## A Group with no Composition Series

### Spookiness in Group Theory

The Jordan-Holder theorem is a group-theoretic analogue of the uniqueness of prime factorizations. A “factorization” of a (finite) group is called a composition series.

Definition. Let $G$ be a group. A composition series of $G$ is a chain of normal subgroups

$1= N_0 \trianglelefteq N_1 \trianglelefteq ... \trianglelefteq N_k = G$

where $N_{i+1}/N_i$ is simple.

There is a theorem stating that composition series of finite groups are unique.

Theorem (Jordan-Holder). Let $G$ be a finite group. Consider two composition series

$1= N_0 \trianglelefteq N_1 \trianglelefteq ... \trianglelefteq N_k = G$

and

$1= M_0 \trianglelefteq M_1 \trianglelefteq ... \trianglelefteq M_l = G$

of $G.$ Then, $k=l$ and the list $\{N_1/N_0,...,N_k/N_k/N_{k-1}\}$ is a permutation of $\{M_1/M_0,...,M_k/M_{k-1}.\}$

So, a composition series tells us a lot about a group. Therefore, we like them. Lucky for us then, there is a theorem stating that all finite groups admit a composition series. But, this is not true for infinite groups.

This bring us to our counterexample: $\mathbb{Z}$ has no composition series. To see this, think about what the first inclusion $N \trianglelefteq \mathbb{Z}$ must be (hint: the quotient must be a finite simple group). Those of us reading this post any time after 2004 have the privilege of being able to say that there is a list of all finite simple groups; only $\mathbb{Z}/p\mathbb{Z}$ will do here.

This yields $p\mathbb{Z} \trianglelefteq \mathbb{Z}.$ Contining forces an infinite regress

$...\mathbb{Z}/p^3\mathbb{Z} \trianglelefteq \mathbb{Z}/p^2\mathbb{Z} \trianglelefteq \mathbb{Z}/p\mathbb{Z} \trianglelefteq \mathbb{Z}$

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## The Failure of Integral Closure

### What Singularities look like to Algebra

Recall that a ring $R$ is integrally closed if it is integral over its field of fractions e.g. any element of Frac$(R)$ which is a root of a monic polynomial with coefficients in $R$ is contained in $R.$ For example, $\mathbb{Z}$ is integrally closed.

Consider $\mathbb{C}[t^2,t^3]$. The polynomial $t = t^3/t^2 \in \mathbb{C}(t)$ is a root of the (monic) polynomial $x^2 - t^2$ and so is integral over $\mathbb{C}[t^2,t^3].$ However, $t \notin \mathbb{C}[t^2,t^3].$ And for those who like pictures: the geometric incarnation of this fact is  that the plane curve $Y^2 - X^3 = 0$ has a singularity.

## Left/Right Noetherian and Left/Right Artinian Rings

### Counterexamples in Upper Triangles Matrix Rings

The ring $\begin{bmatrix} \mathbb{Q} & \mathbb{Q} \\ 0 & \mathbb{Z} \end{bmatrix}$ is left Noetherian. However, it is not right Noetherian, left Artinian or right Artinian.

Similarly, the ring $\begin{bmatrix} \mathbb{R} & \mathbb{R} \\ 0 & \mathbb{Q} \end{bmatrix}$ is left Artinian and left Noetherian, but not right Artinian or right Noetherian.

The details are worked out here under “CIA: Some Upper Triangular Matrix Rings and Noetherian/Artinian Hypotheses.”

## A Nil Ideal that is not Nilpotent

An ideal $I \leq R$ is said to be nil is each of elements is nilpotent. $I$ is said to be nilpotent if there exists some $k$ such that $I^k =0.$ That is, every product of $k$ elements of $I$ is zero.

Consider the ring $R = \mathbb{C}[x_1,x_2,x_3,...]/(x_1,x^2_2,x^3_3,...)$. Then, the ideal $I = (x_1,x_2,x_3,...)$ is nil in $R$ because each of its generators is nilpotent.

Moreover, $I$ is not nilpotent. To see this, assume by contradiction that it is. Then, there exists some $k > 0$ such that $I^k=0.$ This implies that for all $i,$ $x_i^k=0$ in $R$. However, $x_{k+1}^k \neq 0$ in $R$. This is the desired contradiction.

## A monic epi that is not an iso

Consider the ring map $i : \mathbb{Z} \to \mathbb{Q}.$ It is an epimorphism because $\mathbb{Z}$ is initial in Ring. It is a monomorphism because maps into $\mathbb{Z}$ that agree after composition with $i$ agree everywhere.

Another example would be the inclusion $\mathbb{Q} \to \mathbb{R}$ in the category Haus. This follows from the density of $\mathbb{Q}$ in $\mathbb{R}$.

Jargon. A category where all monic epis are isos is called balanced. A famous (and pleasant!) property of toposes is that all toposes are balanced.

## Noetherian Ring with Infinite Krull Dimension

This one is due to Nagata. Consider a polynomial ring  $R$ in countably infinitely many indeterminates over a field $k.$ Let, for each $n$, $\mathfrak{p}_n$ denote the prime ideal $(x_{2^{n-1}},x_{2^{n-1} + 1},...,x_{2^{n}-1})$. Then, define the (multiplicative) subset $S = R - \cup_i \mathfrak{p}_i$. The localization $A = S^{-1}R$ will be our counterexample.

We claim that $A$ is Noetherian. To see this, consider a prime ideal $I \leq A$. First, each $\mathfrak{p}_n$ is maximal and each $f \neq 0$ is contained in only finitely many $\mathfrak{p}_n$. So, $I$ is contained in only finitely many $\mathfrak{p}_n$. For each $n$$I_{\mathfrak{p}_n}$ is finitely generated. Thus, we can obtain a finite list $a_1,...,a_N$ of elements of $I$ that generate each $I_{\mathfrak{p}_n} \subset A_{\mathfrak{p}_n}.$ Note that, by Hilbert’s basis theorem, $A_{\mathfrak{p}_n}$ is Noetherian. Let $J = (a_1,...,a_N).$ Observe that $I_\mathfrak{m} = J_\mathfrak{m}$ for all but finitely many maximal ideals $\mathfrak{m}_1,...,\mathfrak{m}_M,$ but the $\mathfrak{m}_j$ are not any of the $\mathfrak{p}_i$ in which $I$ is contained. Moreover, for all $j, \exists b_i \in I - \mathfrak{m}_j$. Therefore, $(a_1,...,a_N,b_1,...,b_M)$ gives $I$ in each localization and so $I$ is finitely generated. As all prime ideals of $A$ are finitely generated, it is Noetherian.

Lastly, for each $n$ there is a chain $(x_{2^{n-1}}),...,(x_{2^{n-1}},x_{2^{n-1} + 1},...,x_{2^{n}-1})$ of prime ideals. The supremum then of the lengths of chains of prime ideals in $A$ is infinite e.g. $A$ has infinite Krull dimension.

## A Ring Epimorphism that is not a Surjection

Consider the inclusion $i : \mathbb{Z} \to \mathbb{Q}$. It is clearly not surjective. However, given any two ring homomorphisms$f ,g : \mathbb{Q} \to R$ such that$f \circ i = g \circ i$, it holds that $f=g$ because ring homomorphisms out of the rationals that agree on the integers must agree everywhere. So, $i$ is an epimorphism in Ring.