Express the following module as a direct sum of cyclic R-modules











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Now



$dfrac{R}{langle (x-x^2)rangle }cong dfrac{R}{langle (x)rangle }oplus dfrac{R}{langle (1-x)rangle }$ since $x,1-x$ have no common factors so by Chinese Remainder Theorem we can say that the above isomorphism holds.------------------------(1)



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle (x-x^2)rangle }
cong dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}$-----------------------(using (1)).



Again



$dfrac{R}{langle (x)rangle } cong dfrac{R}{langle 1-xrangle }cong dfrac{R}{langle 1-2xrangle }$



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}\ cong dfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Thus



$dfrac{Roplus R }{langle( 1-2x,-x^2),(1-x,x-x^2)rangle}congdfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Is my solution correct?Can someone please tell me ?






abstract-algebra modules







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  • Terrible math notation.
    – Wuestenfux
    Nov 24 at 13:29















up vote
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down vote

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Now



$dfrac{R}{langle (x-x^2)rangle }cong dfrac{R}{langle (x)rangle }oplus dfrac{R}{langle (1-x)rangle }$ since $x,1-x$ have no common factors so by Chinese Remainder Theorem we can say that the above isomorphism holds.------------------------(1)



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle (x-x^2)rangle }
cong dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}$-----------------------(using (1)).



Again



$dfrac{R}{langle (x)rangle } cong dfrac{R}{langle 1-xrangle }cong dfrac{R}{langle 1-2xrangle }$



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}\ cong dfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Thus



$dfrac{Roplus R }{langle( 1-2x,-x^2),(1-x,x-x^2)rangle}congdfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Is my solution correct?Can someone please tell me ?






abstract-algebra modules







share|cite|improve this question
























  • Terrible math notation.
    – Wuestenfux
    Nov 24 at 13:29













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Now



$dfrac{R}{langle (x-x^2)rangle }cong dfrac{R}{langle (x)rangle }oplus dfrac{R}{langle (1-x)rangle }$ since $x,1-x$ have no common factors so by Chinese Remainder Theorem we can say that the above isomorphism holds.------------------------(1)



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle (x-x^2)rangle }
cong dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}$-----------------------(using (1)).



Again



$dfrac{R}{langle (x)rangle } cong dfrac{R}{langle 1-xrangle }cong dfrac{R}{langle 1-2xrangle }$



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}\ cong dfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Thus



$dfrac{Roplus R }{langle( 1-2x,-x^2),(1-x,x-x^2)rangle}congdfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Is my solution correct?Can someone please tell me ?






abstract-algebra modules







share|cite|improve this question















Now



$dfrac{R}{langle (x-x^2)rangle }cong dfrac{R}{langle (x)rangle }oplus dfrac{R}{langle (1-x)rangle }$ since $x,1-x$ have no common factors so by Chinese Remainder Theorem we can say that the above isomorphism holds.------------------------(1)



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle (x-x^2)rangle }
cong dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}$-----------------------(using (1)).



Again



$dfrac{R}{langle (x)rangle } cong dfrac{R}{langle 1-xrangle }cong dfrac{R}{langle 1-2xrangle }$



Hence



$dfrac{R}{langle 1-2xrangle}oplus dfrac{R}{langle (-x^2)rangle } oplus dfrac{R}{langle 1-xrangle }oplus dfrac{R}{langle xrangle}\ cong dfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Thus



$dfrac{Roplus R }{langle( 1-2x,-x^2),(1-x,x-x^2)rangle}congdfrac{R}{langle xrangle}oplus dfrac{R}{langle x^2rangle}$



Is my solution correct?Can someone please tell me ?






abstract-algebra modules




abstract-algebra modules






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edited Nov 24 at 9:21

























asked Nov 24 at 5:53









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  • Terrible math notation.
    – Wuestenfux
    Nov 24 at 13:29


















  • Terrible math notation.
    – Wuestenfux
    Nov 24 at 13:29
















Terrible math notation.
– Wuestenfux
Nov 24 at 13:29




Terrible math notation.
– Wuestenfux
Nov 24 at 13:29















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