~/Blog

Brandon Rozek

Photo of Brandon Rozek

PhD Student @ RPI studying Automated Reasoning in AI and Linux Enthusiast.

Useful Abstract Definitions

Published on

Updated on

4 minute reading time

Warning: This post has not been modified for over 2 years. For technical posts, make sure that it is still relevant.

Chapter 17

By a ring we mean a set $A$ with operations called addition and multiplication which satisfy the following axioms:

  • $A$ with addition alone is an abelian group
  • Multiplication is associative
  • Multiplication is distributive over addition

Since $\langle A, + \rangle$ is an abelian group, there is in $A$ a neutral element for addition. This is called the zero element. Also, every element has an additive inverse called its negative.

If multiplication in a ring $A$ is commutative then we say that $A$ is a commutative ring.

If a ring $A$ has a neutral element for multiplication then we say that the neutral element is the unity of $A$.

If $A$ is a ring with unity, there may be elements in $A$ which have a multiplicative inverse. Such elements are said to be invertible.

If $A$ is a commutative ring with unity in which every nonzero element is invertible, then $A$ is called a field.

In any ring, a nonzero element $a$ is called a divisor of zero if there is a nonzero element $b$ in a ring such that the product $ab$ or $ba$ is equal to zero.

An integral domain is defined to be a commutative ring with unity that has the cancellation property. Another way of saying this is that an integral domain is a commutative ring with unity and has no zero divisors.

Chapter 18

Let $A$ be a ring, and $B$ be a nonempty subset of $A$. $B$ is called a subring if it satisfies the following properties:

  • Closed with respect to addition
  • Closed with respect to negatives
  • Closed with respect to multiplication

Let $B$ be a subring of $A$. We call $B$ an ideal of $A$ if $xb, bx \in B$ for all $b \in B$ and $x \in A$.

The principle ideal generated by $a$, denoted $\langle a \rangle$ is the subring defined by fixing an element $a$ in a subring $B$ of $A$ and multiplying all elements of $A$ by $a$. $$ \langle a \rangle = { xa : x \in A } $$ A homomorphism from a ring $A$ to a ring $B$ is a function $f : A \to B$ satisfying the identities: $$ f(x_1 + x_2 ) = f(x_1) + f(x_2) \ f(x_1x_2) = f(x_1)f(x_2) $$ If there is a homomorphism from $A$ onto $B$, we call $B$ a homomorphic image of $A$.

If $f$ is a homomorphism from a ring $A$ to a ring $B$, the kernel of $f$ is the set of all the elements of $A$ which are carried by $f$ onto the zero element of $B$. In symbols, the kernel of $f$ is the set $$ K = {x \in A: f(x) = 0} $$ If $A$ and $b$ are rings, an isomorphism from $A$ to $B$ is a homomorphism which is a one-to-one correspondence from $A$ to $B$. In other words, it is injective and surjective homomorphism.

If there is an isomorphism from $A$ to $B$ we say that $A$ is isomorphic to $B$, and this fact is expressed by writing $A \cong B$.

Chapter 19

Let $A$ be a ring, and $J$ an ideal of $A$. For any element $a \in A$, the symbol $J + a$ denotes the set of all sums $j + a$ as $a$ remains fixed and $j$ ranges over $J$. That is, $$ J + a = {j + a : j \in J} $$ $J + a$ is called a coset of $J$ in $A$.

Now think of the set which contains all cosets of $J$ in $A$. This set is conventionally denoted by $A / J$ and reads $A$ mod $J$. Then, $A / J$ with coset addition and multiplication is a ring.

An ideal $J$ of a commutative ring is said to be a prime ideal if for any two elements $a$ and $b$ in the ring, $$ ab \in J \implies a \in J \text{ or } b \in J $$ Theorem: If $J$ is a prime ideal of a community ring with unity $A$, then the quotient ring $A / J$ is an integral domain.

An ideal $J$ of $A$ with $J \ne A$ is called a maximal ideal if there exists no proper ideal $K$ of $A$ such that $J \subseteq K$ with $J \ne K$.

Theorem: If $A$ is a commutative ring with unity, then $J$ is a maximal ideal of $A$ iff $A/J$ is a field.

Reply via Email Buy me a Coffee
Was this useful? Feel free to share: Hacker News Reddit Twitter

Published a response to this? :