Knowing that R is a commutative ring and R[X] is a free R-module.
It seems obvious because any R-module can be written as quotient of a free R-module and any of its submodules.
As R[X²] is a submodule of R[X] over R, then it is an ideal of R.
It turn to simply show a bijective homomorphism, but I encounter problems describing the elements in this relations.
Hi, if I correctly understood the problem,
you have a direct sum R[x] =R[x^2]+x*R[x^2] of R-modules, second summand consisting of sums of odd powers of x. Both summands are obviously isomorphic, and both are isomorphic to R[x] as R-modules (substitute t for x^2). The quotient R[x]/R[x^2] is isomorphic as R-module to
x*R[x^2], hence to R[x^2] and consequently to R[x]. Best regards, Yuri
Dear Kelvin,
I have nothing to add to the solution given by Yuri, but I am puzzled by your assertion : " As R[X²] is a submodule of R[X] over R, then it is an ideal of R ". First, R[X²] is not contained in R, so it cannot be an ideal of R. Second, it is not even an ideal of R[X] since it is not stable under multiplication by X. You could say that R[X] and R[X²] are isomorphic as R-modules (see Yuri's argument), and even as R-algebras (same argument), but isomorphism is not equality. Did I misunderstand something ?
Best, Thong
Hello Thong, you haven't misunderstood. Actually I have mistaken.
What you said makes the whole sense, and I built the proof thru the standard way, writing a bijective morphism, and of course reducing the kernel on that quotient.
And also thanks to Yuri, it's easier when we see R[X] as direct sum like you helped me.
Again, nothing to really add to what Yuri said, but more generally you could say that any free R-module with a countable basis is isomorphic to R[x] (as an R-module) and it is fairly straightforward that R[x]/R[x^2] is a free R-module with a countable basis (x,x^3,x^5,...).
The question is essentially equivalent to saying that {1,3,5,..} is in bijective correspondence to {1,2,3...}
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