## 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 $\mathbb{Z}_7$.