# 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 `<span class="math display">     \begin{align*}\begin{align*}     A &\to { \operatorname{End} }(B) \\     a &\mapsto (b\mapsto b\cdot \phi(a))     .\end{align*}\end{align*}     <span>`{=html}

-   $B\otimes_A M$ is a left $B{\hbox{-}}$module via the following action: `<span class="math display">     \begin{align*}\begin{align*}     B &\to { \operatorname{End} }(B\otimes_A M) \\     b_0 &\mapsto (b\otimes m \mapsto b_0 b \otimes m)     .\end{align*}\end{align*}     <span>`{=html}

-   This describes a functor: `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html}

    -   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: `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html}

> 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: `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html} `<span class="math display">     \begin{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*}\end{align*}     <span>`{=html}

-   Check $A{\hbox{-}}$bilinearity: `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html} `<span class="math display">     \begin{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*}\end{align*}     <span>`{=html}

-   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? `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html}

## 1.3.L

If $S\subseteq A$ is multiplicative and $M\in A{\hbox{-}}$Mod, describe a natural isomorphism `<span class="math display"> \begin{align*}\begin{align*} \eta: (S^{-1}A)\otimes_A M \to (S^{-1}M) \end{align*}\end{align*} <span>`{=html} as both $S^{-1}A{\hbox{-}}$modules and $A{\hbox{-}}$modules.

Solution

-   Recall the definition `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html}
-   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 $\alpha(S) \subseteq N^{\times}$: `<span class="math display"> \begin{align*} \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} \end{align*} <span>`{=html}

> 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 `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html}

-   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 `<span class="math display">     \begin{align*}\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*}\end{align*}     <span>`{=html} 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 $T$

```{=tex}
\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 $h_0$: `{=tex}     \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 $X\times Y$: `{=tex}     \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:

```{=tex}
\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: `{=tex}     \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 $R\to W$: `{=tex}     \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

``` {.tikz}
\usepackage{tikz-cd}
\begin{document}
\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{document}
```