Exciting news, I’ll be teaching calculus next quarter!

…kind of.

There is a bit of a catch. I’ll be teaching calculus for business and economics. Commonly referred to as business calc, or (as the TA’s like to call it) busy calc. The catch is that there is no trig (how can you do anything fun in calculus without trig?) and there will be a lot of applications to “business problems”. Regardless, it is still calculus and I am excited!

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Keith Devlin is a professor at Stanford but probably better known as the NPR “math guy”. I really like his latest piece on Weekend Edition.

The Way You Learned Math Is So Old School : NPR

Remember going through your multiplication tables in elementary school? Or crunching through long division? Yeah, it sucked and that’s because it’s lame and useless, literally useless. Especially in today’s society where everyone uses a calculator or a computer to carry out arithmetic calculations. Computers do arithmetic for us, Devlin says, but making computers do the things we want them to do requires algebraic thinking. Elementary schools are starting to notice this and it’s changing the way arithmetic is taught. The emphasis in teaching mathematics today is on getting people to be sophisticated, algebraic thinkers. This doesn’t mean that we should stop teaching arithmetic in school; arithmetic is the foundation for strong algebraic thinking, it is not however, the end goal.

The next time you slice a bagel you might want to try this! You can slice a bagel in a single cut so that the result is two equal halves linked together. Don’t believe it? George W. Hart, sculptor and mathematician, has instructions you can follow here.

The two halves turn out to be mobius strips. For a better visual check out the video.

Theorem. Let $R$ be a commutative ring with unity. The ideal $A$ of $R$ is a maximal ideal if and only if the quotient ring $R/A$ is a field.

Proof. Suppose that $A$ is a maximal ideal of $R$. Let $b\in R\setminus A$. It suffices to show that $b+A$ has a multiplicative inverse. Consider the set $B=\{br+a : r\in R, a\in A\}$. I claim that $B$ is an ideal of $R$ that properly contains $A$. Indeed, we check the following:

Nonempty. Since $B$ contains the element $b$, we have that $B$ is nonempty.

Subtraction. Let $br_1+a_1, br_2+a_2\in B$. Then we have that

$(br_1+a_1)-(br_2+a_2) = b(r_1-r_2)+(a_1-a_2)$.

Since $R$ is a ring and $A$ is an ideal, $r_1-r_2\in R$ and $a_1-a_2\in A$. Thus, $(br_1+a_1)-(br_2+a_2)\in B$.

Multiplication. Let $x\in R$ and $br+a\in B$. Since $R$ is commutative, it suffices to show that $x(br+a)\in B$. We have that

$x(br+a)=bxr+xa$.

Since $R$ is a ring and $A$ is an ideal, $xr\in R$ and $xa\in A$. Thus, $x(br+a)\in B.$

Properly contains $A$. By definition of $B$, $A\subseteq B$. Moreover, notice that $b\in B$ and by assumption, $b\notin A$. Thus, $A$ is a proper subset of $B.$

Therefore, $B$ is as claimed. Now, since $A$ is maximal, we must have that $B=R$. Thus $1\in B$, say, $1=bc+a'$. Then,

$1+A=(bc+a')+A=bc+A=(b+A)(c+A).$

We conclude that $c+A$ is the multiplicative inverse of $b+A$; hence, $R/A$ is a field.

Conversely, suppose that $R/A$ is a field. Let $B$ be an ideal of $R$ that properly contains $A$ (i.e. $A\subset B\subseteq R$) and let $b\in B\setminus A$. Then $b+A$ is a nonzero element of $R/A$ and so there exists an element $c+A\in R/A$ such that $(b+A)(c+A)=1+A$, the multiplicative identity of $R/A$. Since $b\in B$, $c\in R$, and $B$ is an ideal, we have that $bc\in B$. Moreover, since

$1+A=(b+A)(c+A)=bc+A$,

we have that $1-bc\in A$. Thus, $1=(1-bc)+bc\in B$ and since $B$ contains $1$ it must be that $B=R$. We conclude that $A$ is a maximal ideal of $R$. $\Box$

Today is the birthday of my favorite Beatle, George Harrison. He would have been 68. Be sure to listen to your favorite George Harrison song today! This is one of mine

Theorem. A subgroup of a cyclic group is cyclic.

Proof. Let $G$ be a cyclic group generated by $a$ and let $H\leq G$. If $H=\{e\}$ then $H=\langle e \rangle$ is cyclic. If $H\neq\{e\}$, then $a^n\in H$ for $n\in\mathbb{Z}^+$. Let $m$ be the smallest integer in $\mathbb{Z}^+$ such that $a^m\in H$.

I claim that $c=a^m$ generates $H$; that is, $H=\langle a^m \rangle = \langle c \rangle$. We must show that every $b\in H$ is a power of $c$. Let $b\in H$. Then since $H\leq G$, $b=a^n$ for some positive integer $n$. By the division algorithm for $\mathbb{Z}$, there exist $q, r \in \mathbb{Z}$ such that $n=mq+r$ where $0\leq r < m$. Then $a^n=a^{mq+r}=a^{mq}a^r$ so $a^r=a^na^{-mq}$. We must have that $a^r\in H$ because $H$ is a group and $a^n, a^{-mq} \in H$. Since $m$ is the smallest positive integer such that $a^m\in H$ and $0\leq r < m$, it must be the case that $r=0$. Thus, $n=mq$, $b=a^n=a^{mq}=c^q$, and so $b$ is a power of $c$. $\Box$

The first isomorphism theorem is probably the most important result one will learn in a first course on algebra. We will state and prove the theorem for groups. There is, of course, an analogous result for rings. We will state the theorem for rings (the proof is identical to that for groups) and then look at an example using the theorem.

Theorem. If $\varphi : G \to H$ is a homomorphism of groups, then $\text{ker}(\varphi)$ is a normal subgroup of $G$ and $G/\text{ker}(\varphi)\cong\varphi(G)$.

The proof is not terribly complicated; in fact, it is rather “computational”, meaning, we just have to verify a few definitions. It is just a matter of checking that the kernel satisfies the definition of normal and then defining a function from $G/\text{ker}(\varphi)$ to $\varphi(G)$ and checking that the function satisfies the definition of an isomorphism. The “hardest” part of the proof, I think, is actually defining a function from $G/\text{ker}(\varphi)$ to $\varphi(G)$. However, as we will see, the choice of such a function is quite natural.

Proof. First, we will show that the kernel of $\varphi$ is a normal subgroup of $G$. We check the following:

Nonempty. Since $\varphi$ is a homomorphism, it must map the identity in $G$ to the identity in $H$. Thus, $1_G\in\text{ker}(\varphi)$ which shows that the kernel is nonempty.

Closure. Let $a$ and $b$ be in the kernel of $\varphi$. We want to show that $ab^{-1}$ is in the kernel too. We compute that

$\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=(1_H)(1_H)^{-1}=1_H.$

Thus, $ab^{-1}$ is indeed in the kernel of $\varphi$, which shows that $\text{ker}(\varphi)$ is a subgroup of $G$.

Normal. Let $g$ be an element of $G$ and let $k$ be an element in the kernel of $\varphi$. It suffices to show that $gkg^{-1}$ is also in the kernel. We have that

$\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)1_H\varphi(g)^{-1}=1_H.$

Thus, $gkg^{-1}$ is in the kernel of $\varphi$ as desired. This shows that $\text{ker}(\varphi)$ is a normal subgroup of $G$.

Next, we will show that $G/\text{ker}(\varphi)$ is isomorphic to $\varphi(G)$. Define $\psi : G/\text{ker}(\varphi) \to \varphi(G)$ by $\psi(g\text{ker}(\varphi))=\varphi(g)$. I claim that $\psi$ is a well-defined isomorphism. Indeed, we check the following:

Well-Defined. Suppose that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$. Then $a=bk$ for some element $k$ in the kernel of $\varphi$. We wish to show that $\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi)).$ Evaluating $\psi(a\text{ker}(\varphi))$ we see,

$\psi(a\text{ker}(\varphi))=\varphi(a)=\varphi(bk)=\varphi(b)\varphi(k)=\varphi(b)1_H=\psi(b\text{ker}(\varphi)).$

Thus, $\psi$ is well defined.

Homomorphism. Let $a\text{ker}(\varphi)$ and $b\text{ker}(\varphi)$ be cosets in the factor group $G/\text{ker}(\varphi)$. By multiplying cosets we observe that $a\text{ker}(\varphi)b\text{ker}(\varphi)=ab\text{ker}(\varphi).$ Consequentially,

$\psi(a\text{ker}(\varphi)b\text{ker}(\varphi))=\varphi(ab) =\varphi(a)\varphi(b) =\psi(a\text{ker}(\varphi))\psi(b\text{ker}(\varphi)).$

Thus, $\psi$ is a homomorphism.

Injective. Suppose that $\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi))$. We wish to show that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$. Consider the following,

$\varphi(a)=\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi))=\varphi(b).$

Since $\varphi(b)$ is an element of $H$ and $H$ is a group, $\varphi(b)^{-1}$ exists. Therefore,

$\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=1_H,$

which shows that $ab^{-1}$ is in the kernel of $\varphi$. It follows that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$ and so $\psi$ is injective.

Surjective. Let $x$ be an element in the image of $\varphi$. Then there exists an element $g$ in $G$ such that $\varphi(g)=x$. Thus, $\psi(g\text{ker}(\varphi))=\varphi(g)=x$ which shows that $\psi$ is surjective.

We conclude that $\psi$ is a bijective homomorphism, hence an isomorphism. $\Box$

We now state the analogous result for rings.

Theorem. If $\varphi : R \to S$ is a homomorphism of rings, then $\text{ker}(\varphi)$ is an ideal of $R$ and $R/\text{ker}(\varphi)\cong\varphi(R)$.

Example. Let $R$ be a commutative ring with unity. Prove that $R[x] / \langle x \rangle \cong R.$

Define the function $\varphi:R[x]\to R$ to be the evaluation homomorphism $\varphi(f(x))=f(0)$ for any $f(x)$ in $R[x]$. By the first isomorphism theorem, it suffices to check that $\varphi$ is a surjective homomorphism with $\text{ker}\varphi=\langle x \rangle$. We check the following:

Homomorphism. Let $f(x)$ and $g(x)$ be in $R[x]$. We compute that

$\varphi((f+g)(x))=(f+g)(0)=f(0)+g(0)=\varphi(f(x))+\varphi(g(x))$

$\varphi((fg)(x))=(fg)(0)=f(0)g(0)=\varphi(f(x))\varphi(g(x))$.

Thus, $\varphi$ is indeed a ring homomorphism.

Surjective. Let $r$ be an element in $R$. Then $f(x)=x+r$ is in $R[x]$ and $\varphi(f(x))=f(0)=0+r=r$ which shows that $\varphi$ is surjective.

Kernel. Let $f(x)$ be in the kernel of $\varphi$. Then $f(0)=0$ which means that $x$ is a factor of $f(x)$ and so $f(x)$ is in the ideal $\langle x \rangle$. Therefore, $\text{ker}\varphi\subseteq\langle x \rangle$. Now, let $g(x)$ be in the ideal $\langle x \rangle$. Then $g(x)=xf(x)$ for some $f(x)$ in $R[x]$. Thus, $\varphi(g(x))=g(0)=0f(0)=0$ which shows that $g(x)$ is in the kernel of $\varphi$. Therefore, $\langle x \rangle\subseteq\text{ker}\varphi$. We conclude that $\text{ker}\varphi=\langle x \rangle$.

By the first isomorphism theorem we have that $R[x] / \langle x \rangle \cong R.$ $\Box$

My blog doesn’t really have a theme; it’s just a random collection of things that are going on in my life that I want to share. I keep the topics pretty broad so more people will be inclined to read. As a result, the math posts tend to be low. I don’t want my blog to be a “math blog” (you can find a list of such blogs here, or just google “math blogs”). It should be noted that online collaborative mathematics has become quite popular recently. Sites like Math Overflow and Terence Tao’s blog are great examples.

Having said that, I’ve decided that my next few posts will be more math related; in particular, I’ll be posting a couple of algebra theorems each week for the next few weeks. Here’s why. The algebra qualifying exam is in about a month and I’ve been studying really hard for it. I know that some of the other TA’s are studying really hard for it too; and since some of them occasionally glance at my blog (I know, I can’t believe it either), I think it’s worth while to share some of the key algebra theorems.

For the mathematically inclined, please feel free to comment and critique any of the results you see posted. A lot of the proofs are my own, which means there might (will) be errors. Enjoy.

The NBA All-Star Game was this past weekend, which means, slam dunk contest. If you didn’t hear by now, Blake Griffin jumped over a car and won the contest (you can find videos of it all over the internet). But you probably don’t know how the alley-oop was invented. This video should enlighten you.

This is a broadcast from last year that NPR ran for Valentine’s Day. It’s too good to pass up. Be sure to listen to the story, don’t just read the article.

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