So as I study for my comprehensive exams, I will post some of my favorite problems from my studies.
Abstract Algebra
The problem from Abstract Algebra that intrigued me the most this week deals with Noetherian and Artinian modules.
Definition. Let $\left(P,\leq\right)$ be a partially ordered set. We say that $P$ satisfies the Ascending Chain Condition (ACC) with respect to $\leq$ if, for any ascending chain $a_1\leq a_2\leq\cdots$, there exists $n$ such that $a_n = a_{n+1}=\cdots$. We say that $P$ satisfies the Descending Chain Condition (DCC) with respect to $\geq$ if, for any descending chain $a_1\geq a_2\geq\cdots$, there exists $n$ such that $a_n = a_{n+1}=\cdots$.
Definition. Let $M$ be an $R$-module. We say that $M$ is Noetherian if the submodules of $M$ satisfy (ACC) with respect to set-inclusion $\subseteq$, and we say that $M$ is Artinian if the submodules satisfy (DCC) with respect to $\supseteq$.
Definition. Let $M,N$ be $R$-modules. A module homomorphism $f:M\to N$ is a map such that, for all $x,y\in M, a\in R$, we have that $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ and $f\left(ax\right)=af\left(x\right)$. The kernel of a homomorphism is $\ker f=\left\{x\in M\mid f\left(x\right)=0\right\}$, and the image of a homomorphism is $\text{im}\, f=\left\{f\left(x\right)\mid x\in M\right\}$. We call $f:M\to N$ a monomorphism if it is an injective module homomorphism, and this is equivalent to $\ker f=\left\{0\right\}$. We call $f$ an epimorphism if it is a surjective module homomorphism, and this is equivalent to $\text{im}\, f=N$. An isomorphism $f:M\to N$ is a homomorphism which is both injective and surjective.
Problem. (i) Let $M$ be a Noetherian module, and let $f:M\to M$ be an epimorphism. Then, $f$ is an isomorphism.
(ii) Let $M$ be an Artinian module and let $f:M\to M$ be a monomorphism. Then, $f$ is an isomorphism.
Solution.
(i) Consider the ascending chain of submodules
\[
\ker f\subseteq \ker f^2 \subseteq\cdots,
\]
where $f^2$ denotes $f\circ f$, and so on. Then, there exists $n$ so that $\ker f^n=\ker f^{n+1}=\cdots$. Since $f$ is surjective, $f^k$ must also be for all $k\geq1$. Now, let $x\in \ker f^n$, and let $x’\in M$ so that $x=f^n\left(x’\right)$. Then, $0=f^n\left(x\right)=f^{2n}\left(x\right)$, and so $x’\in\ker f^{2n}=\ker f^n$, and thus $x=f^n\left(x’\right)=0$. Therefore, $\left\{0\right\}\subseteq\ker f\subseteq \ker f^n = \left\{0\right\}$, making $f$ injective, as required.
(ii) Consider the descending chain of submodules
\[
\text{im}\,f\supseteq\text{im}\,f^2\supseteq\cdots.
\]
Then, there is $n$ so that $\text{im}\,f^n=\text{im}\,f^{n+1}=\cdots$. Now, let $x\in M$. Then, $f^n\left(x\right)\in\text{im}\,f^n=\text{im}\,f^{n+1}$, and so there is $x’\in M$ so that $f^{n}\left(x\right)=f^{n+1}\left(x’\right)$. Since $f$ is injective, we may apply $f^{-n}$ to both sides to get $x=f\left(x’\right)$, making $f$ surjective, as required.
QED
Real Analysis
The Real Analysis problem here was a fun one from one of my homework sets this semester. Recall from my very first blog post, here, that I show that the primes are essentially “sparse,” because, as you go further down the number line, you see fewer and fewer of them. However, there are infinitely many of them, and this problem shows that there are still “a lot” of them. I like to think of it as an example of the subtleties of “infinity,” and why our intuitions often fail us when we go from the finite to the infinite.
Problem. Show that the sum
\[
\sum_{p\,\text{prime}} \frac{1}{p}
\]
diverges.
Solution.
Given $N>1$, let $p_{1},\ldots,p_{k}$ be primes which divide at least one integer at most $N$, making them precisely the first $k$ primes. Then, we have
\[
\sum_{n=1}^{N}\frac{1}{n}\leq\prod_{j=1}^{k}\sum_{m=0}^{\infty}p_{j}^{-m}=\prod_{j=1}^{k}\left(1-p_{j}^{-1}\right)^{-1}\leq\exp\sum_{j=1}^{k}\frac{2}{p_{j}}.
\]
The first inequality holds as, by the Fundamental Theorem of Arithmetic and for any $n\leq N$, there exist $r_{1},\ldots,r_{k}$ so that $n=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$, and so taking all possible combinations of these prime powers will give at least $\sum_{n=1}^{N}\frac{1}{n}$. The equality in the middle holds as each term in the product is a geometric series. Finally, for the last inequality, I claim that $\left(1-x\right)^{-1}\leq e^{2x}$ for $0\leq x\leq\frac{1}{2}$.
To prove this, consider $f\left(x\right)=\left(1-x\right)e^{2x}$. Then, $f’\left(x\right)=2\left(1-x\right)e^{2x}-e^{2x}=\left(2-2x-1\right)e^{2x}=\left(1-2x\right)e^{2x}>0$ for $0\leq x\leq\frac{1}{2}$. Thus, the minimum value of $f$ on $\left[0,\frac{1}{2}\right]$ is $f\left(0\right)=1$. Therefore, $\left(1-x\right)^{-1}\leq e^{2x}$ for $0\leq x\leq\frac{1}{2}$, giving the last inequality.
Finally, we rearrange terms to get
\[
\frac{1}{2}\log\left(\sum_{n=1}^{N}\frac{1}{n}\right)\leq\sum_{j=1}^{k}\frac{1}{p_{j}}
\]
Since $k$ depends on $N$, and $k\to\infty$ as $N\to\infty$, and since $\frac{1}{2}\log\left(\sum_{n=1}^{N}\frac{1}{n}\right)\to\infty$ also as $N\to\infty$, we have that
\[
\sum_{p\,\text{prime}} \frac{1}{p}
\]
diverges, as required.
QED
As a fun fact, since
\[
\sum_{n\leq x} 1/n = O\left(\log x\right),
\]
this proves, then, that
\[
\sum_{\substack{p\leq x\\p\,\text{prime}}}\frac{1}{p}=O\left(\log\log x\right).
\]
Thus, it would take approximately $e^{e^{10}}\approx9.4\times 10^{9565}$ primes for the sum to finally reach a partial sum exceeding 10. As an analogy to show how large this is, it is said that there are “only” about $10^{80}$ particles in the observable universe. So if each particle in the observable counted up from 0 by 1 for every picosecond ($10^{-12}=1/1000000000000$ of a second) for 100 years, and at the end added up the numbers they ended up at, it would still be less than $10^{102}$. So the number of primes needed for this sum to exceed 10 is beyond astronomically large. Yet, in the end, this sum still diverges to infinity.