Suzie-Q

Posted on August 16, 2011 by

Problem Statement: Let I be an integral domain. Show that I is a field if and only if I has no nontrivial ideals.

Proof: First assume that I is a field and let J be an ideal of I. If 1\in J then it follows that J=I since x1=x\in J for every x\in I. Suppose then that 1\not\in J and let j\in J, then there exists a j^{-1}\in I since by assumption I is a field. But since J is an ideal it follows that jj^{-1}\in J, but since 1\not\in J this is only possible if j=0 and so it follows that J=\{0\}. Thus the only ideals of I are the trivial ideals.

Now assume that the only ideals of I are J=\{0\} and I. Fix a non-zero x\in I and define \phi_{x}:I\rightarrow I by \phi_{x}(y)=xy. \phi_{x} is a homomorphism with \ker\phi_{x}=J=\{0\} since I is an integral domain and so we have no zero divisors. So we have that \phi_{x} is one to one and thus \phi_{x} is onto since it maps I to itself. Now, since 1\in I it follows that there exists y\in I such that \phi_{x}(y)=1\in I. By definition of \phi_{x}(y) this implies that xy=1 and so y=x^{-1}. Thus every non-zero element in I is a unit and so I is a field.

\Box

Reflection: This technique of using a homomorphism to find an inverse is a really helpful tool. This only works since we can force \ker\phi_{x}=\{0\} which is due to the fact that I is an integral domain.

Posted in Algebra, Field, Homomorphism, Ideal, Integral Domain | 1 Comment

UIC Master’s Exam- Spring 2007 A2

Posted on August 14, 2011 by

Problem Statement: (a) State the division algorithm for \mathbb{Q}[x] (b) Show that \mathbb{Q}[x] is a principle ideal domain.

Solutions:

(a) The Division Algorithm: If f(x), q(x)\in\mathbb{Q}[x] with \deg q(x)\leq\deg f(x) then there exist q(x), r(x)\in\mathbb{Q}[x] such that f(x)=g(x)q(x)+r(x) where \deg r(x)<\deg g(x).

(b) Proof: Let I be an ideal of \mathbb{Q}[x]. If I is the trivial ideal then it is principal. So suppose that I is not the trivial ideal. Then there is some non-zero polynomial f\in I such that f has the least degree of any polynomial in I.

Let g\in I, then by the division algorithm we may write

g(x)=f(x)q(x)+r(x)

where q(x), r(x)\in\mathbb{Q}[x] and \deg r(x)<\deg f(x). Now, since I is an ideal and f, g,q\in I it follows that g(x)-f(x)q(x)\in I, so r\in I. But \deg r(x)<\deg f(x) which is the least possible degree in I, so it must be that r(x)=0. And so we have shown that there is a q(x)\in I for every g(x)\in I such that g(x)=f(x)q(x), thus, I=(f(x)) and so \mathbb{Q}[x] is a Principal Ideal Domain.

\Box

Reflection: They key in this proof is that we chose our f to have the least possible degree. This makes the polynomial act kind of like a gcd does for a list of numbers.

Posted in Algebra, Principal Ideal, Principal Ideal Domain | Leave a comment

UIC Master’s Exam- Fall 2007 R2

Posted on August 14, 2011 by

Problem Statement: f is continuous on \mathbb{R}. Given that

g(x)=\dfrac{1}{2\delta}\int\limits_{-\delta}^{\delta}f(x+t)dt

(a) prove g has continuous derivative on \mathbb{R} (Hint: start with the change of variable u=x+t). (b) Given [a,b] and \varepsilon>0 show that there exists a \delta>0 such that \left|g(x)-f(x)\right|<\varepsilon for every x\in [a,b].

Solutions:

(a)Proof: First let us start by making the change of variable suggested in the problem statement.

Let u=x+t, then du=1dt

g(x)=\dfrac{1}{2\delta}\int\limits_{x-\delta}^{x+\delta}f(u)du

Now we may apply the Fundamental Theorem of Calculus and obtain that

g'(x)=\dfrac{1}{2\delta}\left(f(x+\delta)-f(x-\delta)\right)

which is continuous since f is continuous on \mathbb{R}

\Box

(b)Proof: We are given [a,b] and \varepsilon>0. Since f is continuous on a compact set it is uniformly continuous, thus, for our given \varepsilon we may find a \delta>0 such that \left|f(x)-f(y)\right|<\varepsilon whenever \left|x-y\right|<\delta. Now consider the following:

\left|g(x)-f(x)\right|=\left|\dfrac{1}{2\delta}\int\limits_{-\delta}^{\delta}f(x+t)dt-f(x)\right|

=\left|\dfrac{1}{2\delta}\int\limits_{-\delta}^{\delta}\left(f(x+t)-f(x)\right)dt\right|

\leq \dfrac{1}{2\delta}\int\limits_{-\delta}^{\delta}\left|f(x+t)-f(x)\right|dt

And so, for \left|t\right|<\delta it follows that \left|f(x+t)-f(x)\right|<\varepsilon. This allows us to simplify further:

<\dfrac{1}{2\delta}\int\limits_{-\delta}^{\delta}\varepsilon dt

=\dfrac{1}{2\delta}\left(\varepsilon\delta-(-\varepsilon\delta)\right)

=\varepsilon

\Box

Reflection: Always try the easiest method first!!! This is a lesson I keep “learning” but I don’t really seem to be learning it. Part (b) I attempted to do in a very strange way. Hopefully now I have really learned, try the easier method first!! If it doesn’t work then at least you’ll have gained some insight into why it isn’t working, and that can help you with another attempt.

Posted in Analysis, Continuity, Differentiable, Fundamental Theorem of Calculus, Math | Leave a comment

UIC Master’s Exam- Fall 2007 R1

Posted on August 14, 2011 by

Problem Statement: Consider the power series

\sum\limits_{n=1}^{\infty}\dfrac{1}{2^nn}x^n

(a) For which values of x does it converge absolutely? Conditionally? (b) Show that on [-1,1] the convergence is uniform.

Solutions:

(a) Claim: the series converges absolutely on (-2,2) and conditionally on [-2,2).

Consider x\in(-2,2), then it follows that \left|\dfrac{x}{2}\right|<1 and so we may find a rational r\in(\dfrac{x}{2},1). Then it follows that \left|\left(\dfrac{x}{2}\right)^n\dfrac{1}{n}\right|\leq\left|\left(\dfrac{x}{2}\right)^n\right| for every n\in\mathbb{N}. By the geometric series test \sum\limits_{n=1}^{\infty}\left|\left(\dfrac{x}{2}\right)^n\right| converges and so by the comparison test \sum\limits_{n=1}^{\infty}\left|\left(\dfrac{x}{2}\right)^n\dfrac{1}{n}\right| converges. Thus, the series converges absolutely on (-2,2).

When x=-2 then the series becomes \sum\limits_{n=1}^{\infty}(-1)^n\dfrac{1}{n}, the alternating harmonic series, which we know converges by the AST. But when x=2 the series is simply the harmonic series which we know diverges. Thus, the series converges conditionally on [-2,2).

(b) Proof: Consider the series above for x\in[-1,1] it follows that

\left|\dfrac{1}{2^nn}x^n\right|\leq\left|\dfrac{1}{2^nn}\right|\leq\left|\dfrac{1}{2^n}\right|

for all n\in\mathbb{N} and for all x\in[-1,1]

We know that \sum\limits_{n=1}^{\infty}\dfrac{1}{2^n} converges by the p-series test (since \dfrac{1}{2}<1).  And so applying the M-Test we may conclude that \sum\limits_{n=1}^{\infty}\dfrac{1}{2^nn}x^n converges uniformly on [-1,1].

\Box

Reflection: This was a great reminder problem. Whenever I see uniform convergence my first thought is usually to try using the definition, which can be a real pain in the butt. This problem reminded me that when you’re dealing with uniform convergence you should try to use the M-Test if you can because it is usually much much simpler than trying to go straight from the definition. Try the easy way first!!!

Posted in Absolute Convergence, Analysis, Conditional Convergence, M-Test, Math, Series, Series of Functions, Uniform Convergence | Leave a comment

UIC Master’s Exam- Fall 2007 A3

Posted on August 14, 2011 by

Problem Statement: Show that a finite integral domain is a field.

Proof: Fix a\in D, a\neq 0 and consider the function \phi(x)=ax for all x\in D. We wish to show that \phi is injective, and thus a has an inverse.

Suppose that \phi(x)=\phi(y), then ax=ay which implies that a(x-y)=0. Since we are in an integral domain we know there are no zero divisors, so either a=0 or x-y=0. We know that a\neq 0 so it must be that x-y=0 which implies that x=y and so \phi is injective. Furthermore, since \phi maps D to D, and D is finite, we know that \phi is onto D. So, there is some x\in D such that \phi(x)=ax=e, thus, x=a^{-1}.

Thus, D is a field.

\Box

Reflection: This is a classic qual problem. It’s very straightforward and there are many ways to approach the proof. The key rests in the fact that D is finite, otherwise we wouldn’t be able to say that \phi is onto even though it maps D to D and is one-to-one.

Posted in Algebra, Field, Integral Domain, Math | 1 Comment

UIC Master’s Exam- Fall 2007 A2

Posted on August 14, 2011 by

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.

Posted in Algebra, Field, Integral Domain, Math, Principal Ideal | Leave a comment

Gallian- Ch.9 #49

Posted on August 12, 2011 by

Problem Statement: If H\lhd G, prove that C(H)\lhd G.

Proof: Recall that C(H) is the set of all elements of G which commute with all elements of H. Let g\in G, we wish to show that gC(H)c^{-1}\subset C(H). So consider an element s\in gC(H)g^{-1}. Then there is some x\in C(H) such that s=gxg^{-1}. Now, in order to show that s\in C(H) we must show that s commutes with any element h\in H.

Consider gxg^{-1}h. Since H\lhd G it follows that there exists some h'\in H such that g^{-1}h=h'g^{-1}.  Note that this also means that gh'=hg. Using these facts we may rewrite the expression in the following way:

gxg^{-1}h=gxh'g^{-1}

But x commutes with everything from H and so we may again rewrite the expression:

=gh'xg^{-1}

But as stated above, gh'=hg and so we may rewrite one more time:

=hgxg^{-1}

And so we may conclude that gC(H)g^{-1}\subset C(H) for every g\in G and so C(H)\lhd G.

\Box

Reflection: This is one of those proofs that is not entirely difficult, but is very easy to make much more difficult than it needs to be. Thankfully I have Jeremy to straighten me out when I over think these simple problems. The key thing for me while doing this proof was to think in words, and less in symbols. Once we started talking about what it means to be in the centralizer of H then it started to click.

Posted in Algebra, Centralizer, Normal | Leave a comment