# The Unapologetic Mathematician

## The Fundamental Involutory Quandle

As I discussed last time, coloring a knot with any abelian group is secretly using the dihedral quandle associated to that group. This is an involutory quandle with action $a\triangleright b=2a-b$. The reason knot coloring works out so nicely is that the axioms of (involutory) quandles line up with the Reidemeister moves.

But for the moment we’re stuck with picking this or that involutory quandle and counting how many colorings it gives for a given knot. Different quandles give different coloring numbers, and we’d like to find a better way of thinking of them all at once. We’re going to construct a new involutory quandle from a knot that captures all of them.

Take any diagram of the knot we’re interested in. Remember the knot table if you want to pick one out. Now each arc in the diagram has to get some color, no matter what quandle we’re using to color it. Instead of picking a color from a specific quandle, let’s just slap a label like $x$, $y$, or $z$ on each arc. Be sure to use a different label for each different arc.

Now those labels will generate an involutory quandle. We can throw them together with the two quandle compositions to get “words” like $x\triangleright((y\triangleright z)\triangleright z)$. These words, of course, are subject to the normal quandle equivalences, but we need more relations for our purposes. At each crossing the values in a coloring have to satisfy a certain relation, so we’re going to build that right into our quandle. If the arcs labeled $x$ and $z$ meet under the crossing arc labeled $y$, then we must have $z=y\triangleright x$.

This seems to depend on the choice of a diagram, though. Well, it sort of does, but any Reidemeister move gives an isomorphism of quandles relating the two sides. For example, performing the first one splits an arc into two pieces. Say label $x$ becomes $x_1$ and $x_2$. Then the relations we introduce say that $x_1\triangleright x_1=x_2$. But the axioms of quandles say that $x_1\triangleright x_1=x_1$, so $x_1=x_2$ and we can just drop one of these generators and the relation we’ve now “used up”. Try to find the isomorphisms for the other two moves. This justifies calling the quandle we’ve constructed (up to isomorphism) “the” fundamental involutory quandle $Q(K)$ of the knot $K$.

So what’s a coloring? A coloring assigns an element of some quandle to each arc of the knot diagram. But arcs in the diagram are just generators of the fundamental quandle. That is, a coloring is a function that takes generators of the fundamental quandle to a selected target quandle. If it plays nicely with the relations between the generators, it will be a quandle homomorphism. In fact it does, precisely because we picked the relations between the generators to be exactly those required by colorings. A given relation comes from a crossing, and every coloring of a knot obeys the same restrictions at crosings.

In the end we’ve found that the set of all colorings of $K$ by an involutory quandle $Q$ is the set of quandle homomorphisms $\hom_{\mathbf{Quan}}(Q(K),Q)$, so the number of $Q$-colorings is the cardinality of this set. If we have a good understanding of quandles and their homomorphisms, we can read off coloring numbers by involutory quandles from the fundamental involutory quandle.

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May 16, 2007 Posted by | Knot theory, Quandles | 1 Comment

## At last

So I’m up until 4:30 scrawling something out. What’s the only thing that wouldn’t annoy me to no end waking me up at 9:00?

Tulane University.

Finally.

May 16, 2007 Posted by | Uncategorized | 7 Comments

## The lattice of ideals

We know that the collection of all ideals of a given ring form a rig. In fact, they also form a lattice. We put the partial order of inclusion on ideals, so $I$ is below $J$ if $I\subseteq J$.

To show that this poset is a lattice we have to show that pairwise greatest lower bounds and least upper bounds exist. Lower bounds are easy: the intersection $I\cap J$ of two ideals is again an ideal. By definition, any ideal contained in both $I$ and $J$ is contained in $I\cap J$.

Upper bounds are a little trickier, since we can’t just take the union of two ideals. That would work for subsets of a given set, but in general the union of two ideals isn’t an ideal. Instead, we take their sum. Clearly $I\subseteq I+J$ and $J\subseteq I+J$. Also, if $K$ is another ideal containing both $I$ and $J$, then $K$ contains all linear combinations of elements of $I$ and $J$. But $I+J$ is the set of all such linear combinations. Thus $I+J\subseteq K$, and $I+J$ is the least upper bound of $I$ and $J$.

This lattice is related to the divisibility preorder. Given a commutative unital ring $R$ and two elements $a,b\in R$, recall that $a|b$ if there is an $x\in R$ so that $b=ax$. Then every multiple $bk$ of $b$ is also a multiple $axk$ of $a$. Thus we see that the principal ideal $(b)$ is contained in the principal ideal $(a)$. On the other hand, if $(b)\subseteq (a)$ we can see that $b=ax$ for some $x\in R$, so $a|b$. In particular, two elements are associated if and only if they generate the same principal ideal.

Notice that this correspondence reverses the direction of the order. If $a$ is below $b$ in the divisibility ordering, then $(a)$ is above $(b)$ in the ideal ordering. Thus the “greatest common divisor” of two ideals is actually now the least ideal containing both of them. The language of ideals, however, is far more general than that of divisibility. We now need to recast most of what we know about divisibility from our experience with natural numbers into these more general ring-theoretic terms.

May 15, 2007 Posted by | Ring theory | 8 Comments

## Lattices

A poset which has both least upper bounds and greatest lower bounds is called a lattice. In more detail, let’s say we have a poset $(P,\preceq)$ and give it two operations: meet (written $\wedge$) and join (written $\vee$). These satisfy the requirements that

• $x\preceq x\vee y$ and $y\preceq x\vee y$.
• If $x\preceq z$ and $y\preceq z$ then $x\wedge y\preceq z$.
• $x\wedge y\preceq x$ and $x\wedge y\preceq y$.
• If $z\preceq x$ and $z\preceq y$ then $z\preceq x\wedge y$.

Not every poset has a meet and a join operation, but if these operations do exist they are uniquely specified by these requirements. In fact, we can see this sort of like how we saw that direct products of groups are unique up to isomorphism: if we have two least upper bounds for a pair of elements then they must each be below or equal to the other, so they must be the same.

We can derive the following properties of the operations:

• $a\vee(b\vee c)=(a\vee b)\vee c$ and $a\wedge(b\wedge c)=(a\wedge b)\wedge c$.
• $a\vee b=b\vee a$ and $a\wedge b=b\wedge a$.
• $a\vee(a\wedge b)=a$ and $a\wedge(a\vee b)=a$.

from these we see that there’s a sort of duality between the two operations. In fact, we can see that these provide two commutative semigroup structures that happen to interact in a certain nice way.

Actually, it gets even better. If we have two operations on any set satisfying these properties then we can define a partial order: $x\preceq y$ if and only if $x=x\wedge y$. So we can define a lattice either by the order property and get the algebraic properties, or we can define it by the algebraic properties and get the order property from them.

In many cases, a lattice also satisfies $x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)$, or equivalently $x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$. In this case we call it “distributive”. A bit weaker is to require that if $x\preceq z$ then $x\vee(y\wedge z)=(x\vee y)\wedge z$ for all $y$. In this case we call the lattice “modular”.

A lattice may have elements above everything else or below everything else. We call a greatest element of a lattice $1$ and a least element ${0}$. In this case we can define “complements”: $x$ and $y$ are complements if $x\vee y=1$ and $x\wedge y=0$. If the lattice is distributive, then the complement of $x$ is unique if it exists. A distributive lattice where every element has a complement is called “Boolean”.

May 14, 2007 Posted by | Algebra, Lattices, Orders | 10 Comments

## Generators of ideals

Let’s say we’ve got a ring $R$ and an element $x\in R$. What is the smallest left ideal that contains $x$? Well, we have to have all multiples $rx$ for $r\in R$ so it’s closed under left multiplication. If $R$ has a unit, this is all we need. Otherwise, we have to make sure we include all the elements $nx=x+...+x$ with $n$ summands (and their negatives) to make sure it’s an abelian subgroup. Thus the subset $\{rx+nx|r\in R,n\in\mathbb{Z}\}$ is a left ideal in $R$. If $R$ has a unit, we just need the subset $\{rx|R\in R\}$. We call this the principal left ideal generated by $x$, and write $Rx$. We can do something similar for right ideals ($xR$), and for two-sided ideals we get the subset $(x)=\{r_1xr_2+nx|r_1,r_2\in R,n\in\mathbb{Z}\}$.

As for any submodules we can form the sum. If we have elements $x_1,...,x_n\in R$ they generate the left ideal $Rx_1+...+Rx_n$, or a similar right ideal. For two-sided ideals we write $(x_1,...,x_n)$. The term “principal”, however, is reserved for ideals generated by a single element.

Let’s look at these constructions in the ring of integers. Since it’s commutative, every ideal is two-sided. An integer $n$ then generates the principal ideal $(n)=\{kn\}$ of all multiples of $n$. In fact, every ideal in $\mathbb{Z}$ is principal.

If $I\subseteq\mathbb{Z}$ is an ideal, consider the subset of all its (strictly) positive elements. Since this is a subset of the natural numbers it has a least element $n$. I say that every element of $I$ is a multiple of $n$. If not, then there is some $m\geq n$ that $n$ doesn’t divide. If we can apply Euclid’s algorithm to $m$ and $n$, at the first step we get $m=qn+r$ with $r\leq n$. The greatest common divisor $d$ of $m$ and $n$ will thus be less than $n$, and Euclid’s algorithm gives us a linear combination $d=xm+yn$ for integers $m$ and $n$. Thus $d\leq n$ must be in the ideal as well, contradicting the minimality of $n$.

So every ideal of $\mathbb{Z}$ is principal. When this happens for a ring, we call it a “principal ideal ring”, or a “principal ideal domain” if the ring is also an integral domain.

So how do ideals of integers behave under addition and multiplication? The ideal $(m)+(n)$ is the ideal $(m,n)$. This it consists of all the linear combinations $xm+yn$. In particular, the smallest positive such linear combination is the greatest common divisor of $m$ and $n$, as given by Euclid’s algorithm. The product of the ideals $(m)(n)$ is the set of all products of multiples of $m$ and $n$: $k_1mk_2n=k_1k_2mn=kmn$. Thus $(m)(n)=(mn)$.

May 13, 2007 Posted by | Ring theory | 4 Comments

## No, we have no mathematics. We have no mathematics today.

Sorry for the downtime. I had to start the Calc 2 final on my own at 9 in the morning, when I’m usually not even awake yet. Then pushing through to get everything graded by the end of the day. Thus ends the class, and quite possibly my academic career.

Now I can pack up and straighten up my apartment to hit the road tomorrow. I’m spending a couple weeks’ down-time in Maryland, so I’ll be back out at the College Perk in College Park most nights. If you’re in the area, drop by.

May 12, 2007 Posted by | Uncategorized | Leave a comment

## More modules, more ideals

The first construction I want to run through today is related to the amalgamated free product from group theory. Here’s the diagram in modules:

Remember we read it as follows: If we have modules $N$, $M_1$, and $M_2$, and homomorphisms from $N$ into each of $M_1$ and $M_2$, then the “amalgamated direct sum” $M_1\underset{N}\oplus M_2$ is another module with homomorphisms into it from each of $M_1$ and $M_2$ making the square commute. Further, for any other module $X$ and pair of homomorphisms into it, there is a unique homomorphism from $M_1\underset{N}\oplus M_2$ to $X$.

How do we know that such a thing exists? Well, we can take the direct sum of $M_1$ and $M_2$, which comes with homomorphisms into it from $M_1$ and $M_2$. Then we can follow both paths from $N$ to $M_1\oplus M_2$. In general they’re different homomorphisms, since one has an image completely in $M_1$ and the other completely in $M_2$. But since we’re looking at module homomorphisms we can subtract one from the other. To make the square commute we want the image of this difference to be zero, and we can make it be zero by forming the quotient module!

This turns out to be useful right away. Let’s say that $M_1$ and $M_2$ are both submodules of a module $M$. They definitely share the zero element of $M$, but they might share a larger submodule than that. It’s easily verified that their intersection $N$ is a submodule, and it comes with inclusion homomorphisms into each of $M_1$ and $M_2$. Now if we want to “add” the submodules $M_1$ and $M_2$, we had better not treat elements in their intersection differently depending on which submodule module we pick them from, since they’re all just submodules of $M$, so the direct sum isn’t what we want.

Instead, we find that the submodule $M_1+M_2$ of all elements of $M$ of the form $m_1+m_2$ with $m_1\in M_1$ and $m_2\in M_2$ is isomorphic to the direct sum amalgamated over their intersection. In particular, we can apply this to submodules of our base ring $R$ itself — ideals! We define the sum $I+J$ of two ideals as this sum of submodules of $R$.

There’s one more thing we can do for ideals. If we have a left ideal $I\subseteq R$ and an abelian subgroup $A\subseteq R$ then we can form their tensor product over $\mathbb{Z}$: $I\otimes A$. Since $I$ is a left $R$-submodule, this turns out to be a left $R$-submodule of $R\otimes R$. Then we can use the multiplication on $R$ to get a homomorphism $I\otimes A\rightarrow R$. We denote its image as $IA$, and it is a left ideal of $R$. In terms of elements, it’s the set of all sums of products in $R$: $\sum\limits_{k=1}^ni_ka_k$ with $i_k\in I$ and $a_k\in A$. In particular, we could choose $A$ to be another ideal $J$ and get the product of ideals $IJ$.

Now here’s where it gets really fun. Start with a ring $R$ and consider the collection of all its left ideals $I$. There are a bunch of things we can show about these operations on ideals, which I’ll leave as exercises. If it’s easier, use the descriptions in terms of elements, but I think it’s more satisfying to work with the diagrams and universal properties. Here $I$, $J$, and $K$ are ideals, and $\mathbf{0}$ is the ideal consisting of only the zero element.

• $(I+J)+K=I+(J+K)$
• $I+J=J+I$
• $I+\mathbf{0}=I$
• $(IJ)K=I(JK)$
• $I\mathbf{0}=\mathbf{0}=\mathbf{0}I$
• $I(J+K)=IJ+IK$
• $(I+J)K=IK+JK$

What does all this mean? The collection of left ideals of $R$ form a rig, like the natural numbers! Further if $R$ has a unit, then we find $RI=I=IR$, so this rig has a unit. If $R$ is commutative, then $IJ=JI$ so the rig is too.

May 10, 2007 Posted by | Ring theory | 2 Comments

## Submodules, quotient modules, and the First Isomorphism Theorem for modules

Okay, getting a little back down to Earth now. Just like we had for groups and rings, we have an isomorphism theorem for modules.

First off, a submodule of a left $R$-module $M$ is just an abelian subgroup $N\subseteq M$ that’s closed under the action of $R$. That is, for any $r\in R$ and $m\in N$, we have $r\cdot m\in N$. A submodule $N$ comes with an inclusion homomorphism $\iota_{M,N}:N\rightarrow M$.

Now if we take an $R$-module $M$ and a submodule $N\subseteq N$ we can just consider $M$ and $N$ as abelian groups and form the quotient group. Remember that every subgroup of an abelian group is normal, so the quotient is again an abelian group $M/N$ made up of the “slices” $m+N$. It turns out that this is again an $R$-module. Just use the action $r\cdot(m+N)=(r\cdot m)+N$. If we chose a different representative $m+n+N$ of the slice, then we’d get $(r\cdot m+r\cdot n)+N$, which represents the same image, so this $R$ action is well-defined. Quotient modules come with projection homomorphisms $\pi_{M,N}:M\rightarrow M/N$.

Actually there’s one sort of submodule we’ve already seen. Remember that every ring $R$ is a left module over itself. Then the left submodules of $R$ are exactly the left ideals of $R$! If we have a two-sided ideal $I\subseteq R$ then we get a left action of $R$ on the quotient module $R/I$ and a right action as well, since $I$ is also a right submodule. Then we can take the tensor product $R/I\otimes_RR/I$ and get a linear function $R/I\otimes_RR/I\rightarrow R/I$ from multiplying representatives. Presto! Quotient ring!

We can use ideals to give submodules and quotient modules of other $R$-modules too. Take a ring $R$ with left module $M$ and a left ideal $I\subseteq R$. Then we can restrict the action of $R$ on $M$ to the ideal to get $IM=\{im|i\in I,m\in M\}$. This is clearly an abelian subgroup of $M$, and it turns out to be a submodule too. Indeed, we see that $r\cdot(im)=(ri)\cdot m\in IM$. Then we can make the quotient $R$-module $M/IM$ as above. Even better, if $I$ is two-sided this is actually a module over $R/I$: use the right $R$-module structure on $R/I$ and the left $R$-module structure on $M/IM$ and tensor to get $R/I\otimes M/IM$. Then we can get a linear function $R/I\otimes M/IM\rightarrow M/IM$ by choosing representatives and showing that the choice is immaterial.

I promised an isomorphism theorem. Well, I’ll state it, but the proof is pretty much exactly the same as the two we’ve seen before so I’ll leave you to review those. Any homomorphism $M\rightarrow N$ of left (right) $R$-modules factors as the composition $M\rightarrow M/M'\rightarrow N'\rightarrow N$, where the first arrow is the projection homomorphism, the third is the inclusion homomorphism, and the middle arrow is an isomorphism. We call the submodule $M'$ the kernel of the homomorphism and the submodule $N'$ the image of the homomorphism. Notice that there’s no restriction on the sorts of submodules that can be kernels here. For groups a kernel is a normal subgroup, and for rings a kernel is a two-sided ideal, but any submodule can be the kernel of a module homomorphism. This leads to a few more definitions that come in handy. The quotient module $M/M'$ is called the coimage, and the quotient module $N/N'$ is called the cokernel. Thus we see that the coimage and the image of any homomorphism of modules are isomorphic.

May 9, 2007 Posted by | Ring theory | 10 Comments

## Algebras

We have defined a ring as a $\mathbb{Z}$-module (abelian group) $R$ with a linear function $R\otimes R\rightarrow R$ satisfying certain properties. The concept of an algebra takes this definition and extends it to work over more general base rings than $\mathbb{Z}$.

Let $A$ be a module over a commutative ring $R$ with unit. Then $A$ has both a left and a right action by $R$, since $R$ is commutative. Thus, when we take the tensor product $A\otimes_RA$, the result is also an $R$ module. It makes sense, then, to talk about an $R$-module homomorphism $A\otimes_RA\rightarrow A$. Equivalently, this is a “multiplication” function $m:A\times A\rightarrow A$ such that

• $m(a,b_1+b_2)=m(a,b_1)+m(a,b_2)$
• $m(a_1+a_2,b)=m(a_1,b)+m(a_2,b)$
• $m(ra,b)=m(a,rb)=rm(a,b)$

An $R$-module equipped with such a multiplication is called an $R$-algebra. We will often write the multiplication as $m(a,b)=ab$. In many cases of interest, the base ring will be a field $\mathbb{F}$, but any ring is an algebra over $\mathbb{Z}$.

Usually the term “algebra” on its own will refer to an associative algebra. This imposes an additional condition like the one we had in the definition of a ring: $(ab)c=a(bc)$. An algebra may also have a unit $1$ so that $1a=a=a1$ for all $a\in A$. Algebras can also be commutative if $ab=ba$ for all elements $a,b\in A$. There are other kinds of algebras we’ll get to later that are not associative.

Pretty much everything I’ve said about rings works for associative algebras as well, substituting “$R$-module” for “abelian group”. An $R$module $M$ is a left $A$-module if there is an $R$-linear function $A\otimes_RM\rightarrow M$, and a similar definition works for right $A$-modules. We can take direct sums and tensor products of $A$-modules, and we have an $R$-module of homomorphisms $\hom_A(M_1,M_2)$. All these constructions are clear from what we’ve said about modules over rings if we consider that $A$ is a ring, and that an $A$-module is an abelian group with actions of both $R$ and $A$ which commute with each other.

The standard constructions of rings also work for algebras. In particular, we can start with an $R$-module $M$ and build the free $R$-algebra on $M$ like we built the free ring on an abelian group. Just use $\bigoplus_{n\in\mathbb{N}}M^{\otimes_R n}$, where the tensor powers over $R$ make sense because $R$ is commutative.

We can also start with any semigroup $S$ and build the semigroup algebra $R[S]$ just like we did for the semigroup ring $\mathbb{Z}[S]$. As a special case, we can take $S$ to be the free commutative monoid on $n$ generators and get the algebra $R[x_1,...,x_n]$ of polynomials in $n$ variables over $R$. In fact, almost all of “high school algebra” is really about studying the algebra $\mathbb{Q}[x_1,...,x_n]$, where $\mathbb{Q}$ is the field of rational numbers I’m almost ready to define.

Another source of $R$-algebras extends the notion of the ring of endomorphisms. If $M$ is any $R$-module, then ${\rm End}_R(M)=\hom_R(M,M)$ is again an $R$-module, and composition is $R$-bilinear, making this into an $R$-algebra.

Algebras over more general commutative rings than $\mathbb{Z}$ — particularly over fields — are extremely useful objects of study mostly because the linear substrate can often be much simpler. Building everything on abelian groups can get complicated because abelian groups can be complicated, but building everything on vector spaces over a field is generally pretty straightforward since vector spaces and their linear transformations are so simple.

May 8, 2007 Posted by | Ring theory | 11 Comments

## Overpriced journals

In the May issue of the Notices of the American Mathematical Society, there is an article by Allyn Jackson: Jumping Ship: Topology Board Resigns about the mass resignation of the entire editorial board of the Elsevier journal over its exorbitant pricing. This has been a steadily mounting problem to academic mathematics. In 2005 a group set up the Banff Protocol, refusing to have anything to do with excessively-priced journals. If you’re a professional mathematician, go there and sign up.

May 8, 2007 Posted by | Uncategorized | 6 Comments