UIC Master’s Exam- Fall 2007 A2

Jeremy is preparing for the University of Illinois at Chicago Master’s Exam this coming Tuesday and so we have been studying together. His exam covers many topics, including algebra and analysis, and so you will be seeing several posts from the UIC exams in the future.

Problem Statement: Assume that D is an integral domain. (a) Suppose f(x), g(x)\in D[x] are non-zero.  Show that deg fg(x)=deg f(x)+deg g(x). (b) Show that the ideal I=(x+1, 7) of \mathbb{Z}[x] is not principal. (c) Show that \mathbb{Z}[x]/I is a field. How many elements are there in \mathbb{Z}[x]/I?

Solutions:

(a) Let f(x), g(x)\in D[x], then they will have the following form:

f(x)=a_nx^n+\dots+a_1x+a_0 and g(x)=b_mx^m+\dots+b_1x+b_0

Note that we are assuming that a_n\neq 0 and b_m\neq 0. Then it follows that fg(x)=a_nx^nb_mx^m+\dots+a_0b_0 where we know that a_nb_m\neq 0 since both a_n. b_m\neq 0 and D[x] is an integral domain (since D is an integral domain). So it follows that deg fg(x)=n+m=deg f(x)+ deg g(x). \Box

(b) Proof: Suppose that there is an f\in \mathbb{Z}[x] which generates the ideal I. So there must be some polynomial g\in \mathbb{Z}[x] such that 7=gf since 7\in I. Since \mathbb{Z}[x] is an integral domain we may apply part (a) from above and conclude that deg gf=deg g+ deg f. But since gf=7 it follows that deg gf=0 and since the degree must always be positive it follows that deg f=0. So f must be a constant, and it must also divide 7, so f is either 7 or 1.

If f=1 then (f)=\mathbb{Z}[x], but I\neq \mathbb{Z}[x], and so this is a contradiction.

Now suppose that f=7. Then in order for x+1 to be in (f)=I we need that 7(ax+b)=x+1, for some ax+b\in\mathbb{Z}[x]. Distributing and comparing like terms we see that 7b=1 and so it follows that b=\dfrac{1}{7}\not\in\mathbb{Z}, and so this is a contradiction.

Thus, there is no f\in\mathbb{Z}[x] which generates I and so I is not Principal. \Box

(c) Consider the homomorphism \phi:\mathbb{Z}[x]\rightarrow\mathbb{Z}_7 defined by \phi(n)=n\mod 7 and \phi(x)=-1. Then it follows that \phi(x+1)=\phi(x)+\phi(1)=-1+1=0 and \phi(7)=0 so I=\ker\phi. So it follows by the First Isomorphism Theorem that \mathbb{Z}[x]/I\cong\mathbb{Z}_7 which is a field with seven elements. \Box

Reflection: The part we had the most difficult time with was part (c). After seeing the solution it makes sense, but I don’t know how I would have thought to try and make a homomorphism to \mathbb{Z}_7 based on the information we were given. We were trying to show that I was maximal, but that didn’t lead us anywhere solid. Any suggestions anyone has on how to “see” that we should try to make a homomorphism to \mathbb{Z}_7 would be greatly appreciated. I think that I would eventually think to try to use FIT, but I don’t think I would know to make my homomorphism to \mathbb{Z}_7.

Advertisements
This entry was posted in Algebra, Field, Integral Domain, Math, Principal Ideal. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s