Math 153: Course Summary
... multiplication commutes (a × b = b × a) and in which you can do division. Fields might be familiar from linear algebra. Here is a more formal definition of a field. Let R be a ring. Since R is an abelian group, there is an element e such that e + a = a + e = a for all a ∈ R. This element is typicall ...
... multiplication commutes (a × b = b × a) and in which you can do division. Fields might be familiar from linear algebra. Here is a more formal definition of a field. Let R be a ring. Since R is an abelian group, there is an element e such that e + a = a + e = a for all a ∈ R. This element is typicall ...
1 Basic definitions
... Note that, if n > 1, nZ is not a ring with unity. 2. As we saw in Modern Algebra I, Z/nZ is a finite commutative ring with unity for all positive integers n. In the case, the group of units is the multiplicative group (Z/nZ)∗ . The ring Z/nZ is a field ⇐⇒ n = p is a prime number. In this case, we wi ...
... Note that, if n > 1, nZ is not a ring with unity. 2. As we saw in Modern Algebra I, Z/nZ is a finite commutative ring with unity for all positive integers n. In the case, the group of units is the multiplicative group (Z/nZ)∗ . The ring Z/nZ is a field ⇐⇒ n = p is a prime number. In this case, we wi ...
16. Ring Homomorphisms and Ideals Definition 16.1. Let φ: R −→ S
... the set of functions from X to R. We have already seen that F forms a ring, under pointwise addition and multiplication. Let Y be a subset of X and let I be the set of those functions from X to R whose restriction to Y is zero. Then I is an ideal of F . Indeed I is clearly non-empty as the zero func ...
... the set of functions from X to R. We have already seen that F forms a ring, under pointwise addition and multiplication. Let Y be a subset of X and let I be the set of those functions from X to R whose restriction to Y is zero. Then I is an ideal of F . Indeed I is clearly non-empty as the zero func ...
Solutions for the Suggested Problems 1. Suppose that R and S are
... Solution. Let s = ϕ(e), which is an element in S. Since e is an idempotent of R, we have ee = e. Thus, we have ss = ϕ(e)ϕ(e) = ϕ(ee) = ϕ(e) = s . This proves that ss = s and hence that s is an idempotent in the ring S. Now suppose that R = Z and that ϕ : Z → S is a ring homomorphism. Note that 1 is ...
... Solution. Let s = ϕ(e), which is an element in S. Since e is an idempotent of R, we have ee = e. Thus, we have ss = ϕ(e)ϕ(e) = ϕ(ee) = ϕ(e) = s . This proves that ss = s and hence that s is an idempotent in the ring S. Now suppose that R = Z and that ϕ : Z → S is a ring homomorphism. Note that 1 is ...
3. Ring Homomorphisms and Ideals Definition 3.1. Let φ: R −→ S be
... set of functions from X to R. We have already seen that F forms a ring, under pointwise addition and multiplication. Let Y be a subset of X and let I be the set of those functions from X to R whose restriction to Y is zero. Then I is an ideal of F . Indeed I is clearly non-empty as the zero function ...
... set of functions from X to R. We have already seen that F forms a ring, under pointwise addition and multiplication. Let Y be a subset of X and let I be the set of those functions from X to R whose restriction to Y is zero. Then I is an ideal of F . Indeed I is clearly non-empty as the zero function ...
Solutions - NIU Math
... 17. Show that in any ring R the commutative law for addition is redundant, in the sense that it follows from the other axioms for a ring. Solution: The proof has to involve the distributive laws, because they provide the only connection between addition and multiplication in a ring. For any a, b ∈ R ...
... 17. Show that in any ring R the commutative law for addition is redundant, in the sense that it follows from the other axioms for a ring. Solution: The proof has to involve the distributive laws, because they provide the only connection between addition and multiplication in a ring. For any a, b ∈ R ...
Solutions
... (a) Suppose for the sake of contradiction that there exists an open cover {Uα }α∈A of X with no finite subcovers. Then we can find a sequence S of non-empty open T sets {Uα1 , · · · , Uαn , · · · } such that Uαi+1 * j6i Uαj for all i. Let Yi = j6i Uαc j . Then Y1 ⊃ Y2 ⊃ · · · does not satisfy the de ...
... (a) Suppose for the sake of contradiction that there exists an open cover {Uα }α∈A of X with no finite subcovers. Then we can find a sequence S of non-empty open T sets {Uα1 , · · · , Uαn , · · · } such that Uαi+1 * j6i Uαj for all i. Let Yi = j6i Uαc j . Then Y1 ⊃ Y2 ⊃ · · · does not satisfy the de ...
