Modern Algebra Fall 2016
Comments on the first exam
Scores for the first exam have been posted on blackboard. Exams
will be returned in class on Tuesday, 10/4/16. It is important
that you come to class to pick up your exam because you are assigned to
work with a group to produce a complete set of correct solutions that
will be due Tuesday, 10/11.
Exam solutions are a required component for the course portfolio.
Specific directions are
When you get your
exam, please double check that I have added up the points correctly and
also that the score on your paper is the same as the one posted on
blackboard. I try hard not to make errors totaling or recording
scores. But if an error has been made, I will want to correct it
as soon as possible.
The exam scores are shown in the histogram below. Scores have
been rounded up to the nearest integer, so if your scored an 84.5, it
will appear in the histogram as an 85.
The results are a little lower than I expected. Many students
missed probems that I expected to be routine. My conclusion at
this point is that the class as a whole has not mastered the material
so far as well as I would like, possibly because I have been moving
through the material too quickly. I am therefore planning to slow
down a bit, and have modified the class
schedule and assignment
sheet. As a consequence for our class on
Tuesday will not introduce any new
material. so there is no assigned reading for Tuesday.
Instead, students will have an opportunity to work together
on the missed items on the exam, and get started on completing their
exam solutions projects.
Partial Credit System.
I assign partial credit on each part of each problem by thinking
about how I would assign a letter grade to that particular part.
If I think an answer deserves an A I assign be 90% and 100% of the
points; if I think it deserves a B
I assign between 80% and 90% of the points. So if you got
less than 70% of the points, that means I thought that particular
answer was unsatisfactory. If you got less than 60% that means I
thought that particular answer was worthy of a failing grade.
This tends to make my partial credit assignments somewhat more generous
than many other math teachers. On the other hand, I rarely curve an exam. If you got a
total for this exam in the 80's, then your overall grade for the exam
is in the B range.
If you handed in a portfolio for review, please look at the review
sheet I put in the front. If I requested any changes to what you
have put together, please make those changes as soon as
possible, and before the next exam. The review sheet must stay in
the very front of your portfolio. If you wanted to have your
portfolio reviewed but forgot to bring it to the exam, you may bring it
to my office hours by Tuesday, 10/11 and I will review it
there while you wait.
Specific Exam Questions
Problem 2A. Part of the definition of b | a is the condition that b is not 0. This is
consistent with the division algorithm and with your prior
understanding of division: division by zero is not defined. If
your definition says only
that there must exist an integer for which a
= bt then the definition would be satisfied
with a = 0 and b = 0, because 0 = 0t for any integer t. But we do not want 0 | 0
to be true, so we have to include the restriction that b is not zero in the
definition of divides.
This issue arises again in the solution of problem 5. You are
supposed to reach the conclusion that a
| c . It is not enough
to produce an integer t for
which c = at. According to the
definition of divides you
also have to show that c is
Problem 2C. The definition of Integral
Domain includes the assumption that 0R ≠ 1R . What
does that mean, and why is it included? Isn't it obvious
≠ 1R ?
Actually, starting only from the axioms, we know that a ring with
identity must have an element that serves as an addititive identity,
and an element that serves as a multiplicative identity, but the axioms
do not state that they are different elements. So what if there
is a ring where they are the same? In that case we could write,
for any element a of the
ring, a =
1R a =
0R a = 0R , where the final equality is
an application of theorem 3.5(1). So the only way for a ring to
is for the ring to have a single element, and to have addition and
multiplication of that element with itself simply result in that same
element again. This set does satisfy the ring axioms, and is
therefore a commutative ring with identity and without zero divisors,
and for which 0R