Math 441 - Abstract Algebra - Schedule

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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 21 (Adjoining The Square Root Of A Positive Integer To $\mathbb{Z}_p$)

Show that $\mathbb{Z}_7[\sqrt{3}] = \{a+b\sqrt{3}\mid a,b\in \mathbb{Z}_7\} $ is a field. Given a prime $p$ and positive integer $k$, determine conditions that are both necessary and sufficient for the ring $\mathbb{Z}_p[\sqrt{k}] = \{a+b\sqrt{k}\mid a,b\in \mathbb{Z}_p\}$ to be a field.

Click to see a hint.

Why is it enough to just determine conditions that cause $\mathbb{Z}_p[\sqrt{k}]$ to be an integral domain? You should be able to use systems of equations to reduce the problem to solving $$ \begin{bmatrix}a&-bk\\b&a\end{bmatrix} \begin{bmatrix}c\\d\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}. $$ When is the matrix invertible in $\mathbb{Z}_p$? This should give you your necessary and sufficient conditions.

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.

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.
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 $I$.

Problem 23 (The Ideal Generated By A Subset Is An Ideal)

Let $R$ be a ring and $S=\{a_1, a_2, \ldots a_n\}$ be subset of $R$ consisting of $n$ elements (where $n$ is some positive integer. Prove that the ideal generated by $S$ is actually an ideal of $R$.

For more problems, see AllProblems

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