Please Login to access more options.


Today

« January 2014 »

Sun

Mon

Tue

Wed

Thu

Fri

Sat

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

Today we'll get more practice with rings, and introduce ring homomorphisms and factor rings. First, let's show that any time we start with an integral domain, then the ring of polynomials with coefficients in that integral domain will still be an integral domain. In other words, let's show that any time you multiply two polynomials and obtain the zero polynomial, then one of the polynomials must have been zero to start with.

Problem 16 (A Polynomial Ring Over An Integral Domain Is An Integral Domain)

If $D$ is an integral domain, show that $D[x]$, the set of all polynomials with coefficients in $D$, is an integral domain.

Just as we defined a group homomorphism to be a map from one group another that preserved the group operation, we'll define a ring homomorphism to preserve both ring structures.

Definition (Ring Homomorphism And Isomorphism)

A ring homomorphism $\phi$ from a ring $R$ to a ring $S$ is a mapping from $R$ to $S$ that preserves the two ring operations. In other words $\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$. A bijective ring homomorphism is called a ring isomorphism.

The kernel of a group homomorphism is the collection of group elements that map to zero. The same holds true for the kernel of a ring homomorphism.

Definition (Kernel Of A Ring Homomorphism)

Let $\phi:R\to S$ be a ring homomorphism. The kernel of $\phi$ is the set $$\ker \phi = \{r\in R\mid \phi(r)=0\}.$$

When studying groups, we invented the word normal subgroups to parallel the properties of kernel. A subgroup of a group is normal precisely when it is the kernel of a group homomorphism. We then used these properties to create factor groups. We need to do the exact same thing now with rings. We'll first look at some properties of the kernel, and then we'll turn our attention to factor rings.

Problem 18 (Kernels Are Closed Under Multiplication By Arbitrary Elements)

Let $\phi:R\to S$ be ring homomorphism with kernel $K$. Show the following:

  1. The kernel $K$ is a subring of $R$.
  2. If $r\in R$ and $k\in K$, then we have $rk\in K$ and $kr\in K$.

The two properties above are precisely the key for characterizing when we can create a factor ring. We use the word ideal to describe these subrings. The next problem shows that you can create a factor ring, provided you have an ideal.

Definition (Ideal)

A subring $A$ of $R$ is called an ideal if $ra\in A$ and $ar\in A$ whenever $a\in A$ and $r\in R$.

Problem 19 (Ideals Give Us Factor Rings)

Let $R$ be a ring and let $A$ be a subring of $R$. Show the following are equivalent.

  • The set of cosets $\{ r+A\mid r\in R\}$ is a ring under the operations $(s+A)+(t+A) = (s+t)+A$ and $(s+A)(t+A) = st+A$.
  • The subring $A$ is an ideal of $R$.
When you assume the first bullet above, you get to assume that the operations are binary operation. However, when you assume the second bullet, you must prove that the operations are binary operations. If we let $x,y\in R/A$, then there exists $s,t\in R$ such that $x=s+A$ and $y=t+A$. However, the choice of $s$ and $t$ is not unique. Suppose also that $x=s'+A$ and $y=t'+A$. The operation given for computing products then says that $xy=(s+A)(t+A) = st+A$, but it also says that $xy=(s'+A)(t'+A) = s't'+A$. However, if $st+A\neq s't'+A$, then we have a problem, as then $xy$ is not well-defined. This is what you must show. You must prove that if $A$ is an ideal, then you are guaranteed that $(s'+A)(t'+A) = s't'+A$.
Definition (Factor Ring)

Let $R$ be a ring and let $A$ be an ideal of $R$. The set of cosets $\{ r+A\mid r\in R\}$ together with the binary operations $(s+A)+(t+A) = (s+t)+A$ and $(s+A)(t+A) = st+A$ is called the factor ring of $R$ by $A$, or just the factor ring $R/A$.

We'll return to factor rings more next time. They happen to be a key tool in our future study, and we'll spend plenty of time becoming comfortable with them. For the rest of today, the next few problems have you practice with some of the definitions we've been building up over the last week, as well as adding in the characteristic of a field and the ideal generated by something.

Problem 20 (Finite Integral Domains Are Fields)

Suppose that $R$ is a finite integral domain. Prove that $R$ is a field.

Definition (Characteristic Of A Ring)

The characteristic of a ring $R$, written $\text{char } R$, is the smallest positive integer $n$ such that $n\cdot a = a+a+\cdots+a=0$ for all $a\in R$. If no such integer exists, then we say $R$ has characteristic zero.

Problem 22 (Characteristics And Rings With Unity)

Suppose that $R$ is a ring with unity.

  1. Show that the characteristic of $R$ is equal to the least positive integer $n$ such that $n\cdot 1=0$, provided the characteristic is not zero.
  2. Show that the characteristic of an integral domain is either zero or prime.
Definition (Principle Ideal Generated by $a$, or $\left<a\right>$)

Let $R$ be a commutative ring with unity and let $a\in R$. The set $\left<a\right> = \{ra| \ r\in R\}$ is an ideal of $R$ called the principle ideal generated by $a$.

Definition (Ideal Generated By $a_1, \ldots, a_n$, or $\left<a_1,a_2,\ldots,a_n\right>$)

Let $R$ be a commutative ring with unity, and let $a_1, \ldots, a_n \in R$. The set $I=\left<a_1,a_2,\ldots,a_n\right> = \{r_1a_1+\cdots +r_na_n|r_i\in R \}$ is called the ideal generated by $a_1, \ldots, a_n$. Any other ideal that contains $a_1, \ldots, a_n$ must contain $I$.

Was moved to Problem.TheIdealGeneratedByASubsetIsAnIdeal - got rid of the extra "By" in the name.

For more problems, see AllProblems