Math 441 - Abstract Algebra - Schedule

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The first few examples today are designed to help us get a better handle on factor rings. Recall that $\left<a\right>$ is called the principle ideal generated by $a$.

Problem 24 (A Matrix Factor Ring)

Consider the set of two by two matrices $R=M_2(\mathbb{Z})$ with integer coefficients. Let $A$ the set of matrices whose coefficients are multiples of 3.

Show that $A$ is an ideal of $R$.
Find three different elements of $R$ that are in the coset $\begin{bmatrix}5&-2\\7&9\end{bmatrix}+A$.
How many distinct elements are in this factor ring? Justify your answer.
Challenge: This factor ring is isomorphic to another matrix ring we have already seen. What is this ring? You don't need to prove your answer.

Problem 25 (A Factor Ring Of The Gaussian Integers)

Let $R=\mathbb{Z}[i]$, $a=3-i$, and let $A=\left<a\right>$. We would like to examine the factor ring $R/A$, determine how many elements are in this factor ring, and find a simple ring to which $\mathbb{Z}[i]/\left<3-i\right>$ is isomorphic.

Why do we know $3+A=i+A$? Use this to explain why $10+A=0+A$.
Explain why every element of $R/A$ must be of the form $k+A$ where $k\in \{0,1,2,\ldots 9\}$, so there are at most 10 elements.
Suppose the integer $k$ is an element of $A = \left<a\right> = \{ra\mid r\in R\}$. In other words, suppose that there exists an element $r\in R$ such that $k+0i = ra = (c+di)(3-i)$. Use this to show that $k$ must be a multiple of 10.
State a much simpler ring to which $R/A$ is isomorphic, you don't need to prove your answer.

After completing these exercises (or if you get stuck), I strongly recommend that you take a few minutes and read the examples on pages 263-266. They summarize many of the key ideas we need to focus on with factor rings.

Problem 26 (Factor Rings Of $\mathbb{Z}$ And $\mathbb{Z}[x]$)

Answer the following questions as they pertain to the integral domains $\mathbb{Z}$ and $\mathbb{Z}[x]$. You are welcome to rapidly make claims about factor rings, without proof, as we have seen many of these as factor groups in the past.

We know that the ideals of $\mathbb{Z}$ are of the form $n\mathbb{Z} = \left< n\right>$, the set of multiples of $n$.
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ an integral domain?
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ a field?
Now consider the ring of polynomials $\mathbb{Z}[x]$. The ideal $\left<n\right>$ is now the set of polynomials whose coefficients are multiples of $n$. It should not be a surprise that $\mathbb{Z}[x]/\left<n\right> \approx \mathbb{Z}_n[x]$.
- For which $n$ is the factor ring $\mathbb{Z}[x]/\left<n\right>$ an integral domain?
- Show that the factor ring $\mathbb{Z}[x]/\left<n\right>$ is never a field, regardless of which $n$ you pick.
We can look at other ideals of $\mathbb{Z}[x]$.
- Show that $\mathbb{Z}[x]/\left<x\right>$ is an integral domain, but not a field.
- Show that $\mathbb{Z}[x]/\left<3,x\right>$ is a field.

As seen in the previous exercise, sometimes when we create a factor ring, we obtain an integral domain, and sometimes we obtain a field. We would like some words to describe ideals for which the corresponding factor ring is an integral domain, or a field. It would be nice to have a characterization that we could check in the ideal itself, without having to actually look at the factor ring.

Problem 27 ($R/A$ Is An Integral Domain Iff $A$ is prime)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is an integral domain.
If $a,b\in R$ and $ab\in A$ then $a\in A$ or $b\in A$. (We say that $A$ is a prime ideal).

Why do we call an ideal a prime ideal if the second condition above is satisfied? Recall in the integers that if $p$ is a prime, then if $ab$ is a multiple of $p$, then either $a$ or $b$ must be a multiple of $p$. So $ab\in \left<p\right>$ forces either $a\in \left<p\right>$ or $b\in \left<p\right>$. This condition forces the number to be prime, so we extend the notation of prime numbers to prime ideals. Any time a product is divisible by a prime, one of the two factors must be divisible by the prime. We'll extend this to say that any time a product is in a prime ideal, one of the factors must be in the prime ideal.

We now have a simple way to check if $R/A$ is an integral domain. We just have to make sure that if $ab\in A$, then either $a\in A$ or $b\in A$. To obtain a field, we need even more. We need to make sure that if $b\notin A$, then there exists $c\notin A$ such that $bc\in 1+A$, or that $1-bc\in A$, or equivalently $1\in bc+A$. So given any element $b\notin A$, if we can show that $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, then we'd win. Let's first show that this set $B$ is actually an ideal, and then show that $1\in B$ forces $B=R$.

Problem 29 (Obtaining A New Ideal By Adding One Element)

Let $A$ be an ideal of the commutative ring $R$ with unity. We'd like to increase the size of $A$ by adding in a single element $b$. Show the following:

Pick $b\in R$. The set $B=\{bc+a\mid c\in R,a\in A\}$ is an ideal of $R$ that contains both the element $b$ and subset $A$.
We know $1\in A$ if and only if $A=R$.

The set $B$ described above is the smallest ideal of $R$ that contains $b$ and every element in $A$, so we could call it the ideal generated by $b$ and $A$.

Given a proper ideal $A_1$, we can use the idea above to create an ideal $A_2$ that properly contains $A_1$ (add one element not in $A_1$). If this ideal is not the entire ring, we could then continue this process, possibly indefinitely, to obtain a sequence of ideals that properly contain the previous. Such a sequence $A_1\subset A_2\subset A_3\subset \cdots \subset A_n\subset \cdots$ is called an ascending chain of ideals. Once one of these ideals contains the number 1, the sequence terminates. Emmy Noether used this idea to create what we now call Noetherian domains, an integral domain in which any ascending chain of ideals must be finite in length (so the process must stop, no matter what ideal $A$ you start with). We'll come back to Noetherian domains later. See page 275 for more information about Emmy Noether.

Problem 30 (An Ascending Chain Of Ideals In The Integers)

Let $R=\mathbb{Z}$. Consider the ideal $A_1=\left<56\right>.$

Find an ascending chain of ideals consisting of three ideals $A_1\subsetneq A_2\subsetneq A_3$ with $A_3=R$.
What is the largest $n$ such that $A_1\subsetneq A_2\subsetneq A_3\subsetneq \cdots\subsetneq A_n=R$ is an ascending chain of ideals? How many such chains are there of this length?
If $n\in \mathbb{Z}$, describe a process we could use to find a longest possible ascending chain of ideals with $\left<n\right>\subsetneq A_2\subsetneq \cdots\subsetneq A_n=\mathbb{Z}$.

The two facts from the problem Obtaining A New Ideal By Adding One Element will help us characterize the properties of an ideal $A$ that will result in $R/A$ being a field. Remember, the following:

We need to make sure that if $b\notin A$, then there exists $c\in R$ such that $bc\in 1+A$, or equivalently $1\in bc+A$.
So given any element $b\notin A$, if we can show that the ideal $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, which means $B=R$, then we'd win.
This is basically the same as showing that given any ideal $B$ trapped between $A$ and $R$, so $A\subset B\subset R$, then if $B$ is not equal to $A$, it must equal $R$. We call an ideal such as $A$ a maximal ideal.

The next problem has you carefully show that $R/A$ is a field if and only if $A$ is a maximal ideal.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

The previous problems give us the following definitions of prime and maximal ideals. Basically, these are now characteristics of an ideal that we can check to determine if $R/A$ is an integral domain or a field, without having to ever consider elements of the factor ring.

Definition (Prime Ideal And Maximal Ideal)

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a,b\in R$ and $ab\in A$ imply $a\in A$ or $b\in A$.
A maximal ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that, whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then $B=A$ or $B=R$.

Let's end today with another example of some interesting things that can happen in a ring.

Definition (Idempotent And Nilpotent)

Let $R$ be a ring.

We say that $a$ is an idempotent of $R$ if $a^2=a$.
We say that $a$ is nilpotent if $a^n=0$ for some integer $n$.

Problem 32 (Finding Idempotent And Nilpotent Elements In A Matrix Ring)

Complete the following:

In the ring $M_2(\mathbb{Z})$, the matrices $\begin{bmatrix}0&0\\0&0\end{bmatrix}$, $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $\begin{bmatrix}0&0\\0&1\end{bmatrix}$, are idempotents. Find two others.
Show that $\begin{bmatrix}0&t\\0&0\end{bmatrix}$ is nilpotent in $M_2(\mathbb{Z})$ and that $\begin{bmatrix}0&t&0\\0&0&t\\0&0&0\end{bmatrix}$ is nilpotent in $M_3(\mathbb{Z})$, where $t\in\mathbb{Z}$.
Let $X=\begin{bmatrix}0&t&0&0&0\\0&0&t&0&0\\0&0&0&t&0\\0&0&0&0&t\\0&0&0&0&0\end{bmatrix}$. Compute $X^k$ for each positive integer $k$. What patterns do you see? Is $X$ a nilpotent element of $M_5(\mathbb{Z})$?

The matrices above are central to finding the Jordan form of a matrix, and then using the Jordan form to compute the matrix exponential. Come see me out of class if you would like more information on this topic, or consider taking Math 495 next fall, where the topic will be linear algebra.

For more problems, see AllProblems

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