overline 0}=n\Z=(n)\quad Im\phi=\Z_n\)
proposition: \(\ker\phi\) is an ideal of \(R\).
proof:
- \(0\in \ker\phi\) since \(\phi(0)=\phi(0+0)=\phi(0)+\phi(0)\Rightarrow \phi(0)=0\) thus \(ker\phi\) is nonempty.
- \(\forall a,b\in \ker\phi\quad \phi(a-b)=\phi(a)-\phi(b)=0-0=0\Rightarrow a-b\in \ker\phi\)
- \(\forall a\in \ker\phi,b\in R\quad\phi(ba)=b\phi(a)=b\cdot0=0\Rightarrow ba\in \ker\phi\)
proposition: \(Im\phi\) is a subring of \(R\).
proposition: \(\phi\) is injective \(\Leftrightarrow\) \(\ker\phi =0\)
创建日期: 2022年9月5日 23:38:05