A Group with no Composition Series

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


 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}


Failure of Integral Closure

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.

Upper Triangular Matrix Rings and Noetherian/Artinian Hypotheses

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

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

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

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.