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 given here

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.

exam 1 histogram

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.

Portfolios.  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.

Comments About 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 not zero.

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 that  0R1R ?   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 have 0R = 1R  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 = 1R  is true.