Homework 3

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 $I=\ker (f_m)$. (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 ${\mathrm{End}}_{R\text{-}\mathbf{\text{Mod}}}(M)$ 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.

1. Let's denote the functor we are creating by ${\bullet }^S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$. It is already defined on objects, so we need to define it on arrows. To that end, let $f:X\to Y$ be a set map. We can define a set map $f^S:X^S\to Y^S$ by sending each map $h:S\to X$ (which is what the elements of the set $X^S$ are) to the map $f\circ h:S\to Y$ (which is an element of the set $Y^S$). In other words, $f^S$ is the "compose with $f$" map. Observe that if $1_X:X\to X$ is an identity arrow in $\mathbf{\text{Set}}$, then $(1_X{)}^S$ is the "compose with $1_X$" map, which is the identity on $X^S$. (Sanity check: for any $h:S\to X$, the composition $1_X\circ h$ is the same set map $h:X\to S$.)

   It remains to check that this map on arrows is compatible with composition, so suppose $f:X\to Y$ and $g:Y\to Z$ are two set maps. For any $h\in X^S$ (i.e., set map $h:S\to X$), observe that $(g\circ f{)}^S(h)=(g\circ f)\circ h=g\circ (f\circ h)=g^S(f^S(h))$, where the second equality held by the associativity of function composition. Thus, we do indeed have $(g\circ f{)}^S=g^S\circ f^S$ and hence our arrow map is compatible with composition. We therefore have defined a functor ${\bullet }^S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$, as desired.

2. Define the functor ${\bullet }^S\times S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$ by sending each set $X$ to the set $X^S\times S$, and each set map $f:X\to Y$ to the set map $f^S\times 1_S:X^S\times S\to Y^S\times S$. (At the level of elements this is simply $(f^S\times 1_S)(h,s)=(f\circ h,s)$.) It is straightforward to show this does indeed define a functor, with the verification being nearly identical to the steps used in part 1.

3. Recall that a natural transformation $\tau :{\bullet }^S\times S\Rightarrow I$ is the data consisting of a set map ${\tau }_X:X^S\times S\to X$ for every set $X$, such that for every set map $f:X\to Y$ the diagram below commutes:

   In the case of our evaluation map $e_X:X^S\times S\to X$ we are being asked to show that the diagram below is commutative:

   Let's chase some elements. Starting from the top-left corner, let $(h,s)$ be any element of the set $X^S\times S$. Going across and down corresponds to $f(e_X(h,s))=f(h(s))$, while going down and then across corresponds to $e_y((f^S\times 1_S)(h,s))=e_Y(f\circ h,s)=(f\circ h)(s)$. By the definition of function composition in $\mathbf{\text{Set}}$, we always have $f(h(s))=(f\circ h)(s)$, and so the diagram does indeed commute. Thus, evaluation does indeed define a natural transformation. (Did you ever really doubt it?)

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. 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 uniquely 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 $A,B:n\to m$.