20140127

Please Login to access more options.

Today

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 $B=R$.

We'll have Nick, Laura, and Joe share their results related to last time.

We'll introduce the following problems, together with many related ideas.

1. Show that $\mathbb{Z}/\left<n\right>\approx \mathbb{Z}_n$ using the first isomorphism theorem. Make sure you prove that the kernel is what you claim it is. The division algorithm should be useful.
2. Practice long division.
3. Prove that $\mathbb{Z}[x]/\left<x^2+1\right>\approx \mathbb{Z}[i]$, by using the first isomorphism theorem.
4. Prove that $\mathbb{Z}[x]/\left<x^2+1\right>$ is an integral domain using linear algebra.
5. Why is it enough to show that no two polynomials can multiply together to give $(x^2+1)$, instead of showing that no two polynomials can multiply together to give $(x^2+1)p(x)$ for some polynomial $p(x)$?

For more problems, see AllProblems