Thread: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

In Dummit and Foote Ch 9 Polynomial Rings, Proposition 2 (page 296 - see attached) reads as follows:

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Proposition 2: Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with co-efficients in I)

Then

In particular, if I is a prime ideal of R then I[x] is a prime ideal of R[x]

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I am concerned with the proof that .

In this proof (i think) we need the First Isomorphism Theorem. Dmmit and Foote give this theorem as part of Theorem 7 in Ch 7 on page 243 (see attached)

They state the First Isomorphism Theorem as follows: (see attached)


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If is a homomorphism of rings, then the kernel of is an ideal of R, the image of is a subring of S, and is isomorphic as a ring to

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I follow the proof of down to the following statement (see attached page 296 D&F)

... ... , which is to say that ker = I[x] = (I), proving the first part of the proposition.


BUT surely we need to now use the First Isomorphism Theorem to show that we have an isomorphism ( I suspect D&F are assuming we know this bit ... )

However, by the FIT, given that ker = I[x] we have that





So we require


But how do we know that the image of R[x] under is (R/I)[x] and not a subset of (R/I)[x]


Would appreciate some help in clarifying this matter.

Peter

Who is ?

Presumably

With this definition of , it's "easy" to show is a homomorphism with image (R/I)[x] and kernel I[x].

yes, indeed, to apply the FIT, we need to know that φ is onto.

but suppose that f(x) is in (R/I)[x], that is f(x) has coefficients in R/I. recall that the NATURAL homomorphism R-->R/I which sends r-->r+I for each r in R IS onto. so every element b in R/I is a+I for SOME a in R.

so given f(x) = (a₀+I) + (a₁+I)x +...+ (a_n+I)xⁿ in (R/I)[x], we clearly have for

g(x) = a₀ + a₁x +...+ a_nxⁿ, φ(g(x)) = f(x), which shows that φ is surjective.

Thanks Deveno!