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}
Links to this page