Direct Sum of Cyclic Submodules
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From: cwldoc on 17 Mar 2010 09:18 Z[x] = ring of polynomials over the integers. Let R = {all f(x) on Z[x] such that f(x) = a0 + a2 x^2 + ... + an x^n for some nonnegative even integer, n, and the a's are integers}. We are asked to prove that the R-module Z[x] does not decompose into a direct sum of cyclic submodules. However, I do not see why this is true. Why can't we have R-module Z[x] = direct sum of <1> and <x> ? Since <1> = R1 = R and <x> = Rx = {all f(x) on Z[x] such that f(x) = a1 x + a3 x^3 + ... + an x^n for some nonnegative odd integer, n, and the a's are integers}, it would seem that any element of Z[x] is the sum of an element of <1> and an element of <x>, and that the choice of these elements is unique.
From: Arturo Magidin on 22 Mar 2010 15:57 On Mar 17, 12:18 pm, cwldoc <cwl...(a)aol.com> wrote: > Z[x] = ring of polynomials over the integers. > > Let R = > {all f(x) on Z[x] > such that f(x) = a0 + a2 x^2 + ... + an x^n for some nonnegative even integer, n, and the a's are integers}. > > We are asked to prove that the R-module Z[x] does not decompose into a direct sum of cyclic submodules. > > However, I do not see why this is true. Why can't we have > R-module Z[x] = direct sum of <1> and <x> ? > > Since <1> = R1 = R > > and <x> = Rx = {all f(x) on Z[x] > such that f(x) = a1 x + a3 x^3 + ... + an x^n for some nonnegative odd integer, n, and the a's are integers}, > > it would seem that any element of Z[x] is the sum of an element of <1> and an element of <x>, and that the choice of these elements is unique. I agree (at least, I don't see anything wrong right now); are you sure you've got the correct R? If you had R to be the ring of all polynomials with no linear term (that is R = {f(x) : f(x) = a_0 + a_2x^2 + a_3x^3 + ... +a_nx^n}) then it seems to me that the conclusion would follow, for example. -- Arturo Magidin
From: cwldoc on 22 Mar 2010 14:07
> On Mar 17, 12:18 pm, cwldoc <cwl...(a)aol.com> wrote: > > Z[x] = ring of polynomials over the integers. > > > > Let R = > > {all f(x) on Z[x] > > such that f(x) = a0 + a2 x^2 + ... + an x^n for > some nonnegative even integer, n, and the a's are > integers}. > > > > We are asked to prove that the R-module Z[x] does > not decompose into a direct sum of cyclic submodules. > > > > However, I do not see why this is true. Why can't > we have > > R-module Z[x] = direct sum of <1> and <x> ? > > > > Since <1> = R1 = R > > > > and <x> = Rx = {all f(x) on Z[x] > > such that f(x) = a1 x + a3 x^3 + ... + an x^n for > some nonnegative odd integer, n, and the a's are > integers}, > > > > it would seem that any element of Z[x] is the sum > of an element of <1> and an element of <x>, and that > the choice of these elements is unique. > > I agree (at least, I don't see anything wrong right > now); are you sure > you've got the correct R? If you had R to be the ring > of all > polynomials with no linear term (that is R = {f(x) : > f(x) = a_0 + > a_2x^2 + a_3x^3 + ... +a_nx^n}) then it seems to me > that the > conclusion would follow, for example. > > -- > Arturo Magidin You're right on target! Thanks a million! I've been racking my brains for 2 weeks to figure out what was wrong. The stated definition of R was actually {f(x) on Z[x] such that f(x) = a0 + a2 x^2 + ... + an x^n} I mistakenly assumed this to mean all polynomials whose coefficients of odd powers of x are zero. However, looking again at the above definition, it is now obvious that what is intended is all polynomials with no linear term! Thanks again for your help. |