1.
Definition: Give the definition of an ideal in a ring \(R\text{.}\)
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Worksheet Weekly Practice 8
Instructions: You may type up or handwrite your work, but it must be neat, professional, and organized and it must be saved as a PDF file and uploaded to the appropriate Gradescope assignment. Use a scanner or scanning app to convert handwritten work on paper to PDF. I encourage you to type your work using the provided template.
All tasks below must have a complete solution that represents a good-faith attempt at being right to receive engagement credits. If your submission is complete and turned in on time, you will receive full engagement credit for the assignment. All other submissions will receive zero engagement credit. Read the guidelines at Grading Specifications carefully.
To abide by the class academic honesty policy, your work should represent your own understanding in your own words. If you work with other students, you must clearly indicate who you worked with in your submission. The same is true for using tools like generative AI although I strongly discourage you from using such tools since you need to build your own understanding here to do well on exams.
True/False, Multiple Choice, & Fill-In.
For these problems a justification is not required for credit, but it may be useful for your own understanding to include one. True/False problems should be marked True if the statement is always true, and False otherwise. Multiple choice problems may have more than one correct answer if that is indicated in the problem statement; be sure to select all that apply. Fill-in problems require a short answer such as a number, word, or phrase.
2.
Fill-In: The characteristic of an integral domain is either or
Solution.
The characteristic of an integral domain is either 0 or a prime number.
3.
Fill-In: A proper ideal \(I\) is called if \(ab\in I\) implies \(a\in I\) or \(b\in I\text{.}\)
Solution.
Such an ideal is called prime.
Short Response.
Your responses to these questions should be complete solutions with justifications, as per the Grading Specifications.
4.
Find all units, zero-divisors, idempotents, and nilpotent elements in \(\Z_3\times \Z_6\text{.}\)
Solution.
The units of the ring are the elements \((a,b)\) such that \(a\in \Z_3\) is a unit and \(b\in \Z_6\) is a unit. This is true when \(a\in \{1,2\}\) and \(b\in \{1,5\}\text{.}\) Thus the units are
\begin{equation*}
(1,1),(1,5),(2,1),(2,5)\text{.}
\end{equation*}
The zero-divisors of the ring either have the form \((a,x)\) or \((0,b)\) where \(a\in \Z_3\text{,}\) \(b\in\Z_6\text{,}\) and \(x\in \Z_6\) is either 0 or a zero-divisor, since \(\Z_3\) has no zero-divisors. So the zero-divisors are
\begin{gather*}
(1,0),(2,0),(1,2),(2,2),(1,3),(2,3),(1,4),(2,4),\\
(0,1),(0,2),(0,3),(0,4),(0,5)\text{.}
\end{gather*}
The idempotents of the ring are the elements \((a,b)\) such that \(a\in \Z_3\) is an idempotent and \(b\in \Z_6\) is an idempotent. This is true when \(a\in \{0,1\}\) and \(b\in \{0,1,3,4\}\text{.}\) Thus the idempotents are
\begin{equation*}
(0,0),(0,1),(0,3),(0,4),(1,0),(1,1),(1,3),(1,4)\text{.}
\end{equation*}
This ring has no nilpotent elements since \(\Z_3\) has no nilpotent elements and \(\Z_6\) has no nilpotent elements.
5.
In \(\Z_7\text{,}\) give a reasonable interpretation for the expressions \(1/2, -2/3, \sqrt{-3}, -1/6\text{.}\)
Solution.
We have \(1/2\cdot 2= 1\text{,}\) so \(1/2=4\) in \(\Z_7\text{.}\) Similarly, \(1/3\cdot 3 = 1\text{,}\) so \(1/3=5\text{,}\) and thus
\begin{equation*}
-2/3=-2(1/3)=-2(5)=5(5)=4
\end{equation*}
in \(\Z_7.\) Next
\begin{equation*}
\sqrt{-3}\sqrt{-3}=-3=4\text{,}
\end{equation*}
so \(\sqrt{-3}\) could reasonably be interpreted as \(2\) or \(5\) in \(\Z_7\text{.}\) Finally, we have
\begin{equation*}
(-1/6)(-6)=1
\end{equation*}
so \(-1/6=1\) in \(\Z_7\text{.}\)
6.
Find a subring of \(\Z\times \Z\) that is not an ideal of \(\Z\times \Z\text{.}\)
Solution.
Let \(S=\{(n,n)\mid n \in \Z\}\text{.}\) Then \(S\) is nonempty and for \(n,m\in \Z\) we have
\begin{align*}
(n,n)-(m,m) \amp =(n-m,n-m) \in S\\
(n,n)(m,m) \amp = (nm,nm) \in S
\end{align*}
so by the Subring Test, \(S\) is a subring of \(\Z\times \Z\text{.}\) However, \(S\) is not an ideal of \(\Z\times \Z\) since \((1,0)(1,1)=(1,0)\notin S\text{.}\)
7.
How many elements are in \(\Z[i]/\langle 3+ i\rangle\text{?}\) Give reasons for your answer.
Solution.
We claim this quotient has 10 elements. Let \(I=\ideal{3+i}\text{.}\) First, since \(3+i+I=0+I\text{,}\) we have \(3\equiv-i\) as coset representatives and so also \(9\equiv -1\) and \(10\equiv 0\text{.}\) So every coset has a representative \(a\in \{0,1,\dots, 9\}\text{.}\) On the other hand, if
\begin{equation*}
a+I=b+I
\end{equation*}
for \(a\lt b \in \{0,1,\dots,9\}\text{,}\) then \(b-a\in I\) and so \(3+i\) divides \(b-a\text{.}\) But if \(b-a=(3+i)(c+di)\) then we have simultaneously \(3c-d=b-a\) and \(c+3d=0\text{.}\) Substituting \(c=-3d\) into the first equation gives \(-9d-d=b-a\) or \(-10d=b-a\text{.}\) Since \(b-a\in \{1,2,\dots,9\}\text{,}\) this is impossible. Thus all cosets have distinct representatives in \(\{0,1,\dots,9\}\) and so there are exactly 10 elements in the quotient.
8.
Let \(R=\Z_8\times \Z_{30}\text{.}\) Find all maximal ideals of \(R\) and for each maximal ideal \(I\) identify the size of the field \(R/I\text{.}\)
Solution.
An ideal \(I\) of \(R\) is maximal if and only if \(R/I\) is a field. In particular, \(R/I\) must be an integral domain, so it cannot have cosets with representatives of both forms \((a,0)\) and \((0,b)\) for \(a\in \Z_8\) and \(b\in \Z_{30}\text{.}\) Hence, a maximal ideal must be of the form \(I=J\times \Z_{30}\) or \(I=\Z_8\times K\) for some maximal ideal \(J\subseteq \Z_8\) or \(K\subseteq \Z_{30}\text{.}\) The maximal ideal of \(\Z_8\) is only \(\ideal{2}\) and the maximal ideals of \(\Z_{30}\) are \(\ideal{2},\ideal{3},\ideal{5}\text{.}\) Thus the maximal ideals of \(R\) are
\begin{equation*}
\ideal{2}\times \Z_{30}, \Z_8\times \ideal{2}, \Z_8\times \ideal{3}, \Z_8\times \ideal{5}\text{.}
\end{equation*}
