#32: Section 26

Bryana Glasco
Abstract Algebra
Due Date: 5/6/14

Section 26

The material in this section was not too bad. However, I have a couple of questions. Why is if R has unity 1, then ϕ(1) is unity for R’ as one of the conditions in theorem 26.3? Can we go over the definition of ideal? On page 243, the book says that ϕ[N] is an ideal of ϕ[R] although it does not need to be an ideal for R’. Why is that?

I find it interesting that for theorem 26.3 ϕ(1) is not necessarily unity for R’. Since ϕ(0)=0’ is the additive identity for R’, I would have assumed that this would be the same for unity. I also find it interesting that that ϕ[N] is an ideal of ϕ[R] but does not need to be an ideal of R’. It is really neat that we can apply many of the ideas that pertain to factor groups and apply those ideas to rings and have factor rings. 

#31: Rest of Section 23

Bryana Glasco
Abstract Algebra
Due Date: 5/1/14

Section 23: Irreducible Polynomials

When I read the rest of section 23, I understood the material pretty well since we did some problems that dealt with these concepts in class. However, could we go over the proofs for theorem 23.15 and corollary 23.17?

Under the heading Irreducible Polynomials, the book states, “We really are doing high school algebra in a more general setting” which is true in the rest of section 23. On page 214, we learn about a nonconstant polynomial f(x) ∈ F[x] being irreducible over F which means that the polynomial cannot be expressed as a product g(x)h(x) of two polynomials g(x) and h(x). In high school, we learned about this concept, and if a polynomial could not be expressed as a product g(x)h(x) then we knew that the polynomial could not be factorized. Now we use this idea to see if a polynomial can be factorized or not in a certain polynomial ring.  

#30: Section 23

Bryana Glasco
Abstract Algebra
Due Date: 4/29/14

Section 23: The Division Algorithm in F[x]

The material in the first part of section 23 is not too bad. Could we go over the proof of corollary 23.6?

I find it interesting that we can apply the ideas we learned in high school algebra to the ideas we have learned in this course. It is really neat that when we divide polynomials–taught in high school algebra–we are finding the zeros of f(x) ∈ F[x] in this course. I also find corollary 23.5 neat because in high school we knew that a polynomial of degree n had at most n zeros. For this course, we can apply this idea to a polynomial f(x) ∈ F[x] in a field F. 

#29: Rest of Section 22

Bryana Glasco
Abstract Algebra
Due Date: 4/27/14

Section 22: The New Approach

In the rest of section 22, there are only a few definitions and theorems that are not too difficult. Because of this, the material in the rest of section 22 was not too bad. However, could we go over example 22.9?

When we were finding a zero of a polynomial using mappings, I thought this concept was making simple problems more difficult. However, after I read definition 22.10, I saw that we are solving polynomial equations in the language of mappings. So, we are using a concept well known to us and applying it to another, more abstract, concept–we are seeing how solving polynomial equations can be manipulated by other concepts. At first, it was strange to me using mappings to solve polynomial equations but now it is interesting because I can see how one concept from one course can relate to other concepts in another course. 

#28: Section 22

Bryana Glasco
Abstract Algebra
Due Date: 4/23/14

Section 22: Polynomials in an Indeterminate

The material in the first half of section 22 was pretty hard to understand. Can we go over the idea of theorem 22.4? Could we go over more examples that illustrate the evaluation homomorphisms for field theory? What does the ring of polynomials in the n indeterminates x_i with coefficients in R mean?

I find example 22.3 interesting. In high school Algebra, we learn that (x + 1)² ≠ x² + 1 because using foil, or using the distribution property, we get (x + 1)² = x² +2x +1. However, in this course we use the ideas of rings and an indeterminate to help us solve polynomials. So in Z₂(x), (x + 1)² = x² +1. Though I find this interesting, it goes against everything I learned in high school and what I have taught to high school students at my fieldwork placements. When I began reading the example, I kept telling myself that (x + 1)² = x² +1 is not correct. But for this course, I will have to change my mindset to get the correct answers.