前章 環と体の基礎 の内容を、記号を固定する形で確認します。
本記事で環 といえば、乗法の単位元 1 1 1 1 = 0 1 = 0 1 = 0 零環 ）も環として認めます。零環は 1 元集合 { 0 } \{0\} { 0 }
部分環 とは、加法・乗法について閉じており、1 R 1_R 1 R 環準同型 f : R → S f \colon R \to S f : R → S a , b ∈ R a, b \in R a , b ∈ R f ( a + b ) = f ( a ) + f ( b ) f(a+b) = f(a) + f(b) f ( a + b ) = f ( a ) + f ( b ) f ( a b ) = f ( a ) f ( b ) f(ab) = f(a)f(b) f ( ab ) = f ( a ) f ( b ) f ( 1 R ) = 1 S f(1_R) = 1_S f ( 1 R ) = 1 S Definition 7.1［環と体の基礎］ ）。全単射な環準同型を同型 といい、R ≅ S R \cong S R ≅ S
可換環 R R R 整域 であるとは、1 ≠ 0 1 \ne 0 1 = 0 a b = 0 ab = 0 ab = 0 a = 0 a = 0 a = 0 b = 0 b = 0 b = 0
可換環 K K K 体 であるとは、1 ≠ 0 1 \ne 0 1 = 0 a ≠ 0 a \ne 0 a = 0 R R R u u u 単元 （可逆元）であるとは、u v = v u = 1 uv = vu = 1 uv = v u = 1 v ∈ R v \in R v ∈ R
K K K K [ x ] K[x] K [ x ] K K K M n ( K ) M_n(K) M n ( K ) n n n F p = Z / p Z \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z} F p = Z / p Z p p p 環準同型では f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f ( − a ) = − f ( a ) f(-a) = -f(a) f ( − a ) = − f ( a ) f ( 0 ) = f ( 0 + 0 ) = f ( 0 ) + f ( 0 ) f(0) = f(0+0) = f(0)+f(0) f ( 0 ) = f ( 0 + 0 ) = f ( 0 ) + f ( 0 ) f ( 0 ) f(0) f ( 0 ) f ( a ) + f ( − a ) = f ( a + ( − a ) ) = f ( 0 ) = 0 f(a) + f(-a) = f(a + (-a)) = f(0) = 0 f ( a ) + f ( − a ) = f ( a + ( − a )) = f ( 0 ) = 0
§1 の失敗を分析すると、必要な条件が自動的に出てきます。I I I R R R a + I = a ′ + I a + I = a' + I a + I = a ′ + I b + I = b ′ + I b + I = b' + I b + I = b ′ + I i : = a − a ′ ∈ I i := a - a' \in I i := a − a ′ ∈ I j : = b − b ′ ∈ I j := b - b' \in I j := b − b ′ ∈ I a b − a ′ b ′ ∈ I ab - a'b' \in I ab − a ′ b ′ ∈ I
a b − a ′ b ′ = a b − a ′ b + a ′ b − a ′ b ′ = ( a − a ′ ) b + a ′ ( b − b ′ ) = i b + a ′ j ab - a'b' = ab - a'b + a'b - a'b' = (a - a')b + a'(b - b') = ib + a'j ab − a ′ b ′ = ab − a ′ b + a ′ b − a ′ b ′ = ( a − a ′ ) b + a ′ ( b − b ′ ) = ib + a ′ j と分解できます。i , j ∈ I i, j \in I i , j ∈ I b , a ′ b, a' b , a ′ R R R I I I R R R I I I
Definition 3.1 （左イデアル・右イデアル・両側イデアル ）
R R R I I I R R R
I I I R R R 左イデアル であるとは、I I I R R R r ∈ R r \in R r ∈ R a ∈ I a \in I a ∈ I r a ∈ I ra \in I r a ∈ I I I I R R R 右イデアル であるとは、I I I R R R r ∈ R r \in R r ∈ R a ∈ I a \in I a ∈ I a r ∈ I ar \in I a r ∈ I 左イデアルかつ右イデアルであるとき、I I I R R R 両側イデアル 、あるいは単にイデアル といい、I ⊴ R I \trianglelefteq R I ⊴ R
R R R
{ 0 } \{0\} { 0 } R R R 自明なイデアル ）。I ≠ R I \ne R I = R 真のイデアル といいます。以下、断らない限り「イデアル」は両側イデアルを指します。
実際にイデアルであることを確かめるときは、加法部分群の条件をいちいち全部確認するより、次の形が便利です。
Proposition 3.2 （イデアルの判定条件 ）
R R R I I I R R R I I I R R R
I ≠ ∅ I \ne \emptyset I = ∅ 任意の a , b ∈ I a, b \in I a , b ∈ I a − b ∈ I a - b \in I a − b ∈ I
任意の r ∈ R r \in R r ∈ R a ∈ I a \in I a ∈ I r a ∈ I ra \in I r a ∈ I a r ∈ I ar \in I a r ∈ I
Proof（Proposition 3.2） 必要性を示します。I I I I I I 0 ∈ I 0 \in I 0 ∈ I Definition 3.1 の吸収条件そのものです。
十分性を示します。1 より元 a ∈ I a \in I a ∈ I b = a b = a b = a 0 = a − a ∈ I 0 = a - a \in I 0 = a − a ∈ I b ∈ I b \in I b ∈ I ( a , b ) = ( 0 , b ) (a, b) = (0, b) ( a , b ) = ( 0 , b ) − b = 0 − b ∈ I -b = 0 - b \in I − b = 0 − b ∈ I b , c ∈ I b, c \in I b , c ∈ I − c ∈ I -c \in I − c ∈ I ( b , − c ) (b, -c) ( b , − c ) b + c = b − ( − c ) ∈ I b + c = b - (-c) \in I b + c = b − ( − c ) ∈ I I I I 0 0 0 R R R Definition 3.1 の吸収条件そのものなので、I I I
∎ イデアルは部分環とは別物です。実際、1 1 1
Lemma 3.3 （単元を含むイデアル ）
R R R I I I R R R I I I R R R u u u I = R I = R I = R 1 ∈ I 1 \in I 1 ∈ I I = R I = R I = R K K K { 0 } \{0\} { 0 } K K K
Proof（Lemma 3.3） u ∈ I u \in I u ∈ I u − 1 ∈ R u^{-1} \in R u − 1 ∈ R r ∈ R r \in R r ∈ R r = r ⋅ 1 = ( r u − 1 ) u r = r \cdot 1 = (r u^{-1}) u r = r ⋅ 1 = ( r u − 1 ) u u ∈ I u \in I u ∈ I r u − 1 ∈ R ru^{-1} \in R r u − 1 ∈ R ( r u − 1 ) u ∈ I (ru^{-1})u \in I ( r u − 1 ) u ∈ I r ∈ I r \in I r ∈ I r r r R ⊆ I R \subseteq I R ⊆ I I ⊆ R I \subseteq R I ⊆ R I = R I = R I = R
1 1 1 1 ⋅ 1 = 1 1 \cdot 1 = 1 1 ⋅ 1 = 1 1 ∈ I 1 \in I 1 ∈ I I = R I = R I = R
最後に K K K I ⊴ K I \trianglelefteq K I ⊴ K I ≠ { 0 } I \ne \{0\} I = { 0 } a ∈ I a \in I a ∈ I a ≠ 0 a \ne 0 a = 0 a a a I = K I = K I = K
∎ 次に、与えられた元からイデアルを作る方法を用意します。以下この項では R R R
Definition 3.4 （生成されるイデアルと単項イデアル ）
R R R S S S R R R S S S R R R ( S ) (S) ( S ) S S S 生成するイデアル という。S = { a 1 , … , a n } S = \{a_1, \ldots, a_n\} S = { a 1 , … , a n } ( a 1 , … , a n ) (a_1, \ldots, a_n) ( a 1 , … , a n ) a a a
( a ) = a R = { r a : r ∈ R } (a) = aR = \{ ra : r \in R \} ( a ) = a R = { r a : r ∈ R } を a a a 単項イデアル という。
共通部分がイデアルであることはProposition 3.2 で確かめられます。イデアルの族 { I λ } \{I_\lambda\} { I λ } 0 0 0 a , b a, b a , b I λ I_\lambda I λ a − b a - b a − b r a ra r a a r ar a r ( S ) (S) ( S ) S S S
( S ) (S) ( S ) R R R
J : = { ∑ k = 1 n r k s k ∣ n ≥ 0 , r k ∈ R , s k ∈ S } J := \Big\{ \textstyle\sum_{k=1}^{n} r_k s_k \ \Big|\ n \ge 0,\ r_k \in R,\ s_k \in S \Big\} J := { ∑ k = 1 n r k s k n ≥ 0 , r k ∈ R , s k ∈ S } とおきます（n = 0 n = 0 n = 0 0 0 0 J J J 0 ∈ J 0 \in J 0 ∈ J ∑ r k s k + ∑ ( − r l ′ ) s l ′ \sum r_k s_k + \sum (-r'_l) s'_l ∑ r k s k + ∑ ( − r l ′ ) s l ′ r ∑ r k s k = ∑ ( r r k ) s k ∈ J r \sum r_k s_k = \sum (r r_k) s_k \in J r ∑ r k s k = ∑ ( r r k ) s k ∈ J s ∈ S s \in S s ∈ S 1 ⋅ s ∈ J 1 \cdot s \in J 1 ⋅ s ∈ J S ⊆ J S \subseteq J S ⊆ J S S S I I I r k s k r_k s_k r k s k J ⊆ I J \subseteq I J ⊆ I J J J S S S J = ( S ) J = (S) J = ( S ) S = { a } S = \{a\} S = { a } ∑ k r k a = ( ∑ k r k ) a \sum_k r_k a = \big(\sum_k r_k\big) a ∑ k r k a = ( ∑ k r k ) a { r a : r ∈ R } = a R \{ ra : r \in R \} = aR { r a : r ∈ R } = a R
Example 3.5 （イデアルの例と、イデアルでない例 ）
n Z n\mathbb{Z} n Z n ∈ Z n \in \mathbb{Z} n ∈ Z n Z = ( n ) n\mathbb{Z} = (n) n Z = ( n ) Z \mathbb{Z} Z n a − n b = n ( a − b ) na - nb = n(a-b) na − nb = n ( a − b ) r ⋅ n a = n ( r a ) r \cdot na = n(ra) r ⋅ na = n ( r a )
部分環だがイデアルでない例。 Z \mathbb{Z} Z Q \mathbb{Q} Q 0 , 1 0, 1 0 , 1 1 2 ∈ Q \tfrac12 \in \mathbb{Q} 2 1 ∈ Q 1 ∈ Z 1 \in \mathbb{Z} 1 ∈ Z 1 2 ⋅ 1 = 1 2 ∉ Z \tfrac12 \cdot 1 = \tfrac12 \notin \mathbb{Z} 2 1 ⋅ 1 = 2 1 ∈ / Z
消える条件で定まるイデアル。 I = { f ∈ R [ x ] : f ( 0 ) = 0 } I = \{ f \in \mathbb{R}[x] : f(0) = 0 \} I = { f ∈ R [ x ] : f ( 0 ) = 0 } f ( 0 ) = g ( 0 ) = 0 f(0) = g(0) = 0 f ( 0 ) = g ( 0 ) = 0 ( f − g ) ( 0 ) = 0 (f-g)(0) = 0 ( f − g ) ( 0 ) = 0 h ∈ R [ x ] h \in \mathbb{R}[x] h ∈ R [ x ] ( h f ) ( 0 ) = h ( 0 ) f ( 0 ) = 0 (hf)(0) = h(0) f(0) = 0 ( h f ) ( 0 ) = h ( 0 ) f ( 0 ) = 0 I = ( x ) I = (x) I = ( x ) f ∈ ( x ) f \in (x) f ∈ ( x ) f = x g f = xg f = xg f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f = ∑ k = 0 n c k x k f = \sum_{k=0}^{n} c_k x^k f = ∑ k = 0 n c k x k f ( 0 ) = c 0 = 0 f(0) = c_0 = 0 f ( 0 ) = c 0 = 0 f = x ∑ k = 1 n c k x k − 1 ∈ ( x ) f = x\sum_{k=1}^{n} c_k x^{k-1} \in (x) f = x ∑ k = 1 n c k x k − 1 ∈ ( x )
左イデアルだが右イデアルでない例。 R = M 2 ( R ) R = M_2(\mathbb{R}) R = M 2 ( R ) L L L P = ( p q r s ) ∈ R P = \begin{pmatrix} p & q \\ r & s\end{pmatrix} \in R P = ( p r q s ) ∈ R A = ( a 0 b 0 ) ∈ L A = \begin{pmatrix} a & 0 \\ b & 0\end{pmatrix} \in L A = ( a b 0 0 ) ∈ L
P A = ( p a + q b 0 r a + s b 0 ) ∈ L PA = \begin{pmatrix} pa + qb & 0 \\ ra + sb & 0\end{pmatrix} \in L P A = ( p a + q b r a + s b 0 0 ) ∈ L なので、L L L E 11 = ( 1 0 0 0 ) ∈ L E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \in L E 11 = ( 1 0 0 0 ) ∈ L E 12 = ( 0 1 0 0 ) E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} E 12 = ( 0 0 1 0 )
E 11 E 12 = ( 0 1 0 0 ) ∉ L E_{11}E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \notin L E 11 E 12 = ( 0 0 1 0 ) ∈ / L となるからです。非可換環では左右の区別が実際に効きます。
イデアルが手に入ったので、§1 で試した構成をきちんと実行します。まず商集合を用意します。
Definition 4.1 （法 I による合同と剰余類 ）
R R R I I I R R R a , b ∈ R a, b \in R a , b ∈ R
a ≡ b ( m o d I ) ⟺ a − b ∈ I a \equiv b \pmod{I} \iff a - b \in I a ≡ b ( mod I ) ⟺ a − b ∈ I と定める。これは R R R a a a a + I = { a + i : i ∈ I } a + I = \{ a + i : i \in I \} a + I = { a + i : i ∈ I } a a a 剰余類 といい、商集合を R / I R/I R / I
同値関係であることを確かめます。a − a = 0 ∈ I a - a = 0 \in I a − a = 0 ∈ I a − b ∈ I a - b \in I a − b ∈ I b − a = − ( a − b ) ∈ I b - a = -(a-b) \in I b − a = − ( a − b ) ∈ I a − b ∈ I a - b \in I a − b ∈ I b − c ∈ I b - c \in I b − c ∈ I a − c = ( a − b ) + ( b − c ) ∈ I a - c = (a-b) + (b-c) \in I a − c = ( a − b ) + ( b − c ) ∈ I 関係と同値関係 のDefinition 4.1［関係と同値関係］ を参照してください）。同値類が a + I a+I a + I b ∼ a ⟺ b − a ∈ I ⟺ b ∈ a + I b \sim a \iff b - a \in I \iff b \in a + I b ∼ a ⟺ b − a ∈ I ⟺ b ∈ a + I
Theorem 4.2 （剰余環 ）
R R R I I I R R R R / I R/I R / I
( a + I ) + ( b + I ) : = ( a + b ) + I , ( a + I ) ( b + I ) : = a b + I (a + I) + (b + I) := (a+b) + I, \qquad (a+I)(b+I) := ab + I ( a + I ) + ( b + I ) := ( a + b ) + I , ( a + I ) ( b + I ) := ab + I によって定めると、これらは代表元の取り方によらずに定まる。さらに R / I R/I R / I 0 + I = I 0 + I = I 0 + I = I 1 + I 1 + I 1 + I a + I a+I a + I ( − a ) + I (-a) + I ( − a ) + I π : R → R / I \pi \colon R \to R/I π : R → R / I π ( a ) = a + I \pi(a) = a + I π ( a ) = a + I ker π = I \ker \pi = I ker π = I R R R R / I R/I R / I R / I R/I R / I R R R I I I 剰余環 （商環）といい、π \pi π 自然な全射 という。
Proof（Theorem 4.2） a + I = a ′ + I a + I = a' + I a + I = a ′ + I b + I = b ′ + I b + I = b' + I b + I = b ′ + I Definition 4.1 より i : = a − a ′ ∈ I i := a - a' \in I i := a − a ′ ∈ I j : = b − b ′ ∈ I j := b - b' \in I j := b − b ′ ∈ I
加法の well-defined 性。 ( a + b ) − ( a ′ + b ′ ) = ( a − a ′ ) + ( b − b ′ ) = i + j (a+b) - (a'+b') = (a-a') + (b-b') = i + j ( a + b ) − ( a ′ + b ′ ) = ( a − a ′ ) + ( b − b ′ ) = i + j I I I i + j ∈ I i + j \in I i + j ∈ I ( a + b ) + I = ( a ′ + b ′ ) + I (a+b) + I = (a'+b') + I ( a + b ) + I = ( a ′ + b ′ ) + I
乗法の well-defined 性。 §3 冒頭の分解を使います。
a b − a ′ b ′ = ( a − a ′ ) b + a ′ ( b − b ′ ) = i b + a ′ j . ab - a'b' = (a - a')b + a'(b - b') = ib + a'j. ab − a ′ b ′ = ( a − a ′ ) b + a ′ ( b − b ′ ) = ib + a ′ j . i ∈ I i \in I i ∈ I b ∈ R b \in R b ∈ R I I I Definition 3.1 ）から i b ∈ I ib \in I ib ∈ I a ′ ∈ R a' \in R a ′ ∈ R j ∈ I j \in I j ∈ I I I I a ′ j ∈ I a'j \in I a ′ j ∈ I i b + a ′ j ∈ I ib + a'j \in I ib + a ′ j ∈ I a b − a ′ b ′ ∈ I ab - a'b' \in I ab − a ′ b ′ ∈ I a b + I = a ′ b ′ + I ab + I = a'b' + I ab + I = a ′ b ′ + I ここで両側イデアルであることが本質的に使われました。
環の公理。 演算が R R R R R R
( ( a + I ) ( b + I ) ) ( c + I ) = ( a b + I ) ( c + I ) = ( a b ) c + I = a ( b c ) + I = ( a + I ) ( ( b + I ) ( c + I ) ) , \big((a+I)(b+I)\big)(c+I) = (ab+I)(c+I) = (ab)c + I = a(bc) + I = (a+I)\big((b+I)(c+I)\big), ( ( a + I ) ( b + I ) ) ( c + I ) = ( ab + I ) ( c + I ) = ( ab ) c + I = a ( b c ) + I = ( a + I ) ( ( b + I ) ( c + I ) ) , 分配律は
( a + I ) ( ( b + I ) + ( c + I ) ) = ( a + I ) ( ( b + c ) + I ) = a ( b + c ) + I = ( a b + a c ) + I = ( a + I ) ( b + I ) + ( a + I ) ( c + I ) (a+I)\big((b+I)+(c+I)\big) = (a+I)\big((b+c)+I\big) = a(b+c) + I = (ab + ac) + I = (a+I)(b+I) + (a+I)(c+I) ( a + I ) ( ( b + I ) + ( c + I ) ) = ( a + I ) ( ( b + c ) + I ) = a ( b + c ) + I = ( ab + a c ) + I = ( a + I ) ( b + I ) + ( a + I ) ( c + I ) です。加法の結合律・可換律も同様です。( 0 + I ) + ( a + I ) = a + I (0+I) + (a+I) = a + I ( 0 + I ) + ( a + I ) = a + I ( a + I ) + ( ( − a ) + I ) = 0 + I (a+I) + ((-a)+I) = 0 + I ( a + I ) + (( − a ) + I ) = 0 + I ( 1 + I ) ( a + I ) = a + I = ( a + I ) ( 1 + I ) (1+I)(a+I) = a + I = (a+I)(1+I) ( 1 + I ) ( a + I ) = a + I = ( a + I ) ( 1 + I ) R R R a b + I = b a + I ab + I = ba + I ab + I = ba + I R / I R/I R / I
自然な全射。 π ( a + b ) = ( a + b ) + I = ( a + I ) + ( b + I ) = π ( a ) + π ( b ) \pi(a+b) = (a+b)+I = (a+I)+(b+I) = \pi(a) + \pi(b) π ( a + b ) = ( a + b ) + I = ( a + I ) + ( b + I ) = π ( a ) + π ( b ) π ( a b ) = a b + I = π ( a ) π ( b ) \pi(ab) = ab + I = \pi(a)\pi(b) π ( ab ) = ab + I = π ( a ) π ( b ) π ( 1 ) = 1 + I \pi(1) = 1 + I π ( 1 ) = 1 + I R / I R/I R / I π \pi π R / I R/I R / I a + I a+I a + I
ker π = { a ∈ R : a + I = 0 + I } = { a ∈ R : a − 0 ∈ I } = I \ker \pi = \{ a \in R : a + I = 0 + I \} = \{ a \in R : a - 0 \in I \} = I ker π = { a ∈ R : a + I = 0 + I } = { a ∈ R : a − 0 ∈ I } = I です。
∎ I = R I = R I = R R / R R/R R / R 1 + R = 0 + R 1 + R = 0 + R 1 + R = 0 + R I = { 0 } I = \{0\} I = { 0 } π \pi π R / { 0 } ≅ R R/\{0\} \cong R R / { 0 } ≅ R
Example 4.4 （6 を法とする剰余環の乗法表 ）
I = 6 Z = ( 6 ) I = 6\mathbb{Z} = (6) I = 6 Z = ( 6 ) Theorem 4.2 より Z / 6 Z \mathbb{Z}/6\mathbb{Z} Z /6 Z 0 ˉ , 1 ˉ , 2 ˉ , 3 ˉ , 4 ˉ , 5 ˉ \bar{0}, \bar{1}, \bar{2}, \bar{3}, \bar{4}, \bar{5} 0 ˉ , 1 ˉ , 2 ˉ , 3 ˉ , 4 ˉ , 5 ˉ a ˉ : = a + 6 Z \bar{a} := a + 6\mathbb{Z} a ˉ := a + 6 Z 6 q + r 6q + r 6 q + r 0 ≤ r ≤ 5 0 \le r \le 5 0 ≤ r ≤ 5 a ˉ = r ˉ \bar{a} = \bar{r} a ˉ = r ˉ
× \times × 0 ˉ \bar 0 0 ˉ 1 ˉ \bar 1 1 ˉ 2 ˉ \bar 2 2 ˉ 3 ˉ \bar 3 3 ˉ 4 ˉ \bar 4 4 ˉ 5 ˉ \bar 5 5 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 0 ˉ \bar 0 0 ˉ 1 ˉ \bar 1 1 ˉ 0 ˉ \bar 0 0 ˉ 1 ˉ \bar 1 1 ˉ 2 ˉ \bar 2 2 ˉ 3 ˉ \bar 3 3 ˉ 4 ˉ \bar 4 4 ˉ 5 ˉ \bar 5 5 ˉ 2 ˉ \bar 2 2 ˉ 0 ˉ \bar 0 0 ˉ 2 ˉ \bar 2 2 ˉ 4 ˉ \bar 4 4 ˉ 0 ˉ \bar 0 0 ˉ 2 ˉ \bar 2 2 ˉ 4 ˉ \bar 4 4 ˉ 3 ˉ \bar 3 3 ˉ 0 ˉ \bar 0 0 ˉ 3 ˉ \bar 3 3 ˉ 0 ˉ \bar 0 0 ˉ 3 ˉ \bar 3 3 ˉ 0 ˉ \bar 0 0 ˉ 3 ˉ \bar 3 3 ˉ 4 ˉ \bar 4 4 ˉ 0 ˉ \bar 0 0 ˉ 4 ˉ \bar 4 4 ˉ 2 ˉ \bar 2 2 ˉ 0 ˉ \bar 0 0 ˉ 4 ˉ \bar 4 4 ˉ 2 ˉ \bar 2 2 ˉ 5 ˉ \bar 5 5 ˉ 0 ˉ \bar 0 0 ˉ 5 ˉ \bar 5 5 ˉ 4 ˉ \bar 4 4 ˉ 3 ˉ \bar 3 3 ˉ 2 ˉ \bar 2 2 ˉ 1 ˉ \bar 1 1 ˉ
2 ˉ ⋅ 3 ˉ = 6 ˉ = 0 ˉ \bar 2 \cdot \bar 3 = \bar 6 = \bar 0 2 ˉ ⋅ 3 ˉ = 6 ˉ = 0 ˉ Z / 6 Z \mathbb{Z}/6\mathbb{Z} Z /6 Z 1 ˉ \bar 1 1 ˉ 1 ˉ , 5 ˉ \bar 1, \bar 5 1 ˉ , 5 ˉ 5 ˉ ⋅ 5 ˉ = 25 ‾ = 1 ˉ \bar 5 \cdot \bar 5 = \overline{25} = \bar 1 5 ˉ ⋅ 5 ˉ = 25 = 1 ˉ Z / 5 Z \mathbb{Z}/5\mathbb{Z} Z /5 Z 1 ˉ ⋅ 1 ˉ = 2 ˉ ⋅ 3 ˉ = 4 ˉ ⋅ 4 ˉ = 1 ˉ \bar 1 \cdot \bar 1 = \bar 2 \cdot \bar 3 = \bar 4 \cdot \bar 4 = \bar 1 1 ˉ ⋅ 1 ˉ = 2 ˉ ⋅ 3 ˉ = 4 ˉ ⋅ 4 ˉ = 1 ˉ 1 ˉ , 2 ˉ , 3 ˉ , 4 ˉ \bar 1, \bar 2, \bar 3, \bar 4 1 ˉ , 2 ˉ , 3 ˉ , 4 ˉ Corollary 5.5［環と体の基礎］ ）。6 6 6 5 5 5
§4 で「イデアルを与えると全射準同型 π : R → R / I \pi \colon R \to R/I π : R → R / I
Theorem 5.1 （環の準同型定理（第一同型定理） ）
f : R → S f \colon R \to S f : R → S K : = ker f = { a ∈ R : f ( a ) = 0 } K := \ker f = \{ a \in R : f(a) = 0 \} K := ker f = { a ∈ R : f ( a ) = 0 } Im f : = f ( R ) \operatorname{Im} f := f(R) Im f := f ( R )
K K K R R R Im f \operatorname{Im} f Im f S S S 写像 f ˉ : R / K → Im f \bar f \colon R/K \to \operatorname{Im} f f ˉ : R / K → Im f f ˉ ( a + K ) : = f ( a ) \bar f(a + K) := f(a) f ˉ ( a + K ) := f ( a ) R / ker f ≅ Im f R/\ker f \cong \operatorname{Im} f R / ker f ≅ Im f
さらに π : R → R / K \pi \colon R \to R/K π : R → R / K ι : Im f → S \iota \colon \operatorname{Im} f \to S ι : Im f → S f = ι ∘ f ˉ ∘ π f = \iota \circ \bar f \circ \pi f = ι ∘ f ˉ ∘ π
Proof（Theorem 5.1） (1) Proposition 3.2 の 3 条件を確かめます。f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 0 ∈ K 0 \in K 0 ∈ K K ≠ ∅ K \ne \emptyset K = ∅ a , b ∈ K a, b \in K a , b ∈ K f ( a − b ) = f ( a ) − f ( b ) = 0 − 0 = 0 f(a-b) = f(a) - f(b) = 0 - 0 = 0 f ( a − b ) = f ( a ) − f ( b ) = 0 − 0 = 0 a − b ∈ K a - b \in K a − b ∈ K r ∈ R r \in R r ∈ R a ∈ K a \in K a ∈ K
f ( r a ) = f ( r ) f ( a ) = f ( r ) ⋅ 0 = 0 , f ( a r ) = f ( a ) f ( r ) = 0 ⋅ f ( r ) = 0 f(ra) = f(r)f(a) = f(r) \cdot 0 = 0, \qquad f(ar) = f(a)f(r) = 0 \cdot f(r) = 0 f ( r a ) = f ( r ) f ( a ) = f ( r ) ⋅ 0 = 0 , f ( a r ) = f ( a ) f ( r ) = 0 ⋅ f ( r ) = 0 より r a , a r ∈ K ra, ar \in K r a , a r ∈ K K K K
(2) f ( 1 R ) = 1 S f(1_R) = 1_S f ( 1 R ) = 1 S 1 S ∈ Im f 1_S \in \operatorname{Im} f 1 S ∈ Im f Im f \operatorname{Im} f Im f f ( a ) , f ( b ) f(a), f(b) f ( a ) , f ( b ) f ( a ) − f ( b ) = f ( a − b ) ∈ Im f f(a) - f(b) = f(a-b) \in \operatorname{Im} f f ( a ) − f ( b ) = f ( a − b ) ∈ Im f f ( a ) f ( b ) = f ( a b ) ∈ Im f f(a)f(b) = f(ab) \in \operatorname{Im} f f ( a ) f ( b ) = f ( ab ) ∈ Im f Im f \operatorname{Im} f Im f S S S
(3) まず well-defined 性です。a + K = b + K a + K = b + K a + K = b + K a − b ∈ K a - b \in K a − b ∈ K f ( a ) − f ( b ) = f ( a − b ) = 0 f(a) - f(b) = f(a-b) = 0 f ( a ) − f ( b ) = f ( a − b ) = 0 f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) f ˉ \bar f f ˉ
f ˉ \bar f f ˉ Theorem 4.2 の演算の定義より
f ˉ ( ( a + K ) + ( b + K ) ) = f ˉ ( ( a + b ) + K ) = f ( a + b ) = f ( a ) + f ( b ) = f ˉ ( a + K ) + f ˉ ( b + K ) , f ˉ ( ( a + K ) ( b + K ) ) = f ˉ ( a b + K ) = f ( a b ) = f ( a ) f ( b ) = f ˉ ( a + K ) f ˉ ( b + K ) \begin{aligned}
\bar f\big((a+K) + (b+K)\big) &= \bar f\big((a+b)+K\big) = f(a+b) = f(a) + f(b) = \bar f(a+K) + \bar f(b+K), \\
\bar f\big((a+K)(b+K)\big) &= \bar f(ab + K) = f(ab) = f(a)f(b) = \bar f(a+K)\,\bar f(b+K)
\end{aligned} f ˉ ( ( a + K ) + ( b + K ) ) f ˉ ( ( a + K ) ( b + K ) ) = f ˉ ( ( a + b ) + K ) = f ( a + b ) = f ( a ) + f ( b ) = f ˉ ( a + K ) + f ˉ ( b + K ) , = f ˉ ( ab + K ) = f ( ab ) = f ( a ) f ( b ) = f ˉ ( a + K ) f ˉ ( b + K ) であり、f ˉ ( 1 + K ) = f ( 1 R ) = 1 S \bar f(1 + K) = f(1_R) = 1_S f ˉ ( 1 + K ) = f ( 1 R ) = 1 S Im f \operatorname{Im} f Im f
全射性。Im f \operatorname{Im} f Im f f ( a ) = f ˉ ( a + K ) f(a) = \bar f(a + K) f ( a ) = f ˉ ( a + K ) f ˉ \bar f f ˉ Im f \operatorname{Im} f Im f
単射性。f ˉ ( a + K ) = f ˉ ( b + K ) \bar f(a+K) = \bar f(b+K) f ˉ ( a + K ) = f ˉ ( b + K ) f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) f ( a − b ) = 0 f(a-b) = 0 f ( a − b ) = 0 a − b ∈ K a - b \in K a − b ∈ K Definition 4.1 より a + K = b + K a + K = b + K a + K = b + K f ˉ \bar f f ˉ
分解。 任意の a ∈ R a \in R a ∈ R ( ι ∘ f ˉ ∘ π ) ( a ) = ι ( f ˉ ( a + K ) ) = ι ( f ( a ) ) = f ( a ) (\iota \circ \bar f \circ \pi)(a) = \iota(\bar f(a+K)) = \iota(f(a)) = f(a) ( ι ∘ f ˉ ∘ π ) ( a ) = ι ( f ˉ ( a + K )) = ι ( f ( a )) = f ( a ) f = ι ∘ f ˉ ∘ π f = \iota \circ \bar f \circ \pi f = ι ∘ f ˉ ∘ π
∎ flowchart LR
R["R"] -->|"π: 自然な全射"| Q["R / Ker f"]
Q -->|"誘導される同型"| M["Im f"]
M -->|"包含写像"| S["S"]
R -->|"f"| S 任意の環準同型は「全射・同型・単射」の 3 段に一意に分解される この定理の使い方は決まっています。R / I R/I R / I I I I f f f
Example 5.2 （複素数体を多項式環の剰余環として作る ）
φ : R [ x ] → C \varphi \colon \mathbb{R}[x] \to \mathbb{C} φ : R [ x ] → C φ ( f ) = f ( i ) \varphi(f) = f(i) φ ( f ) = f ( i ) i i i ( f + g ) ( i ) = f ( i ) + g ( i ) (f+g)(i) = f(i)+g(i) ( f + g ) ( i ) = f ( i ) + g ( i ) ( f g ) ( i ) = f ( i ) g ( i ) (fg)(i) = f(i)g(i) ( f g ) ( i ) = f ( i ) g ( i ) 1 1 1 1 1 1 Example 7.4［環と体の基礎］ ）。
全射性。 任意の a + b i ∈ C a + bi \in \mathbb{C} a + bi ∈ C a , b ∈ R a, b \in \mathbb{R} a , b ∈ R φ ( a + b x ) = a + b i \varphi(a + bx) = a + bi φ ( a + b x ) = a + bi
核。 ker φ = ( x 2 + 1 ) \ker \varphi = (x^2+1) ker φ = ( x 2 + 1 ) ⊇ \supseteq ⊇ f = ( x 2 + 1 ) g f = (x^2+1)g f = ( x 2 + 1 ) g φ ( f ) = ( i 2 + 1 ) g ( i ) = 0 ⋅ g ( i ) = 0 \varphi(f) = (i^2+1)g(i) = 0 \cdot g(i) = 0 φ ( f ) = ( i 2 + 1 ) g ( i ) = 0 ⋅ g ( i ) = 0 ⊆ \subseteq ⊆ f ∈ ker φ f \in \ker\varphi f ∈ ker φ x 2 + 1 x^2+1 x 2 + 1 1 1 1 R [ x ] \mathbb{R}[x] R [ x ] f = ( x 2 + 1 ) q + r f = (x^2+1)q + r f = ( x 2 + 1 ) q + r r = a + b x r = a + bx r = a + b x a , b ∈ R a, b \in \mathbb{R} a , b ∈ R i i i 0 = f ( i ) = 0 + r ( i ) = a + b i 0 = f(i) = 0 + r(i) = a + bi 0 = f ( i ) = 0 + r ( i ) = a + bi a , b a, b a , b a = b = 0 a = b = 0 a = b = 0 r = 0 r = 0 r = 0 f = ( x 2 + 1 ) q ∈ ( x 2 + 1 ) f = (x^2+1)q \in (x^2+1) f = ( x 2 + 1 ) q ∈ ( x 2 + 1 )
結論。 Theorem 5.1 より
R [ x ] / ( x 2 + 1 ) ≅ Im φ = C . \mathbb{R}[x]/(x^2+1) \;\cong\; \operatorname{Im}\varphi = \mathbb{C}. R [ x ] / ( x 2 + 1 ) ≅ Im φ = C . 剰余環 R [ x ] / ( x 2 + 1 ) \mathbb{R}[x]/(x^2+1) R [ x ] / ( x 2 + 1 ) a + b x + ( x 2 + 1 ) a + bx + (x^2+1) a + b x + ( x 2 + 1 ) x ˉ 2 = − 1 ˉ \bar x^2 = -\bar 1 x ˉ 2 = − 1 ˉ x 2 = − 1 x^2 = -1 x 2 = − 1 i i i
Example 4.4 で、Z / 6 Z \mathbb{Z}/6\mathbb{Z} Z /6 Z Z / 5 Z \mathbb{Z}/5\mathbb{Z} Z /5 Z ( 6 ) (6) ( 6 ) ( 5 ) (5) ( 5 ) R R R 1 ≠ 0 1 \ne 0 1 = 0 可換 環とします。
Definition 6.1 （素イデアルと極大イデアル ）
R R R 1 ≠ 0 1 \ne 0 1 = 0
真のイデアル P P P P ≠ R P \ne R P = R 素イデアル であるとは、任意の a , b ∈ R a, b \in R a , b ∈ R a b ∈ P ab \in P ab ∈ P a ∈ P a \in P a ∈ P b ∈ P b \in P b ∈ P
真のイデアル M M M 極大イデアル であるとは、M ⊆ J ⊆ R M \subseteq J \subseteq R M ⊆ J ⊆ R R R R J J J J = M J = M J = M J = R J = R J = R
素イデアルの条件は、R = Z R = \mathbb{Z} R = Z P = ( p ) P = (p) P = ( p ) p ∣ a b p \mid ab p ∣ ab p ∣ a p \mid a p ∣ a p ∣ b p \mid b p ∣ b
Proposition 6.2 （素イデアルと整域 ）
R R R 1 ≠ 0 1 \ne 0 1 = 0 P P P R R R P P P R / P R/P R / P
Proof（Proposition 6.2） 準備として、P ≠ R P \ne R P = R R / P R/P R / P 1 + P ≠ 0 + P 1 + P \ne 0 + P 1 + P = 0 + P 1 + P = 0 + P 1 + P = 0 + P 1 + P = 0 + P 1 ∈ P 1 \in P 1 ∈ P Lemma 3.3 よりこれは P = R P = R P = R R R R Theorem 4.2 より R / P R/P R / P
( ⇒ ) (\Rightarrow) ( ⇒ ) P P P P ≠ R P \ne R P = R R / P R/P R / P 1 + P ≠ 0 + P 1 + P \ne 0 + P 1 + P = 0 + P ( a + P ) ( b + P ) = 0 + P (a+P)(b+P) = 0 + P ( a + P ) ( b + P ) = 0 + P Theorem 4.2 の乗法の定義より a b + P = 0 + P ab + P = 0 + P ab + P = 0 + P a b ∈ P ab \in P ab ∈ P P P P a ∈ P a \in P a ∈ P b ∈ P b \in P b ∈ P a + P = 0 + P a + P = 0 + P a + P = 0 + P b + P = 0 + P b + P = 0 + P b + P = 0 + P R / P R/P R / P R / P R/P R / P
( ⇐ ) (\Leftarrow) ( ⇐ ) R / P R/P R / P 1 + P ≠ 0 + P 1 + P \ne 0 + P 1 + P = 0 + P P ≠ R P \ne R P = R P P P a b ∈ P ab \in P ab ∈ P ( a + P ) ( b + P ) = a b + P = 0 + P (a+P)(b+P) = ab + P = 0 + P ( a + P ) ( b + P ) = ab + P = 0 + P R / P R/P R / P a + P = 0 + P a + P = 0+P a + P = 0 + P b + P = 0 + P b+P = 0+P b + P = 0 + P a ∈ P a \in P a ∈ P b ∈ P b \in P b ∈ P P P P
∎ Theorem 6.3 （極大イデアルと体 ）
R R R 1 ≠ 0 1 \ne 0 1 = 0 M M M R R R M M M R / M R/M R / M
Proof（Theorem 6.3） ( ⇒ ) (\Rightarrow) ( ⇒ ) M M M M ≠ R M \ne R M = R R / M R/M R / M 1 + M ≠ 0 + M 1 + M \ne 0 + M 1 + M = 0 + M a + M ≠ 0 + M a + M \ne 0 + M a + M = 0 + M a ∉ M a \notin M a ∈ / M
J : = M + ( a ) = { m + r a : m ∈ M , r ∈ R } J := M + (a) = \{ m + ra : m \in M,\ r \in R \} J := M + ( a ) = { m + r a : m ∈ M , r ∈ R } とおきます。J J J 0 = 0 + 0 ⋅ a ∈ J 0 = 0 + 0\cdot a \in J 0 = 0 + 0 ⋅ a ∈ J ( m + r a ) − ( m ′ + r ′ a ) = ( m − m ′ ) + ( r − r ′ ) a ∈ J (m + ra) - (m' + r'a) = (m - m') + (r - r')a \in J ( m + r a ) − ( m ′ + r ′ a ) = ( m − m ′ ) + ( r − r ′ ) a ∈ J s ( m + r a ) = s m + ( s r ) a ∈ J s(m+ra) = sm + (sr)a \in J s ( m + r a ) = s m + ( sr ) a ∈ J s m ∈ M sm \in M s m ∈ M M M M Proposition 3.2 より J ⊴ R J \trianglelefteq R J ⊴ R m = m + 0 ⋅ a m = m + 0\cdot a m = m + 0 ⋅ a M ⊆ J M \subseteq J M ⊆ J a = 0 + 1 ⋅ a ∈ J a = 0 + 1 \cdot a \in J a = 0 + 1 ⋅ a ∈ J a ∉ M a \notin M a ∈ / M M ⊊ J M \subsetneq J M ⊊ J M M M Definition 6.1 ）より J = R J = R J = R
したがって 1 ∈ J 1 \in J 1 ∈ J 1 = m + r a 1 = m + ra 1 = m + r a m ∈ M m \in M m ∈ M r ∈ R r \in R r ∈ R
( r + M ) ( a + M ) = r a + M = ( 1 − m ) + M = 1 + M (r + M)(a + M) = ra + M = (1 - m) + M = 1 + M ( r + M ) ( a + M ) = r a + M = ( 1 − m ) + M = 1 + M です（最後の等号は ( 1 − m ) − 1 = − m ∈ M (1-m) - 1 = -m \in M ( 1 − m ) − 1 = − m ∈ M a + M a + M a + M R / M R/M R / M R / M R/M R / M 1 ≠ 0 1 \ne 0 1 = 0 0 0 0
( ⇐ ) (\Leftarrow) ( ⇐ ) R / M R/M R / M 1 + M ≠ 0 + M 1 + M \ne 0 + M 1 + M = 0 + M M ≠ R M \ne R M = R M M M M ⊆ J ⊆ R M \subseteq J \subseteq R M ⊆ J ⊆ R J J J J ≠ M J \ne M J = M a ∈ J ∖ M a \in J \setminus M a ∈ J ∖ M a + M ≠ 0 + M a + M \ne 0 + M a + M = 0 + M R / M R/M R / M b + M b + M b + M ( b + M ) ( a + M ) = 1 + M (b+M)(a+M) = 1 + M ( b + M ) ( a + M ) = 1 + M b a − 1 ∈ M ba - 1 \in M ba − 1 ∈ M M ⊆ J M \subseteq J M ⊆ J b a − 1 ∈ J ba - 1 \in J ba − 1 ∈ J a ∈ J a \in J a ∈ J J J J b a ∈ J ba \in J ba ∈ J
1 = b a − ( b a − 1 ) ∈ J 1 = ba - (ba - 1) \in J 1 = ba − ( ba − 1 ) ∈ J となり、Lemma 3.3 から J = R J = R J = R M M M
∎ Corollary 6.4 （極大イデアルは素イデアル ）
R R R 1 ≠ 0 1 \ne 0 1 = 0 R R R
Proof（Corollary 6.4） M M M Theorem 6.3 より R / M R/M R / M Theorem 5.2［環と体の基礎］ ）。実際、体では 1 ≠ 0 1 \ne 0 1 = 0 a b = 0 ab = 0 ab = 0 a ≠ 0 a \ne 0 a = 0 a − 1 a^{-1} a − 1 b = a − 1 ( a b ) = a − 1 ⋅ 0 = 0 b = a^{-1}(ab) = a^{-1} \cdot 0 = 0 b = a − 1 ( ab ) = a − 1 ⋅ 0 = 0 R / M R/M R / M Proposition 6.2 から M M M
逆が成り立たない例を挙げます。R = Z R = \mathbb{Z} R = Z P = ( 0 ) P = (0) P = ( 0 ) Z / ( 0 ) ≅ Z \mathbb{Z}/(0) \cong \mathbb{Z} Z / ( 0 ) ≅ Z Proposition 6.2 より ( 0 ) (0) ( 0 ) ( 0 ) ⊊ ( 2 ) ⊊ Z (0) \subsetneq (2) \subsetneq \mathbb{Z} ( 0 ) ⊊ ( 2 ) ⊊ Z 2 ∈ ( 2 ) ∖ ( 0 ) 2 \in (2) \setminus (0) 2 ∈ ( 2 ) ∖ ( 0 ) 1 ∉ ( 2 ) 1 \notin (2) 1 ∈ / ( 2 ) ( 0 ) (0) ( 0 ) ( 0 ) (0) ( 0 ) R = Z [ x ] R = \mathbb{Z}[x] R = Z [ x ] P = ( x ) P = (x) P = ( x ) Exercise 8.3 ）。
∎ 真のイデアル I （I ≠ R） 例: Z の (4) 素イデアル P 例: Z[x] の (x) 極大イデアル M 例: Z の (7) R/I は零環でない可換環 R/P は整域 R/M は体 イデアルの階層と剰余環の性質の対応。R は単位元をもつ可換環で 1 ≠ 0 とする Z \mathbb{Z} Z n Z n\mathbb{Z} n Z
Definition 7.1 （単項イデアル整域 ）
整域 R R R R R R 単項イデアル整域 （principal ideal domain, PID）という。
Proposition 7.2 （整数環と多項式環は単項イデアル整域 ）
Z \mathbb{Z} Z n ≥ 0 n \ge 0 n ≥ 0 ( n ) = n Z (n) = n\mathbb{Z} ( n ) = n Z K K K K [ x ] K[x] K [ x ] { 0 } \{0\} { 0 } I I I I I I 0 0 0 g g g I = ( g ) I = (g) I = ( g ) Proof（Proposition 7.2） (1) Z \mathbb{Z} Z I ⊴ Z I \trianglelefteq \mathbb{Z} I ⊴ Z I = { 0 } I = \{0\} I = { 0 } I = ( 0 ) I = (0) I = ( 0 ) I ≠ { 0 } I \ne \{0\} I = { 0 } a ∈ I a \in I a ∈ I a ≠ 0 a \ne 0 a = 0 − a ∈ I -a \in I − a ∈ I I I I n : = min { m ∈ I : m > 0 } n := \min\{ m \in I : m > 0 \} n := min { m ∈ I : m > 0 } n ∈ I n \in I n ∈ I ( n ) = n Z ⊆ I (n) = n\mathbb{Z} \subseteq I ( n ) = n Z ⊆ I
逆の包含を示します。a ∈ I a \in I a ∈ I a = q n + r a = qn + r a = q n + r 0 ≤ r < n 0 \le r < n 0 ≤ r < n q n ∈ I qn \in I q n ∈ I r = a − q n ∈ I r = a - qn \in I r = a − q n ∈ I r > 0 r > 0 r > 0 r ∈ I r \in I r ∈ I 0 < r < n 0 < r < n 0 < r < n n n n r = 0 r = 0 r = 0 a = q n ∈ ( n ) a = qn \in (n) a = q n ∈ ( n ) I = ( n ) I = (n) I = ( n ) Z \mathbb{Z} Z
(2) まず K [ x ] K[x] K [ x ] f , g ∈ K [ x ] f, g \in K[x] f , g ∈ K [ x ] 0 0 0 a m ≠ 0 a_m \ne 0 a m = 0 b n ≠ 0 b_n \ne 0 b n = 0 m = deg f m = \deg f m = deg f n = deg g n = \deg g n = deg g f g fg f g x m + n x^{m+n} x m + n a m b n a_m b_n a m b n K K K a m b n ≠ 0 a_m b_n \ne 0 a m b n = 0 f g ≠ 0 fg \ne 0 f g = 0 K [ x ] K[x] K [ x ] deg ( f g ) = deg f + deg g \deg(fg) = \deg f + \deg g deg ( f g ) = deg f + deg g
次に K [ x ] K[x] K [ x ] g ≠ 0 g \ne 0 g = 0 m : = deg g m := \deg g m := deg g b m b_m b m K K K b m − 1 b_m^{-1} b m − 1 f f f f = q g + r f = qg + r f = q g + r r = 0 r = 0 r = 0 deg r < m \deg r < m deg r < m q , r q, r q , r f = 0 f = 0 f = 0 deg f < m \deg f < m deg f < m q = 0 q = 0 q = 0 r = f r = f r = f n : = deg f ≥ m n := \deg f \ge m n := deg f ≥ m n n n f f f a n a_n a n
f 1 : = f − a n b m − 1 x n − m g f_1 := f - a_n b_m^{-1} x^{\,n-m} g f 1 := f − a n b m − 1 x n − m g とおくと、x n x^n x n a n − a n b m − 1 b m = 0 a_n - a_n b_m^{-1} b_m = 0 a n − a n b m − 1 b m = 0 f 1 = 0 f_1 = 0 f 1 = 0 deg f 1 < n \deg f_1 < n deg f 1 < n f 1 = q 1 g + r f_1 = q_1 g + r f 1 = q 1 g + r r = 0 r = 0 r = 0 deg r < m \deg r < m deg r < m f = ( a n b m − 1 x n − m + q 1 ) g + r f = (a_n b_m^{-1} x^{\,n-m} + q_1) g + r f = ( a n b m − 1 x n − m + q 1 ) g + r
さて I ⊴ K [ x ] I \trianglelefteq K[x] I ⊴ K [ x ] I ≠ { 0 } I \ne \{0\} I = { 0 } I I I 0 0 0 g ∈ I g \in I g ∈ I g ∈ I g \in I g ∈ I ( g ) ⊆ I (g) \subseteq I ( g ) ⊆ I f ∈ I f \in I f ∈ I f = q g + r f = qg + r f = q g + r q g ∈ I qg \in I q g ∈ I r = f − q g ∈ I r = f - qg \in I r = f − q g ∈ I r ≠ 0 r \ne 0 r = 0 deg r < deg g \deg r < \deg g deg r < deg g g g g r = 0 r = 0 r = 0 f = q g ∈ ( g ) f = qg \in (g) f = q g ∈ ( g ) I = ( g ) I = (g) I = ( g ) K [ x ] K[x] K [ x ]
∎ PID では、極大イデアルの判定が「割り切れなさ」の判定に還元されます。整域 R R R p p p 既約元 であるとは、p ≠ 0 p \ne 0 p = 0 p p p p = a b p = ab p = ab a a a b b b
主張。 R R R p ∈ R p \in R p ∈ R p ≠ 0 p \ne 0 p = 0 ( p ) (p) ( p ) p p p
まず補助的な事実を確認します。Lemma 3.3 より、( p ) = R (p) = R ( p ) = R 1 ∈ ( p ) 1 \in (p) 1 ∈ ( p ) 1 = r p 1 = rp 1 = r p r r r p p p
( ⇐ ) (\Leftarrow) ( ⇐ ) p p p p p p ( p ) ≠ R (p) \ne R ( p ) = R ( p ) ⊆ J ⊆ R (p) \subseteq J \subseteq R ( p ) ⊆ J ⊆ R R R R J = ( d ) J = (d) J = ( d ) p ∈ ( d ) p \in (d) p ∈ ( d ) p = d c p = dc p = d c c ∈ R c \in R c ∈ R p p p d d d c c c d d d J = ( d ) = R J = (d) = R J = ( d ) = R c c c d = p c − 1 ∈ ( p ) d = pc^{-1} \in (p) d = p c − 1 ∈ ( p ) J = ( d ) ⊆ ( p ) J = (d) \subseteq (p) J = ( d ) ⊆ ( p ) J = ( p ) J = (p) J = ( p ) ( p ) (p) ( p )
( ⇒ ) (\Rightarrow) ( ⇒ ) ( p ) (p) ( p ) ( p ) ≠ R (p) \ne R ( p ) = R p p p p = a b p = ab p = ab p ∈ ( a ) p \in (a) p ∈ ( a ) ( p ) ⊆ ( a ) ⊆ R (p) \subseteq (a) \subseteq R ( p ) ⊆ ( a ) ⊆ R ( a ) = R (a) = R ( a ) = R ( a ) = ( p ) (a) = (p) ( a ) = ( p ) a a a a ∈ ( p ) a \in (p) a ∈ ( p ) a = p t a = pt a = pt t t t p = a b = p t b p = ab = ptb p = ab = pt b p ( 1 − t b ) = 0 p(1 - tb) = 0 p ( 1 − t b ) = 0 R R R p ≠ 0 p \ne 0 p = 0 1 = t b 1 = tb 1 = t b b b b p p p
この主張とTheorem 6.3 を組み合わせると、PID では既約元による剰余環がちょうど体になります 。Z \mathbb{Z} Z ± p \pm p ± p p p p Z / p Z = F p \mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p Z / p Z = F p K [ x ] K[x] K [ x ]
Example 7.3 （4 元体の構成 ）
K = F 2 = { 0 , 1 } K = \mathbb{F}_2 = \{0, 1\} K = F 2 = { 0 , 1 } f = x 2 + x + 1 ∈ F 2 [ x ] f = x^2 + x + 1 \in \mathbb{F}_2[x] f = x 2 + x + 1 ∈ F 2 [ x ] f f f f f f F 2 \mathbb{F}_2 F 2
f ( 0 ) = 0 + 0 + 1 = 1 ≠ 0 , f ( 1 ) = 1 + 1 + 1 = 1 ≠ 0 f(0) = 0 + 0 + 1 = 1 \ne 0, \qquad f(1) = 1 + 1 + 1 = 1 \ne 0 f ( 0 ) = 0 + 0 + 1 = 1 = 0 , f ( 1 ) = 1 + 1 + 1 = 1 = 0 （F 2 \mathbb{F}_2 F 2 1 + 1 = 0 1 + 1 = 0 1 + 1 = 0 f f f ( f ) (f) ( f ) F 2 [ x ] \mathbb{F}_2[x] F 2 [ x ] Theorem 6.3 より F : = F 2 [ x ] / ( f ) F := \mathbb{F}_2[x]/(f) F := F 2 [ x ] / ( f )
F F F Proposition 7.2 の証明で使った除法定理より、任意の g ∈ F 2 [ x ] g \in \mathbb{F}_2[x] g ∈ F 2 [ x ] g = q f + r g = qf + r g = q f + r r = 0 r = 0 r = 0 deg r ≤ 1 \deg r \le 1 deg r ≤ 1 g + ( f ) = r + ( f ) g + (f) = r + (f) g + ( f ) = r + ( f ) r , r ′ r, r' r , r ′ r + ( f ) = r ′ + ( f ) r + (f) = r' + (f) r + ( f ) = r ′ + ( f ) r − r ′ ∈ ( f ) r - r' \in (f) r − r ′ ∈ ( f ) r − r ′ r - r' r − r ′ f f f r − r ′ = 0 r - r' = 0 r − r ′ = 0 F F F a + b α a + b\alpha a + b α a , b ∈ F 2 a, b \in \mathbb{F}_2 a , b ∈ F 2 α : = x + ( f ) \alpha := x + (f) α := x + ( f ) F = { 0 , 1 , α , α + 1 } F = \{0, 1, \alpha, \alpha+1\} F = { 0 , 1 , α , α + 1 }
乗法は α 2 + α + 1 = 0 \alpha^2 + \alpha + 1 = 0 α 2 + α + 1 = 0 α 2 = α + 1 \alpha^2 = \alpha + 1 α 2 = α + 1 − 1 = 1 -1 = 1 − 1 = 1
α ⋅ α = α + 1 , α ( α + 1 ) = α 2 + α = ( α + 1 ) + α = 1 , ( α + 1 ) ( α + 1 ) = α 2 + 2 α + 1 = α 2 + 1 = ( α + 1 ) + 1 = α . \begin{aligned}
\alpha \cdot \alpha &= \alpha + 1, \\
\alpha(\alpha+1) &= \alpha^2 + \alpha = (\alpha+1) + \alpha = 1, \\
(\alpha+1)(\alpha+1) &= \alpha^2 + 2\alpha + 1 = \alpha^2 + 1 = (\alpha+1) + 1 = \alpha.
\end{aligned} α ⋅ α α ( α + 1 ) ( α + 1 ) ( α + 1 ) = α + 1 , = α 2 + α = ( α + 1 ) + α = 1 , = α 2 + 2 α + 1 = α 2 + 1 = ( α + 1 ) + 1 = α . × \times × 1 1 1 α \alpha α α + 1 \alpha+1 α + 1 1 1 1 1 1 1 α \alpha α α + 1 \alpha+1 α + 1 α \alpha α α \alpha α α + 1 \alpha+1 α + 1 1 1 1 α + 1 \alpha+1 α + 1 α + 1 \alpha+1 α + 1 1 1 1 α \alpha α
各行に 1 1 1 0 0 0 1 , α , α + 1 1, \alpha, \alpha+1 1 , α , α + 1 α − 1 = α + 1 \alpha^{-1} = \alpha+1 α − 1 = α + 1 F 4 \mathbb{F}_4 F 4
F 4 \mathbb{F}_4 F 4 Z / 4 Z \mathbb{Z}/4\mathbb{Z} Z /4 Z Z / 4 Z \mathbb{Z}/4\mathbb{Z} Z /4 Z 2 ˉ ⋅ 2 ˉ = 0 ˉ \bar 2 \cdot \bar 2 = \bar 0 2 ˉ ⋅ 2 ˉ = 0 ˉ p p p f ∈ F p [ x ] f \in \mathbb{F}_p[x] f ∈ F p [ x ] deg f = n \deg f = n deg f = n p n p^n p n ガロア理論への招待 のRemark 2.2［ガロア理論への招待］ ）。
Example 7.4 （単項イデアル整域でない環の例 ）
R = Z [ x ] R = \mathbb{Z}[x] R = Z [ x ] I = ( 2 , x ) I = (2, x) I = ( 2 , x )
I I I I = { h ∈ Z [ x ] : h ( 0 ) が偶数 } I = \{ h \in \mathbb{Z}[x] : h(0) \text{ が偶数} \} I = { h ∈ Z [ x ] : h ( 0 ) が偶数 } ⊆ \subseteq ⊆ h = 2 f + x g h = 2f + xg h = 2 f + xg h ( 0 ) = 2 f ( 0 ) + 0 = 2 f ( 0 ) h(0) = 2f(0) + 0 = 2f(0) h ( 0 ) = 2 f ( 0 ) + 0 = 2 f ( 0 ) ⊇ \supseteq ⊇ h ( 0 ) = 2 m h(0) = 2m h ( 0 ) = 2 m h − h ( 0 ) h - h(0) h − h ( 0 ) 0 0 0 h − 2 m = x g h - 2m = xg h − 2 m = xg h = 2 m + x g = 2 ⋅ m + x ⋅ g ∈ I h = 2m + xg = 2 \cdot m + x \cdot g \in I h = 2 m + xg = 2 ⋅ m + x ⋅ g ∈ I
I I I I = ( d ) I = (d) I = ( d ) 2 ∈ I = ( d ) 2 \in I = (d) 2 ∈ I = ( d ) 2 = d e 2 = de 2 = d e e ∈ Z [ x ] e \in \mathbb{Z}[x] e ∈ Z [ x ] Z \mathbb{Z} Z Z [ x ] \mathbb{Z}[x] Z [ x ] deg d + deg e = deg 2 = 0 \deg d + \deg e = \deg 2 = 0 deg d + deg e = deg 2 = 0 deg d = 0 \deg d = 0 deg d = 0 d d d d ∣ 2 d \mid 2 d ∣ 2 d ∈ { ± 1 , ± 2 } d \in \{\pm 1, \pm 2\} d ∈ { ± 1 , ± 2 }
d = ± 1 d = \pm 1 d = ± 1 ( d ) = Z [ x ] (d) = \mathbb{Z}[x] ( d ) = Z [ x ] 1 ( 0 ) = 1 1(0) = 1 1 ( 0 ) = 1 1 ∉ I 1 \notin I 1 ∈ / I I ≠ Z [ x ] I \ne \mathbb{Z}[x] I = Z [ x ] d = ± 2 d = \pm 2 d = ± 2 x ∈ I = ( d ) x \in I = (d) x ∈ I = ( d ) x = ± 2 g x = \pm 2 g x = ± 2 g g ∈ Z [ x ] g \in \mathbb{Z}[x] g ∈ Z [ x ] x x x 1 = ± 2 g 1 1 = \pm 2 g_1 1 = ± 2 g 1 g 1 ∈ Z g_1 \in \mathbb{Z} g 1 ∈ Z g 1 g_1 g 1 よって I I I Z [ x ] \mathbb{Z}[x] Z [ x ]
I I I ψ : Z [ x ] → F 2 \psi \colon \mathbb{Z}[x] \to \mathbb{F}_2 ψ : Z [ x ] → F 2 ψ ( h ) = h ( 0 ) ‾ \psi(h) = \overline{h(0)} ψ ( h ) = h ( 0 ) h ( 0 ) h(0) h ( 0 ) 2 2 2 ψ \psi ψ Z [ x ] → Z \mathbb{Z}[x] \to \mathbb{Z} Z [ x ] → Z Z → F 2 \mathbb{Z} \to \mathbb{F}_2 Z → F 2 0 , 1 0, 1 0 , 1 ker ψ = { h : h ( 0 ) が偶数 } = I \ker \psi = \{h : h(0) \text{ が偶数}\} = I ker ψ = { h : h ( 0 ) が偶数 } = I Theorem 5.1 より Z [ x ] / I ≅ F 2 \mathbb{Z}[x]/I \cong \mathbb{F}_2 Z [ x ] / I ≅ F 2 F 2 \mathbb{F}_2 F 2 Theorem 6.3 から I I I
生成元が 2 個必要な極大イデアルが実在する、というのがこの例の要点です。PID の外に出ると「1 個の元で潰す」だけでは足りなくなります。
Exercise 8.1 標準
R R R I , J I, J I , J R R R
I + J : = { a + b : a ∈ I , b ∈ J } I + J := \{ a + b : a \in I,\ b \in J \} I + J := { a + b : a ∈ I , b ∈ J } I ∩ J I \cap J I ∩ J I J : = { ∑ k = 1 n a k b k : n ≥ 0 , a k ∈ I , b k ∈ J } IJ := \{ \sum_{k=1}^{n} a_k b_k : n \ge 0,\ a_k \in I,\ b_k \in J \} I J := { ∑ k = 1 n a k b k : n ≥ 0 , a k ∈ I , b k ∈ J } R R R R = Z R = \mathbb{Z} R = Z I = ( m ) I = (m) I = ( m ) J = ( n ) J = (n) J = ( n ) m , n ≥ 1 m, n \ge 1 m , n ≥ 1 I + J I+J I + J I ∩ J I \cap J I ∩ J I J IJ I J Solution 1. Proposition 3.2 の 3 条件を確かめます。
I + J I+J I + J 0 = 0 + 0 ∈ I + J 0 = 0 + 0 \in I+J 0 = 0 + 0 ∈ I + J ( a + b ) − ( a ′ + b ′ ) = ( a − a ′ ) + ( b − b ′ ) (a+b) - (a'+b') = (a-a') + (b-b') ( a + b ) − ( a ′ + b ′ ) = ( a − a ′ ) + ( b − b ′ ) a − a ′ ∈ I a - a' \in I a − a ′ ∈ I b − b ′ ∈ J b - b' \in J b − b ′ ∈ J r ∈ R r \in R r ∈ R r ( a + b ) = r a + r b r(a+b) = ra + rb r ( a + b ) = r a + r b r a ∈ I ra \in I r a ∈ I r b ∈ J rb \in J r b ∈ J
I ∩ J I \cap J I ∩ J 0 ∈ I 0 \in I 0 ∈ I 0 ∈ J 0 \in J 0 ∈ J a , b ∈ I ∩ J a, b \in I \cap J a , b ∈ I ∩ J a − b a - b a − b I I I J J J r ∈ R r \in R r ∈ R r a ra r a I I I J J J
I J IJ I J 0 0 0 I J IJ I J ∑ a k b k − ∑ a l ′ b l ′ = ∑ a k b k + ∑ ( − a l ′ ) b l ′ \sum a_k b_k - \sum a'_l b'_l = \sum a_k b_k + \sum (-a'_l) b'_l ∑ a k b k − ∑ a l ′ b l ′ = ∑ a k b k + ∑ ( − a l ′ ) b l ′ − a l ′ ∈ I -a'_l \in I − a l ′ ∈ I r ∑ a k b k = ∑ ( r a k ) b k r \sum a_k b_k = \sum (r a_k) b_k r ∑ a k b k = ∑ ( r a k ) b k r a k ∈ I ra_k \in I r a k ∈ I
2. Proposition 7.2 より Z \mathbb{Z} Z ( d ) (d) ( d ) d ≥ 0 d \ge 0 d ≥ 0
I + J = ( gcd ( m , n ) ) I + J = (\gcd(m,n)) I + J = ( g cd( m , n )) I + J = ( d ) I+J = (d) I + J = ( d ) d ≥ 0 d \ge 0 d ≥ 0 m = m + 0 ∈ I + J = ( d ) m = m + 0 \in I+J = (d) m = m + 0 ∈ I + J = ( d ) d ∣ m d \mid m d ∣ m d ∣ n d \mid n d ∣ n d d d m , n m, n m , n d ∈ ( d ) = I + J d \in (d) = I+J d ∈ ( d ) = I + J d = u m + v n d = um + vn d = u m + v n u , v ∈ Z u, v \in \mathbb{Z} u , v ∈ Z m , n m, n m , n e e e d d d d d d
I ∩ J = ( lcm ( m , n ) ) I \cap J = (\operatorname{lcm}(m,n)) I ∩ J = ( lcm ( m , n )) I ∩ J = ( l ) I \cap J = (l) I ∩ J = ( l ) l ≥ 0 l \ge 0 l ≥ 0 l ∈ ( m ) l \in (m) l ∈ ( m ) l ∈ ( n ) l \in (n) l ∈ ( n ) m ∣ l m \mid l m ∣ l n ∣ l n \mid l n ∣ l l l l m , n ≥ 1 m, n \ge 1 m , n ≥ 1 I ∩ J ≠ { 0 } I \cap J \ne \{0\} I ∩ J = { 0 } l ≥ 1 l \ge 1 l ≥ 1 c c c ( m ) ∩ ( n ) = ( l ) (m) \cap (n) = (l) ( m ) ∩ ( n ) = ( l ) l ∣ c l \mid c l ∣ c l l l
I J = ( m n ) IJ = (mn) I J = ( mn ) ∑ ( a k m ) ( b k n ) = m n ∑ a k b k \sum (a_k m)(b_k n) = mn \sum a_k b_k ∑ ( a k m ) ( b k n ) = mn ∑ a k b k I J ⊆ ( m n ) IJ \subseteq (mn) I J ⊆ ( mn ) m n = m ⋅ n ∈ I J mn = m \cdot n \in IJ mn = m ⋅ n ∈ I J ( m n ) ⊆ I J (mn) \subseteq IJ ( mn ) ⊆ I J
なお I ∩ J = I J I \cap J = IJ I ∩ J = I J m = n = 2 m = n = 2 m = n = 2 I ∩ J = ( 2 ) I \cap J = (2) I ∩ J = ( 2 ) I J = ( 4 ) IJ = (4) I J = ( 4 ) gcd ( m , n ) = 1 \gcd(m,n) = 1 g cd( m , n ) = 1
Exercise 8.2 標準
準同型定理を使って Z / 6 Z ≅ Z / 2 Z × Z / 3 Z \mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} Z /6 Z ≅ Z /2 Z × Z /3 Z
Solution f : Z → Z / 2 Z × Z / 3 Z f \colon \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} f : Z → Z /2 Z × Z /3 Z f ( a ) = ( a + 2 Z , a + 3 Z ) f(a) = (a + 2\mathbb{Z},\ a + 3\mathbb{Z}) f ( a ) = ( a + 2 Z , a + 3 Z ) Theorem 4.2 ）で、直積環の演算は成分ごとなので、f f f f ( 1 ) = ( 1 + 2 Z , 1 + 3 Z ) f(1) = (1+2\mathbb{Z},\ 1+3\mathbb{Z}) f ( 1 ) = ( 1 + 2 Z , 1 + 3 Z )
核。 f ( a ) = 0 f(a) = 0 f ( a ) = 0 2 ∣ a 2 \mid a 2 ∣ a 3 ∣ a 3 \mid a 3 ∣ a a = 2 b a = 2b a = 2 b 3 ∣ 2 b 3 \mid 2b 3 ∣ 2 b 3 3 3 3 ∤ 2 3 \nmid 2 3 ∤ 2 3 ∣ b 3 \mid b 3 ∣ b b = 3 c b = 3c b = 3 c a = 6 c a = 6c a = 6 c 6 ∣ a 6 \mid a 6 ∣ a 2 ∣ a 2 \mid a 2 ∣ a 3 ∣ a 3 \mid a 3 ∣ a ker f = 6 Z \ker f = 6\mathbb{Z} ker f = 6 Z
適用。 Theorem 5.1 より Z / 6 Z ≅ Im f \mathbb{Z}/6\mathbb{Z} \cong \operatorname{Im} f Z /6 Z ≅ Im f 6 6 6 Z / 2 Z × Z / 3 Z \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} Z /2 Z × Z /3 Z 2 × 3 = 6 2 \times 3 = 6 2 × 3 = 6 Im f \operatorname{Im} f Im f 6 6 6 6 6 6 Im f = Z / 2 Z × Z / 3 Z \operatorname{Im} f = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} Im f = Z /2 Z × Z /3 Z Z / 6 Z ≅ Z / 2 Z × Z / 3 Z \mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} Z /6 Z ≅ Z /2 Z × Z /3 Z
これは中国剰余定理の環論版で、一般には互いに素なイデアル I + J = R I + J = R I + J = R R / ( I ∩ J ) ≅ R / I × R / J R/(I \cap J) \cong R/I \times R/J R / ( I ∩ J ) ≅ R / I × R / J Exercise 8.1 で見た ( 2 ) + ( 3 ) = ( gcd ( 2 , 3 ) ) = ( 1 ) = Z (2)+(3) = (\gcd(2,3)) = (1) = \mathbb{Z} ( 2 ) + ( 3 ) = ( g cd( 2 , 3 )) = ( 1 ) = Z
Exercise 8.3 標準
Z [ x ] \mathbb{Z}[x] Z [ x ] ( x ) (x) ( x ) ( x ) ⊊ ( 2 , x ) (x) \subsetneq (2,x) ( x ) ⊊ ( 2 , x ) ( 2 , x ) (2,x) ( 2 , x )
Solution 代入写像 φ : Z [ x ] → Z \varphi \colon \mathbb{Z}[x] \to \mathbb{Z} φ : Z [ x ] → Z φ ( h ) = h ( 0 ) \varphi(h) = h(0) φ ( h ) = h ( 0 ) Example 3.5 の 3 と同じ議論（定数項が 0 0 0 x x x ker φ = ( x ) \ker \varphi = (x) ker φ = ( x ) Theorem 5.1 より
Z [ x ] / ( x ) ≅ Z . \mathbb{Z}[x]/(x) \cong \mathbb{Z}. Z [ x ] / ( x ) ≅ Z . Z \mathbb{Z} Z Proposition 6.2 から ( x ) (x) ( x ) Z \mathbb{Z} Z 2 2 2 1 / 2 1/2 1/2 Theorem 6.3 から ( x ) (x) ( x )
( x ) (x) ( x ) x g xg xg x g = 2 ⋅ 0 + x ⋅ g xg = 2 \cdot 0 + x \cdot g xg = 2 ⋅ 0 + x ⋅ g ( 2 , x ) (2,x) ( 2 , x ) ( x ) ⊆ ( 2 , x ) (x) \subseteq (2,x) ( x ) ⊆ ( 2 , x ) 2 ∈ ( 2 , x ) 2 \in (2,x) 2 ∈ ( 2 , x ) φ ( 2 ) = 2 ≠ 0 \varphi(2) = 2 \ne 0 φ ( 2 ) = 2 = 0 2 ∉ ( x ) 2 \notin (x) 2 ∈ / ( x ) ( x ) ⊊ ( 2 , x ) (x) \subsetneq (2,x) ( x ) ⊊ ( 2 , x ) ( 2 , x ) (2,x) ( 2 , x ) Example 7.4 で示したとおり、Z [ x ] / ( 2 , x ) ≅ F 2 \mathbb{Z}[x]/(2,x) \cong \mathbb{F}_2 Z [ x ] / ( 2 , x ) ≅ F 2
( x ) (x) ( x ) ( 2 , x ) (2,x) ( 2 , x )
Exercise 8.4 難
K K K R R R 1 R ≠ 0 R 1_R \ne 0_R 1 R = 0 R f : K → R f \colon K \to R f : K → R f f f
Solution Theorem 5.1 の 1 より ker f \ker f ker f K K K K K K Lemma 3.3 より ker f = { 0 } \ker f = \{0\} ker f = { 0 } ker f = K \ker f = K ker f = K
ker f = K \ker f = K ker f = K 1 K ∈ ker f 1_K \in \ker f 1 K ∈ ker f f ( 1 K ) = 0 R f(1_K) = 0_R f ( 1 K ) = 0 R f ( 1 K ) = 1 R f(1_K) = 1_R f ( 1 K ) = 1 R 1 R = 0 R 1_R = 0_R 1 R = 0 R R R R ker f = { 0 } \ker f = \{0\} ker f = { 0 }
単射性を導きます。f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) f ( a − b ) = f ( a ) − f ( b ) = 0 f(a-b) = f(a) - f(b) = 0 f ( a − b ) = f ( a ) − f ( b ) = 0 a − b ∈ ker f = { 0 } a - b \in \ker f = \{0\} a − b ∈ ker f = { 0 } a = b a = b a = b f f f
言い換えると、体 K K K K K K I = { 0 } I = \{0\} I = { 0 } I = K I = K I = K Theorem 6.3 で極大イデアルが登場したのは、これ以上潰せない状態に到達した合図だったわけです。
松坂和夫『代数系入門』岩波書店、1976 — 環・イデアル・剰余環を扱う章。定義の積み上げが丁寧で、群の商との対応を確認しながら読めます。
雪江明彦『代数学2 環と体とガロア理論』日本評論社、2010 — 環論の基礎を扱う章。素イデアル・極大イデアルから可換環論への接続がよく整理されています。
D. S. Dummit, R. M. Foote, Abstract Algebra , 3rd ed., Wiley, 2004 — Chapter 7 (Introduction to Rings), Chapter 8 (Euclidean Domains, Principal Ideal Domains, and Unique Factorization Domains)。演習の量が圧倒的です。
M. F. Atiyah, I. G. Macdonald, Introduction to Commutative Algebra , Addison-Wesley, 1969 — Chapter 1 (Rings and Ideals)。対応定理・素イデアル・極大イデアルが最短距離でまとまっています。
S. Lang, Algebra , 3rd ed., Graduate Texts in Mathematics 211, Springer, 2002 — Chapter II (Rings)。可換とは限らない環の両側イデアルの扱いを含みます。
主張（クルルの定理）。 R R R 1 ≠ 0 1 \ne 0 1 = 0 R R R R R R I I I I I I
証明の筋。 Σ \Sigma Σ I I I R R R I ∈ Σ I \in \Sigma I ∈ Σ Σ \Sigma Σ { J λ } \{J_\lambda\} { J λ } Σ \Sigma Σ J : = ⋃ λ J λ J := \bigcup_\lambda J_\lambda J := ⋃ λ J λ J J J a ∈ J λ 1 a \in J_{\lambda_1} a ∈ J λ 1 b ∈ J λ 2 b \in J_{\lambda_2} b ∈ J λ 2 a , b a, b a , b J λ J_\lambda J λ a − b ∈ J λ ⊆ J a - b \in J_\lambda \subseteq J a − b ∈ J λ ⊆ J J λ J_\lambda J λ J λ J_\lambda J λ Lemma 3.3 より 1 ∉ J λ 1 \notin J_\lambda 1 ∈ / J λ 1 ∉ J 1 \notin J 1 ∈ / J J J J J ∈ Σ J \in \Sigma J ∈ Σ Σ \Sigma Σ M M M I I I
注意。 この証明はツォルンの補題、すなわち選択公理に依存します。Z \mathbb{Z} Z K [ x ] K[x] K [ x ] Proposition 7.2 のように生成元を明示して極大イデアルを直接作れるので選択公理は要りません。抽象的な環に対して「存在は言えるが構成はできない」という状況になるのが、この定理の特徴です。系として、可換環 R R R a a a ( a ) (a) ( a ) a a a