# More Exercises

## 1.3.K

Note: I think this is an exercise about base change.

Part a: For $$M$$ an $$A{\hbox{-}}$$module and $$\phi: A\to B$$ a morphism of rings, give $$B\otimes_A M$$ the structure of a $$B{\hbox{-}}$$module and show that it describes a functor $$A{\hbox{-}}\text{Mod}\to B{\hbox{-}}\text{Mod}$$.

Solution

• $$B\otimes_A M$$ makes sense: $$B$$ is a $$(B, A){\hbox{-}}$$bimodule with the usual multiplication on the left and the right action

\begin{align*}
A &\to \endo(B) \\
a &\mapsto (b\mapsto b\cdot \phi(a))
.\end{align*}

• $$B\otimes_A M$$ is a left $$B{\hbox{-}}$$module via the following action:

\begin{align*}
B &\to \endo(B\otimes_A M) \\
b_0 &\mapsto (b\otimes m \mapsto b_0 b \otimes m)
.\end{align*}

• This describes a functor:

\begin{align*}
F: A{\hbox{-}}\text{Mod} &\to B{\hbox{-}}\text{Mod} \\
X &\mapsto B\otimes_A X \\
(X\xrightarrow{f} Y) &\mapsto (B\otimes_A X \xrightarrow{\operatorname{id}_B \otimes f} B\otimes_A Y)
.\end{align*}

• Need to check:
• Preserves identity morphism, i.e. $$X\in A{\hbox{-}}$$Mod implies $$F(\operatorname{id}_X) = \operatorname{id}_{F(X)}$$ in $$B{\hbox{-}}$$Mod.
• Preserves composition: $$F(f\circ g) = F(f) \circ F(g)$$.
• Preserving identity morphisms:

• By construction $$X{\circlearrowleft}_{\operatorname{id}_X}$$ maps to $$B\otimes_A X \xrightarrow{\operatorname{id}_B \otimes\operatorname{id}_X} B\otimes_A X$$, can argue that this is the identity map for $$B{\hbox{-}}$$modules.
• Preserving composition:

\begin{align*}
(X\xrightarrow{f} Y \xrightarrow{g} Z) \mapsto (B\otimes_A X \xrightarrow{ \operatorname{id}_B\otimes f} B\otimes_A Y \xrightarrow{\operatorname{id}_B \otimes g} B\otimes_A Z) = (B\otimes_A X \xrightarrow{\operatorname{id}_B \otimes(g\circ f)} B\otimes_A Z )
.\end{align*}

Note: not sure if there’s anything to show here.

Part b: If $$\psi: A\to C$$ is another ring morphism, show that $$B\otimes_A C$$ has a ring structure.

Solution:

• Note $$B\otimes_A C$$ makes sense, since $$C$$ is a left $$A{\hbox{-}}$$module via $$a\mapsto (c\mapsto \psi(a)c)$$.

• Need to define $$(B\otimes_A C, P, M)$$ such that it’s an abelian group under $$P$$ (plus), a monoid under $$M$$ (multiplication), and left/right distributivity.

• Start by defining on cartesian products:

\begin{align*}
P: \qty{B\otimes_A C}^{\times 2} &\to B\otimes_A C \\
P\qty{ (b_1 \otimes c_1), (b_2\otimes c_2)} &= (b_1 +_B b_2) \otimes(c_1 +_C c_2)
,\end{align*}
\begin{align*}
M: \qty{B\otimes_A C}^{\times 2} &\to B\otimes_A C \\
M\qty{ (b_1 \otimes c_1),  (b_2\otimes c_2)} &= (b_1 \cdot_B b_2) \otimes(c_1 \cdot_C c_2)
.\end{align*}

• Check $$A{\hbox{-}}$$bilinearity:

\begin{align*}
P(a\cdot (b_1\otimes c_1),\, (b_2\otimes c_2))
&\coloneqq\qty{ a \cdot (b_1 + b_2)} \otimes(c_1 + c_2)  \\
&= \qty{ (b_1 + b_2)} \otimes a\cdot (c_1 + c_2) \quad\text{since $C$ is a left $A{\hbox{-}}$module} \\
&\coloneqq P((b_1\otimes c_1),\, a\cdot (b_2\otimes c_2))
.\end{align*}
\begin{align*}
M(a\cdot (b_1\otimes c_1),\, (b_2\otimes c_2))
&\coloneqq\qty{a\cdot (b_1 \cdot b_2)} \otimes(c_1 \cdot c_2) \\
&= (b_1 \cdot b_2) \otimes\qty{ a\cdot (c_1 \cdot c_2) } \quad\text{since $C$ is a left $A{\hbox{-}}$module} \\
&\coloneqq M((b_1\otimes c_1),\, a\cdot (b_2\otimes c_2))
.\end{align*}

• So these lift to maps out of $$(B\otimes_A C)^{\otimes 2}$$.

• $$P$$ forms an abelian group: clear because $$+_B, +_C$$ do, and commuting is just done within each factor.

• $$M$$ forms a monoid: clear for some reason.

• Checking distributivity, claim: it suffices to check on elementary tensors and extend by linearity?

\begin{align*}
(b_0 \otimes c_0) \cdot \qty{(b_1 \otimes c_1) + (b_2\otimes c_2) }
&= (b_0\otimes c_0) \cdot \qty{ (b_1 + b_2) \otimes(c_1 + c_2) } \\
&= (b_0(b_1 + b_2)) \otimes( c_0(c_1 + c_2)) \\
&= (b_0 b_1 + b_0 b_2) \otimes(c_0 c_1 + c_0 c_2) \\
&= \cdots
.\end{align*}

## 1.3.L

If $$S\subseteq A$$ is multiplicative and $$M\in A{\hbox{-}}$$Mod, describe a natural isomorphism

\begin{align*}
\eta: (S^{-1}A)\otimes_A M \to (S^{-1}M)
\end{align*}
as both $$S^{-1}A{\hbox{-}}$$modules and $$A{\hbox{-}}$$modules.

Solution

• Recall the definition
\begin{align*}
S^{-1}A\coloneqq\left\{{ {a\over s} {~\mathrel{\Big\vert}~}a\in A, s\in S}\right\} / \sim \\
{a_1 \over s_1} \sim {a_2 \over s_2} \iff \exists s\in S \text{ such that } s\qty{ s_2 a_1 - s_1 a_2 } = 0_A
.\end{align*}
• Similarly $$S^{-1}M = \left\{{{m\over s}}\right\}/\sim$$.

The universal property: in $$A{\hbox{-}}$$Mod, $$M\to S^{-1}M$$ is initial among all morphisms $$\alpha: M\to N$$ such that $$\alpha(S) \subseteq N^{\times}$$:

\begin{center}
\begin{tikzcd}
& S^{-1}M \ar[d, dotted, "{\exists ! \tilde\alpha}"] \\
M\ar[ru, "S^{-1}\cdot"]\ar[r, "\alpha"] & N
\end{tikzcd}
\end{center}


Strategy: define a map $$M\to S^{-1}A \otimes_A M$$ such that $$S$$ is invertible in the image to obtain a map? Show they satisfy the same universal property?

• Since $$M \in A{\hbox{-}}$$Mod, we have an action $$a\cdot m$$, so define

\begin{align*}
\eta: (S^{-1}A)\times M &\to (S^{-1}M) \\
\qty{ {a\over s}, m } &\mapsto {a\cdot m \over s }
.\end{align*}

• The tensor product $$S^{-1}A \otimes_A M$$ makes sense.

• $$S^{-1}A$$ is a right $$A{\hbox{-}}$$module by $$a_0 \mapsto \qty{ {a\over s} \mapsto {a_0 a \over s}}$$.
• $$S^{-1}M$$ is a left $$A{\hbox{-}}$$module by $$a_0 \mapsto (m \mapsto a_0 \cdot m)$$ where the action comes from the $$A{\hbox{-}}$$module structure of $$M$$.
• The map makes sense as an $$A{\hbox{-}}$$module morphism

• $$S^{-1}A \otimes_A M$$ is a left $$A{\hbox{-}}$$module by $$a_0 \mapsto \qty{{a\over s}\otimes m \mapsto {a_0 a \over s} \otimes m}$$
• $$S^{-1}M$$ is a left $$A{\hbox{-}}$$module by $$a_0 \mapsto \qty{ {m\over s } \mapsto {a_0 \cdot m \over s}}$$ using the $$A{\hbox{-}}$$module structure on $$M$$.
• The map makes sense as an $$S^{-1}A{\hbox{-}}$$module morphism

• $$S^{-1}A \otimes_A M$$ is a left $$S^{-1}A{\hbox{-}}$$module by $${a_0\over s_0} \mapsto \qty{ {a\over s}\otimes m \mapsto {a_0 a \over s_0 s} \otimes m }$$
• $$S^{-1}M$$ is a left $$S^{-1}A{\hbox{-}}$$module by $${a_0\over s_0} \mapsto \qty{{m \over s} \mapsto {a_0 \cdot m \over s_0 s} }$$ by the $$A{\hbox{-}}$$module structure on $$M$$.
• Well-defined: ?

• $$A{\hbox{-}}$$bilinear: let $$r\in A$$, then

\begin{align*}
\eta\qty{r \cdot {a\over s}, m}
&\coloneqq\eta\qty{{r\cdot a\over s}, m}  \\
&\coloneqq{\psi(r\cdot a)(m) \over s} \\
&= {r\cdot \psi(a)(m) \over s} \quad\text{since $\psi$ is a ring morphism} \\
&= {\psi(a)(r\cdot m) \over s} \quad\text{since $\psi(a)$ is a ring morphism} \\
&\coloneqq\eta\qty{ {a\over s}, r\cdot m}
.\end{align*}
So this lifts to a map out of the tensor product.

• $$S^{-1}A{\hbox{-}}$$bilinear?

## 1.3.P

Show that the fiber product over the terminal object is the cartesian product.

Solution:

• Recall definition: $$T$$ is terminal iff every object $$X$$ admits a morphism $$X\to T$$.
• Strategy: use both universal products to produce an isomorphism
• Let $${\operatorname{pr}}_X, {\operatorname{pr}}_Y$$ by the cartesian product projections, and $${\operatorname{pr}}_X^T, {\operatorname{pr}}_Y^T$$ be the fiber product projections
• Let $$T_X, T_Y$$ be the maps $$X\to T, Y\to T$$.
• Since $$X\times Y$$ is an object in this category, it admits one unique map to $$T$$
\begin{center}
\begin{tikzcd}
X\times Y\ar[r, "{\operatorname{pr}}_Y"]\ar[d, "{\operatorname{pr}}_X"']\ar[dr, "T_{X\times Y}"] & Y\ar[d, "T_Y"] \\
X\ar[r, "T_X"'] & T
\end{tikzcd}
\end{center}

• But now $$T_Y \circ {\operatorname{pr}}_Y: X\times Y \to T$$ is another such map, so it must equal $$T_{X\times Y}$$.
• Similarly $$T_X \circ {\operatorname{pr}}_X$$ is equal to $$T_{X\times Y}$$.
• Thus $$T_Y \circ {\operatorname{pr}}_Y = T_X \circ {\operatorname{pr}}_Y$$, which is part of the universal property for $$X\times_T Y$$.
• By the universal property of $$X\times Y$$, for every $$W$$ admitting maps to $$X, Y$$ we get the following $$h_0$$:
\begin{center}
\begin{tikzcd}
W \ar[drr, bend left]\ar[rdd, bend right]\ar[dr, dotted, "\exists ! h_0"] & & \\
& X\times Y\ar[r, "{\operatorname{pr}}_Y"]\ar[d, "{\operatorname{pr}}_X"] & Y\ar[d, "T_Y"] \\
& X\ar[r, "T_X"] & T
\end{tikzcd}
\end{center}


Note that $$T$$ doesn’t matter in this particular diagram.

• This gives us the LHS diagram, the RHS comes from the universal property of $$X\times Y$$:
\begin{center}
\begin{tikzcd}
X\times_T Y\ar[drr, "{\operatorname{pr}}_Y^T", bend left]\ar[rdd, "{\operatorname{pr}}_X^T", bend right]\ar[dr, dotted, "\exists ! h_0"] & & \\
& X\times Y\ar[r, "{\operatorname{pr}}_Y"]\ar[d, "{\operatorname{pr}}_X"] & Y\ar[d, "T_Y"] \\
& X\ar[r, "T_X"] & T
\end{tikzcd}
\begin{tikzcd}
X\times Y\ar[drr, "{\operatorname{pr}}_Y", bend left]\ar[rdd, "{\operatorname{pr}}_X", bend right]\ar[dr, dotted, "\exists ! h_1"] & & \\
& X\times_T Y\ar[r, "{\operatorname{pr}}_Y^T"]\ar[d, "{\operatorname{pr}}_X^T"] & Y\ar[d, "T_Y"] \\
& X\ar[r, "T_X"] & T
\end{tikzcd}
\end{center}

• By commutativity, $$h_0 \circ h_1 = \operatorname{id}_{X\times Y}$$ and vice-versa?

## 1.3.Q

Show that if the two squares in this diagram are cartesian, then then outer square is also cartesian:

\begin{center}
\begin{tikzcd}
U \ar[r]\ar[d] & V\ar[d] \\
W \ar[r]\ar[d] & X\ar[d] \\
Y \ar[r] & Z
\end{tikzcd}
\end{center}


Solution:

• Need to show that given two maps $$R\to V$$ and $$R\to Y$$ such that $$(V\to Z) \circ (U\to V) = (Y\to Z) \circ (R\to Y)$$, then there is a unique map $$R\to U$$ giving a commuting diagram:
\begin{center}
\begin{tikzcd}
U\ar[drr, bend left] \ar[rdd, bend right] & & \\
& R\ar[ul, dotted, "\exists !\, ?"] \ar[r]\ar[d] & V\ar[d] \\
& Y \ar[r] & Z
\end{tikzcd}
\end{center}

• Applying the bottom square:
• Need to produce maps $$R\to X$$ and $$R\to Y$$
• We’re given a map $$R\to Y$$ by assumption.
• We can build a map $$R\to X$$ by taking $$(V\to X) \circ (R\to V)$$.
• We then get a map $$R\to W$$:
\begin{center}
\begin{tikzcd}
W\ar[drr, bend left] \ar[rdd, bend right] & & \\
& R\ar[ul, dotted, "\exists !"] \ar[r]\ar[d] & X\ar[d] \\
& Y \ar[r] & Z
\end{tikzcd}
\end{center}

• Applying the top square:
• We have a map $$R\to V$$ by assumption
• We have a map $$R\to W$$ from step 1
• We have maps $$V\to X$$ and $$W\to X$$ from the top square
• We thus obtain
\begin{center}
\begin{tikzcd}
U\ar[drr, bend left] \ar[rdd, bend right] & & \\
& R\ar[ul, dotted, "\exists !"] \ar[r]\ar[d] & V\ar[d] \\
& W \ar[r] & X
\end{tikzcd}
\end{center}