Difference between revisions of "Exercises:Rings and Modules - 2016 - 1/Problem 1"
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===Problem 1=== | ===Problem 1=== | ||
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| − | {{ | + | ====Part A==== |
| − | ====Solution==== | + | Let {{M|R}} be a [[u-ring]]. Fix an {{M|a\in R}} and define a {{link|homomorphism|u-ring}}: |
| − | {{ | + | * {{M|\varphi_a:R[T]\rightarrow R}} by {{M|\varphi_a:P(T)\mapsto P(a)}} - evaluation at {{M|a}}. |
| + | By restriction of scalars every {{M|\varphi_a}} gives the target {{M|R}} the structure of an [[R-module|{{M|R[T]}}-module]], which we denote {{M|R_a}}. | ||
| + | |||
| + | Show that for {{M|a,b\in R}} that: | ||
| + | * there is an {{M|R[T]}}-module {{link|isomorphism|module}} between {{M|R_a}} and {{M|R_b}} | ||
| + | {{iff}} | ||
| + | * {{M|1=a=b}} | ||
| + | =====Solution===== | ||
| + | ====Part B==== | ||
| + | Let {{M|M}} be an {{M|R}}-module. Show that there is a [[surjection]] from a [[free module|free {{M|R}}-module]] ''onto'' {{M|M}}. | ||
| + | =====Solution===== | ||
| + | ====Part C==== | ||
| + | Show that the [[Z-Module|{{M|\mathbb{Z} }}-module]], {{M|\mathbb{Q} }}, is ''not'' free. | ||
| + | =====Solution===== | ||
<noinclude> | <noinclude> | ||
==Notes== | ==Notes== | ||
Latest revision as of 16:05, 19 October 2016
Contents
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]
- [ilmath]a=b[/ilmath]
Solution
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].
Solution
Part C
Show that the [ilmath]\mathbb{Z} [/ilmath]-module, [ilmath]\mathbb{Q} [/ilmath], is not free.
Solution
Notes
References
