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6.6.4 Subring, Ideal and Quotient ring
Definition 29: A subring of a ring R is a nonempty subset S of R which is also a ring under the same operations. Example :
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Theorem 6.34: A subset S of a ring R is a subring if and only if for a, bS:
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Example: Let [R;+,·] be a ring
Example: Let [R;+,·] be a ring. Then C={x|xR, and a·x=x·a for all aR} is a subring of R. Proof: For x,yC, x+y,-x?C, x·y?C i.e. aR,a·(x+y)=?(x+y)·a,a·(-x)=?(-x)·a,a·(x·y) =?(x·y)·a
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2.Ideal(理想) Definition 30:. Let [R; + , * ] be a ring. A subring S of R is called an ideal of R if rs S and srS for any rR and sS. To show that S is an ideal of R it is sufficient to check that (a) [S; +] is a subgroup of [R; + ]; (b) if rR and sS, then rsS and srS.
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Example: [R;+,. ] is a commutative ring with identity element
Example: [R;+,*] is a commutative ring with identity element. For aR,(a)={a*r|rR},then [(a);+,*] is an ideal of [R;+,*]. If [R;+,*] is a commutative ring, For a R, (a)={a*r+na|rR,nZ}, then [(a);+,*] is an ideal of [R;+,*].
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Principle ideas Definition 31: If R is a commutative ring and aR, then (a) ={a*r+na|rR} is the principle ideal defined generated by a. Example: Every ideal in [Z;+,*] is a principle. Proof: Let D be an ideal of Z. If D={0}, then it holds. Suppose that D{0}. Let b=minaD{|a| | a0,where a D}.
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3. Quotient ring Theorem 6.35: Let [R; + ,*] be a ring and let S be an ideal of R. If R/S ={S+a|aR} and the operations and on the cosets are defined by (S+a)(S+b)=S+(a + b) ; (S+a)(S+b) =S+(a*b); then [R/S; , ] is a ring. Proof: Because [S;+] is a normal subgroup of [R;+], [R/S;] is a group. Because [R;+] is a commutative group, [R/S;] is also a commutative group. Need prove [R/S;] is an algebraic system, a sumigroup, distributive laws
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Definition 32: Under the conditions of Theorem 6
Definition 32: Under the conditions of Theorem 6.35, [R/S; , ] is a ring which is called a quotient ring. Example:Let Z(i)={a+bi|a,bZ}, E(i)={2a+2bi|a,bZ}. [E(i);+,*] is an ideal of ring [Z(i);+,*]. Is Quotient ring [Z(i)/E(i); , ] a field? zero-divistor!
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Example: Let R be a commutative ring, and H be an ideal of R
Example: Let R be a commutative ring, and H be an ideal of R. Prove that quotient ring R/H is an integral domain For any a,bR, if abH, then aH or bH. Proof: (1)If quotient ring R/H is an integral domain, then aH or bH when abH for any a,bR. (2)R is a commutative ring, and H be an ideal of R. If aH or bH when abH for any a,bR, then quotient ring R/H is an integral domain.
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Definition 33: Let be a ring homomorphism from ring [R;+,
Definition 33: Let be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. The kernel of is the set ker={xR|(x)=0S}. Theorem 6.36: Let be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. Then (1)[(R);+’,*’] is a subring of [S;+’,*’] (2)[ker;+,*] is an ideal of [R;+,*].
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Theorem 6.37(fundamental theorem of homomorphism for rings): Let be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. Then [R/ker;,] [(R);+’,*’]
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Exercise: 1. Let f : R→S be a ring homomorphism, with a subring A of R. Show that f(A) is a subring of S. 2. Let f: R→S be a ring homomorphism, with an ideal A of R. Does it follow that f(A) is an ideal of S? 3.Prove Theorem 6.36 4. Let f : R→S be a ring homomorphism, and S be an ideal of f (R). Prove: (1)f -1(S) an ideal of R, where f -1(S)={xR|f (x)S} (2)R/f -1(S) f (R)/S
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