Flashcards
Rings: Ideal Homomorphisms
Cards in this deck (20)
- 1If a and b are in the kernel of f, i.e. f(a)=f(b)=0 , and r is any element of R, thena) f(a-b) = f(a)-f(b)=0
b) f(ar)=f(a)f(r)=0
c) f(ra)=f(r)f(a)=0
\(\Rightarrow\) ker(f) is an ideal - 2The homomorphism of an ideal is one-to-one ifThe kernel consists only of the identity element
- 3Prove: If the only proper ideal of R is the trivial ideal {0}, then if f is a homomorphism from R to S, and it does not map all elements of R to the identity in S, then it is injective.The kernel of the homomorphism must be an ideal, and since the only ideals are R and the trivial ideal, one of these two must be the kernel. However, since not all elements of R map to the identity of S, R is not the kernel, so the trivial ideal must be
- 4If I is an ideal of a ring R and a ∈ R then a coset of I is a set of the forma + I = {a + s | s I }
- 5The set of all cosets is denoted byR/I
- 6the cosets partition the ring intodisjoint subsets
- 7The element a is called acoset representative of the coset
- 8The ideal I is itself a coset :0 + I
- 9Prove: If I is an ideal of a ring R, the set R/I is a ring under the operations (a + I) + (b + I) = (a + b) + I and (a + I) . (b + I) = (ab) + IWe need to check that the operations are "well-defined". That is if \(a_1\) and \(a_2\) are representatives of the same coset and \(b_1\) and \(b_2\) represent the same coset then \(a_1+b_1\) and \(a_2+b_2\) represent the same coset and so do \(a_1b_1\) and \(a_2b_2\).
We have \(a_1-a_2I\) and \(b_1-b_2I\) and so adding these shows that \((a_1 + b_1) - (a_2 + b_2) ∈ I\) and so these do represent the same coset.
Similarly, for the product, observe that \(a_1b_1 - a_2b_2 = (a_1 - a_2)b_1 + a_2(b_1 - b_2)\) and the result follows from the properties of the ideal.
Once you know that the operations are well-defined the ring axioms follow easily.
Note that the zero of the factor ring is the coset 0 + I or the ideal I itself. - 10Prove: If f :R→ S is a ring homomorphism the factor ring R/ker(f) is isomorphic to im(f)Using the factor theorem, we can find a homomorphism from R/Ker(f) to S, and since the kernel is the same as the ideal used in forming the quotient group, and since the f is an epimorphism over its image, this homomorphism is an isomorphism.
- 11Let I and J be ideals of a ring R. Then I + J and I ∩ J are also ideals and the factor rings (I + J)/J and I/(I ∩ J) are isomorphicTo prove the result, Define θ : I→ (I + J)/J by i ↦ i + J.
Since every coset in (I + J)/J has a representative of this form θ is onto.
If i ∈ ker(θ) then i + J = J and so i ∈ J and hence is in I ⊆ J.
The result now follows from the First Isomorphism Theorem. - 12Let I and J be ideals of a ring R with I ⊆ J. Then J/I is an ideal of R/I and (R/I)/(J/I) \(\cong\) R/JIt is easy to verify that J/I is an ideal of R/I.
Define a map θ : R/I→ R/J by a + I ↦ a + J for any coset a + I of I in R. This is clearly onto.
An element a + I is in the kernel of θ if a + I = J which (since I ⊆ J) will only happen if a ∈ J. Thus ker(θ) = J/I ⊆ R/I and the result follows from the the First Isomorphism Theorem. - 13Let I be an ideal of the commutative ring R:a) The natural projection mapping f: R \(\to\) R/I defined by f(a) = a+I for all a in R is a ring homomorphism, and ker(f) = I
b) There is a one-to-one correspondence between the ideals of R/I and the ideals of R that contain I - 14Let I be a proper ideal of the commutative ring R. Then I is said to be a prime ideal of R iffor all a,b in R it is true that if ab is in I then a is in I or b is in I
- 15The ideal I is said to be a maximal ideal of R iffor all ideals J of R such that I \(\subseteq\) J \(\subseteq\) R, either J = I or J = R
- 16Let I be a proper ideal of the commutative ring R with identity:a) The factor ring R/I is a field if and only if I is a maximal ideal of R
b) The factor ring R/I is a integral domain if and only if I is a prime ideal of R
c) If I is maximal, then it is a prime ideal - 17Prove: Every nonzero prime ideal of a principal ideal domain is maximalhttps://www.youtube.com/watch?v=r7HsgdP1l8E
- 18Prove: Every non-invertible element of a ring is contained in some maximal idealIf a is a noninvertible element, then the generated ideal (a) is not the whole ring. If it were, then 1∈(a)1, implying ab=1 for some b, a contradiction. As a proper ideal, (a), and hence a, must then be contained in a maximal ideal.
- 19Prove: A commutative ring R is an Integral Domain if and only if (0) is a prime idealThis follows simply because \(x=0 \iff x \in (0) \)
- 20Prove: A commutative ring R is a Field if and only if (0) is a maximal ideal (that is there are no proper ideals)We only need to show that every element \( 0 \neq x \in R \) is invertible.
Since every non-invertible element of a ring is contained in some maximal ideal, x is contained in some (proper) maximal ideal, a contradiction
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- Undergrad Math Flashcards on Rings: Ideal Homomorphisms, created by Vincent Tembo on 26/07/2016.
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