Algebra question - rings and ideals

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SUMMARY

The discussion focuses on the relationship between an ideal I in a commutative ring R and the ideal (I) generated in the polynomial ring R[x]. It is established that (I) consists of polynomials with coefficients in I, defined as (I) = {a*p(x): a is in I and p(x) in R[x]}. The participants clarify that since R is commutative, the ideal (I) is indeed a subset of the set A, which contains polynomials with coefficients in I. The conclusion emphasizes that A is an ideal of R that contains I, thus confirming the properties of (I).

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[SOLVED] Algebra question - rings and ideals

Homework Statement


Let R be a (nonzero) commutative ring with identity and I be an ideal of I. Denote (I) the ideal of R[x] generated by I. The book says that (I) is the set of polynomials with coefficients in I. Why is that?

The Attempt at a Solution



Call A the set of polynomials with coefficients in I.

R is commutative and so is R[x], therefor (I) is simply given by

(I) = {a*p(x): a is in I and p(x) in R[x]}

So clearly (I) is a subset of A.

But for the other inclusion, given b_0+...+b_nx^n in A, we need to find an element a in I and a set {a_0,...,a_n} in R such that a*a_i = b_i for all i=1,...,n.

How is this achieved??
 
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If b_0+...+b_nx^n is in A, then b_0, ..., b_n are all in I, and thus in (I). The latter is an ideal, so b_kx^k is certainly in (I), and so is the sum of such things.

Edit:
R is commutative and so is R[x], therefor (I) is simply given by

(I) = {a*p(x): a is in I and p(x) in R[x]}
I don't really see how this follows. I would instead prove that A is an ideal of R that contains I and hence contains (I) by definition.
 
Last edited:
Of course it does not follows! I was confused.
 

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