Exercises:Rings and Modules - 2016 - 1

Problems

Problem 1

Part A

Let [ilmath]R[/ilmath] be a u-ring. Fix an [ilmath]a\in R[/ilmath] and define a homomorphism:

• [ilmath]\varphi_a:R[T]\rightarrow R[/ilmath] by [ilmath]\varphi_a:P(T)\mapsto P(a)[/ilmath] - evaluation at [ilmath]a[/ilmath].

By restriction of scalars every [ilmath]\varphi_a[/ilmath] gives the target [ilmath]R[/ilmath] the structure of an [ilmath]R[T][/ilmath]-module, which we denote [ilmath]R_a[/ilmath].

Show that for [ilmath]a,b\in R[/ilmath] that:

• there is an [ilmath]R[T][/ilmath]-module isomorphism between [ilmath]R_a[/ilmath] and [ilmath]R_b[/ilmath]

if and only if

• [ilmath]a=b[/ilmath]

Part B

Let [ilmath]M[/ilmath] be an [ilmath]R[/ilmath]-module. Show that there is a surjection from a free [ilmath]R[/ilmath]-module onto [ilmath]M[/ilmath].

Part C

Show that the [ilmath]\mathbb{Z} [/ilmath]-module, [ilmath]\mathbb{Q} [/ilmath], is not free.

Problem 2

Part A

Compute the homology groups at [ilmath]\mathbb{Q}^5[/ilmath] and [ilmath]\mathbb{Z}^5[/ilmath] of the the following complexes:

1. [ilmath]\xymatrix{ \mathbb{Q}^3 \ar[r]^{f_1} & \mathbb{Q}^5 \ar[r]^{f_2} & \mathbb{Q}^2 } [/ilmath] with [ilmath]f_1:=\begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}[/ilmath] and [ilmath]f_2:=\begin{pmatrix} 1 & 1 & 0 & -1 & -1 \\ 2 & 2 & 0 & -2 & -2\end{pmatrix}[/ilmath]
2. [ilmath]\xymatrix{ \mathbb{Z}^3 \ar[r]^{f_1} & \mathbb{Z}^5 \ar[r]^{f_2} & \mathbb{Z}^2 } [/ilmath] with [ilmath]f_1:=\begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}[/ilmath] and [ilmath]f_2:=\begin{pmatrix} 1 & 1 & 0 & -1 & -1 \\ 2 & 2 & 0 & -2 & -2\end{pmatrix}[/ilmath]

Part B

Let:

[ilmath]\xymatrix{ M_1 \ar[d]_{f_1} \ar[r] & M_2 \ar[r] \ar[d]_{f_2} & M_3 \ar[r] \ar[d]_{f_3} & M_4 \ar[r] \ar[d]_{f_4} & M_5 \ar[d]_{f_5} \\ N_1 \ar[r] & N_2 \ar[r] & N_3 \ar[r] & N_4 \ar[r] & N_5 }[/ilmath]

be a commutative diagram of [ilmath]R[/ilmath]-modules in which the rows are exact sequences. Show the Five lemma:

• If [ilmath]f_1,f_2,f_4[/ilmath] and [ilmath]f_5[/ilmath] are isomorphism then so is [ilmath]f_3[/ilmath]

Problem 3

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Solution

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Problem 4

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TODO: Solution

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