Prove that a given function is a ring homomorphism and describe its kernel
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.7 Solution: Let $A,B \in R$ be arbitrary, with $A = \begin{bmatrix} a_1 & b_1 \\ 0…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.7 Solution: Let $A,B \in R$ be arbitrary, with $A = \begin{bmatrix} a_1 & b_1 \\ 0…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.5 Describe all ring homomorphisms from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Z}$. In each case, describe the kernel…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.4 Find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}/(30)$. In each case describe the kernel and the…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.3 Find all homomorphic images of $\mathbb{Z}$. Solution: Recall that every additive subgroup of $\mathbb{Z}$ is cyclic,…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.2 Show that the additive group $\mathbb{Z}$ acts on itself by $z \cdot a = z+a$ for…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.13 An element $x \in R$, $R$ a ring, is called nilpotent if $x^m = 0$ for some…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.15 Prove that a quotient of a divisible abelian group by any proper subgroup is also divisible.…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.5 Prove for all $n > 1$ that $\mathbb{Z}/(n)$ is not a group under multiplication of residue…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.4 Prove that multiplication of residue classes in $\mathbb{Z}/(n)$ is associative. (You may assume that it is…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.3 Prove that addition of residue classes in $\mathbb{Z}/(n)$ is associative. (You may assume that it is…