Homework 2

Following the hint, consider the morphism $p:S_3\to S_2$ given by composition of the projection morphism $S_3\to S_3/A_3$ with the isomorphism $S_3/A_3\cong S_2$. One can readily verify that the composition of this morphism $p:S_3\to S_2$ with the inclusion morphism $i:S_2\to S_3$ gives the identity morphism $S_2\to S_2$.

Now suppose there were some functor $Z:\mathbf{\text{Grp}}\to \mathbf{\text{Ab}}$ that mapped each group $G$ to its center $\mathrm{Z}(G)$. Since $\mathrm{Z}(S_2)=S_2$ and $\mathrm{Z}(S_3)=\{e\}$, the functor $Z$ must map the morphism $i:S_2\to S_3$ to the trivial morphism $Z(i):S_2\to \{e\}$. This forces the composition $Z(p)\circ Z(i)$ to be the trivial morphism $S_2\to S_2$ that maps everything to the identity. On the other hand, since $p\circ i$ is the identity morphism on $S_2$, by the assumption that $Z$ is a functor we must have that $Z(p\circ i)$ is the identity morphism on $Z(S_2)=S_2$. Thus, we cannot have $Z(p)\circ Z(i)=Z(p\circ i)$, contradicting the assumption that $Z$ was a functor (and hence must be compatible with composition).

Another way to say this all very succinctly is that group morphisms don't respect membership in centers. Our example is that of the element $(1,2)$ that is in the center of $S_2$ but maps (under the usual inclusion of $S_2$ into $S_3$) to an element not in the center of $S_3$.

1. First note that in the category ${\mathbf{\text{Matr}}}_F$ every commutative diagram of the form

   corresponds to an $n\times l$ matrix $D$ such that $AD=BD$. This latter equality is equivalent to the condition that $(A-B)D=0$, which in turn is equivalent to the condition that $(A-B){\mathbf{d}}_i=\mathbf{0}$ for every column ${\mathbf{d}}_i$ of the matrix $D$. In other words, $AD=BD$ exactly when the columns of $D$ are in the null space of $A-B$.

   With that in mind, let $\{{\mathbf{e}}_1,\dots ,{\mathbf{e}}_k\}$ be a basis for the null space of $A-B$ and let $E$ be the $n\times k$ matrix with those vectors as its columns. We claim $E:k\to n$ is an equalizer of the arrows $A,B:n\to m$ in ${\mathrm{Matr}}_F$. First note that by construction (and our observation above) we certainly have $AE=BE$, i.e., the diagram below commutes:

   Now suppose $D$ is an $n\times l$ matrix such that $AD=BD$, i.e., we have a commutative diagram

   By our earlier observation, this implies that every column ${\mathbf{d}}_i$ for $D$ is in the null space of $A-B$. Since $\{{\mathbf{e}}_1,\dots ,{\mathbf{e}}_k\}$ is a basis for that null space, there are unique $h_{j,i}\in F$ such that ${\mathbf{d}}_i=h_{1,i}{\mathbf{e}}_1+\cdots +h_{k,i}{\mathbf{e}}_k$. In other words, if we let ${\mathbf{h}}_i$ be the corresponding column vector, then $E{\mathbf{h}}_i={\mathbf{d}}_i$. If we let $H$ be the matrix with columns ${\mathbf{h}}_i$, then we have produced a $k\times l$ matrix $H$ such that $EH=D$. (Note that this matrix $H$ is also unique determined, since each column ${\mathbf{h}}_i$ was also uniquely determined.) We've thus produced a unique arrow $H:l\to k$ through which the arrow $D$ factors:

   This proves $E:k\to n$ is an equalizer for the parallel arrows $A,B:n\to m$.

2. This is the dual situation to the previous part. We first simply note that for an $l\times m$ matrix $D$ we have $DA=DB$ exactly when $D(A-B)=0$, or equivalently when $(A-B{)}^{⊺}{D}^{⊺}=0$. This in turn is equivalent to every column of $D^{⊺}$ (i.e., every row of $D$) being in the null space of $(A-B{)}^{⊺}$. So if we let $C$ be a $k\times n$ matrix whose rows are a basis for the null space of $(A-B{)}^{⊺}$, then the same argument as above shows that $DA=DB$ exactly when $D$ factors uniquely as $D=HC$. In other words, we have the dual commutative diagram

This proves $C:m\to k$ is a coequalizer of the parallel arrows $A,B:n\to m$.

Recall that an equivalence relation on $X$ is a subset $E\subseteq X\times X$ satisfying various properties (e.g., transitive, etc.). Two elements $x_1,x_2\in X$ are said to be equivalent exactly when $(x_1,x_2)\in E$. As such, the set $E$ comes with projection maps ${\pi }_i:E\to X$ given by ${\pi }_i(x_1,x_2)=x_i$. The set $X/E$ is defined to be the collection of equivalence classes in $X$. Each element in $X/E$ is a subset $S\subseteq X$ consisting of all elements equivalent to each other. Each such subset $S$ is usually denoted $[x]$, where $x\in X$ is any representative of that subset. In other words, we have $[x_1]=[x_2]$ exactly when $(x_1,x_2)\in E$. The quotient set $X/E$ also comes with a projection map $\pi :X\to X/E$, defined by $x\mapsto [x]$. We claim this last map is a coequalizer of the parallel set maps ${\pi }_1,{\pi }_2:E\to X$.

To prove this, first note that for every $(x_1,x_2)\in E$ we have $\pi ({\pi }_1(x_1,x_2))=\pi (x_1)=[x_1]=[x_2]=\pi ({\pi }_2(x_1,x_2))$. In other words, $\pi \circ {\pi }_1=\pi \circ {\pi }_2$ and so we do indeed have a commutative diagram

Next suppose $f:X\to Y$ is a set map with $f\circ {\pi }_1=f\circ {\pi }_2$, i.e., we have a commutative diagram

Define a set map $h:X/E\to Y$ be mapping $[x]\mapsto f(x)$. First observe that this map is actually well defined: if $[x_1]=[x_2]$ in $X/E$, then $(x_1,x_2)\in E$ and $f(x_1)=f({\pi }_1(x_1,x_2))=(f\circ {\pi }_1)(x_1,x_2)=(f\circ {\pi }_2)(x_1,x_2)=f(x_2)$. Also observe that $h$ is defined precisely so that $h\circ \pi =f$. Moreover, if $\tilde{h}:X/E\to Y$ is any set map such that $\tilde{h}\circ \pi =f$, then for every $[x]\in X/E$ we must have $\tilde{h}([x])=\tilde{h}(\pi (x))=(\tilde{h}\circ \pi )(x)=f(x)=h([x])$. This proves our map $h$ is the unique set map such that $h\circ \pi =f$. We've thus proven $f$ factors uniquely through $\pi$:

This proves $\pi :X\to X/E$ is a coequalizer of the parallel arrows ${\pi }_1,{\pi }_2:E\to X$.

1. Suppose $G$ is a finite abelian group, say of order $n$. By Lagrange's Theorem we then have $ng=0_G$ for every $g\in G$. But this exactly says that every $g\in G$ is torsion when $G$ is viewed as a $\mathbf{Z}$-module, and hence $G$ is torsion as a $\mathbf{Z}$-module.

2. One nice example is that of the group $U$ of roots of unity in $\mathbf{C}$, although that group is the rare abelian group that uses multiplicative notation. In that group we have (by definition!) that for each $\zeta \in U$ there is some positive integer $n$ with ${\zeta }^n=1$. Noting that the $\mathbf{Z}$-action on $U$ is by $m\star \zeta ={\zeta }^m$ and the "zero element" (i.e., identity) in $U$ is 1, we thus have that every $\zeta \in U$ is torsion when $U$ is viewed as a $\mathbf{Z}$-module.

   Here's another example. Suppose $F$ is an infinite field of positive characteristic; e.g., $F={\mathbf{Z}}_p(t)$. Considered as a $\mathbf{Z}$-module, every element in $F$ is killed by the characteristic of the field, and hence is torsion.

   A third example: the direct product of an infinite number of copies of a fixed finite abelian group, say $G={\displaystyle \prod _{i=1}^{\mathrm{\infty }}{\mathbf{Z}}_p}$.

1. First suppose a nonzero $R$-module $M$ is irreducible. Take any nonzero element $m\in M$ and define an $R$-module morphism $f_m:R\to M$ by $1_R\mapsto m$. (Note that the fact this is an $R$-module morphism implies we must have $f_m(r)=f_m(r\cdot 1_R)=r\cdot f_m(1_R)=r\cdot m$ for every $r\in R$.) The image of $f_m$ is a nonzero submodule of $M$, so by the irreducibility of $M$ the image must be all of $M$, i.e., $f_m$ is surjective. By the First Isomorphism Theorem for Modules we then have an $R$-module isomorphism $R/I\simeq M$, where $I=\ker (f_m)$. ([Examples of submodules#Submodules and ideals|Recall]] that when $R$ is viewed as an $R$-module, the submodules of $R$ correspond exactly to the ideals of the ring $R$.) Now, by the Fourth Isomorphism Theorem for Modules there is a bijection between the submodules of $R/I$ and the submodules of $R$ that contain $I$. Since $M$ is irreducible so is $R/I$, and hence the only submodules of $R/I$ are the zero submodule and the entire module $R/I$; by the correspondence this implies that the only submodules (i.e., ideals) of $R$ that contain $I$ are $I$ itself and $R$. Thus, $I$ is a maximal ideal.

   The converse direction is almost identical. Suppose $M$ is an $R$-module and $M\simeq R/I$ for some maximal ideal $I$ of $R$. (Note that this implies $M$ is nonzero, since maximal ideals are proper.) Since $I$ is a maximal ideal of $R$, the only ideals in $R/I$ are the zero ideal and the entire ring $R/I$. It follows that the only submodules of $M$ are the zero submodule and the entire module $M$ itself. Thus, $M$ is irreducible.

2. Suppose $f:M_1\to M_2$ is a nonzero morphism between irreducible $R$-modules. The image of $f$ is a nonzero submodule of $M_2$, so by the irreducibility of $M_2$ it must equal all of $M_2$, i.e., $f$ is surjective. The kernel of $f$ is a proper submodule of $M_1$, so by the irreducibility of $M_1$ it must be trivial, i.e., $f$ is injective. Thus, $f$ is an isomorphism.

3. We already know that ${\mathrm{End}}_R(M)$ is a ring, so we just need to verify it is a division ring, i.e., every nonzero element is invertible. So take any nonzero endomorphism $f:M\to M$. Since $M$ is irreducible, by the previous part we have that $f$ must be an isomorphism. Thus, there is an inverse morphism $f^{-1}:M\to M$, i.e., an inverse endomorphism. That's all we needed.