20140124

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Today

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:

1. 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 $A$.
2. 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 $A_1=\left<56\right>.$

1. Find an ascending chain of ideals consisting of three ideals $A_1\subsetneq A_2\subsetneq A_3$ with $A_3=R$.
2. 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?
3. 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 $B=R$.

In class we looked at the factor ring $\mathbb{Z}[x]/\left<x^2\right>$, showed as a set that it was equal to $\{a_1x+a_0\mid a_1,a_0\in \mathbb{Z}\}$, and showed that it is not an integral domain as $$(x+\left<x^2\right>)(x+\left<x^2\right>) = (x^2+\left<x^2\right>)=(0+\left<x^2\right>).$$ We also showed that $\mathbb{Z}[x]/\left<x^2-1\right>$ was equal to the same set, however multiplication is different in this field, yet it is not an integral domain because $(x^2-1)=(x-1)(x+1)$. The next problem has you examine this question if we instead use the polynomial $x^2+1$.

Problem 39 (Some Polynomial Factor Rings)

Complete the following:

1. Use the first isomorphism theorem to prove that $\mathbb{Z}[x]/\left<x^2+1\right>$ is isomorphic to $\mathbb{Z}[i]$. From this we know that $\mathbb{Z}[x]/\left<x^2+1\right>$ is an integral domain and not a field.
2. Is $\mathbb{Q}[x]/\left<x^2+1\right>$ an integral domain? a field?
3. Is $\mathbb{R}[x]/\left<x^2+1\right>$ an integral domain? a field?
4. Is $\mathbb{C}[x]/\left<x^2+1\right>$ an integral domain? a field?

Problem 35 (When Is A Polynomial Factor Ring An Integral Domain)

Let $R=\mathbb{Z}[x]$. Suppose that $A=\left<p(x)\right>$ is a principle ideal generated by a single polynomial.

1. Find a polynomial $p(x)$ of degree 2 such that $R/A$ is not an integral domain. Then find one of degree 3 and degree 4.
2. Find a polynomial $p(x)$ of degree 3 such that $R/A$ is an integral domain.
3. If you know that $R/A$ is not an integral domain, what do you know about $p(x)$?
4. State a simple condition on $p(x)$ that is equivalent to $R/A$ being an integral domain.

The last two problems for today are designed as review problems, to help you practice with the definitions. We may not prove all of these in class. They should be very similar to problems we completed last semester.

Problem 36 (The Ideal Test)

Suppose that $A$ is a nonempty subset $A$ of a ring $R$. Prove that $A$ is an ideal of $R$ if

1. $a-b\in A$ whenever $a,b\in A$, and
2. $ra$ and $ar$ are in $A$ whenever $a\in A$ and $r\in R$.

In other words, prove that an ideal is a nonempty subset that is closed under subtraction and multiplication by an arbitrary ring element.

Problem 37 (Properties Of Ring Homomorphisms)

Let $\phi:R\to S$ be a ring homomorphism. Prove the following properties.

1. For any $r\in R$ and any positive integer $n$, $\phi(n\cdot r) = n\phi(r)$ and $\phi(r^n) = (\phi(r))^n$.
2. If $A$ is a subring of $R$, then $\phi (A) = \{\phi(a)|a\in A\}$ is a subring of $S$.
3. If $A$ is an ideal, then $\phi(A)$ is an ideal of $\phi(R)$.
4. If $B$ is an ideal of $S$, then $\phi^{-1}(B) = \{r\in R|\phi(r)\in B\}$ is an ideal of $R$.
5. If $R$ is commutative, then $\phi(R)$ is commutative.
6. If $R$ has a unity 1, if $S\neq \{0\}$, and if $\phi$ is onto, then $\phi(1)$ is the unity of $S$.
7. We know $\phi$ is an isomorphism if and only if $\phi$ is onto and $\ker\phi = \{r\in R|\phi(r)=0\} = \{0\}$.
8. If $\phi$ is an isomorphism from $R$ onto $S$, then $\phi^{-1}$ is an isomorphism from $S$ onto $R$.

For more problems, see AllProblems