# Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

• Feb 8th 2013, 05:58 PM
Bernhard
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 $\displaystyle R[x]/I[x] \cong (R/I)[x]$

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 $\displaystyle R[x]/I[x] \cong (R/I)[x]$.

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 $\displaystyle \phi : R \rightarrow S$ is a homomorphism of rings, then the kernel of $\displaystyle \phi$ is an ideal of R, the image of $\displaystyle \phi$ is a subring of S, and $\displaystyle R / ker \phi$ is isomorphic as a ring to $\displaystyle \phi (R)$

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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 $\displaystyle \phi$ = 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 $\displaystyle \phi$ = I[x] we have that

$\displaystyle R[x]/ ker \phi = R[x] / I[x][ \cong \phi [ R[x] ]$

So we require $\displaystyle \phi [R[x]] = (R/I) [x]$

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

Would appreciate some help in clarifying this matter.

Peter
• Feb 8th 2013, 06:19 PM
johng
Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2
Who is $\displaystyle \phi$?

Presumably $\displaystyle \phi(a_kx^k+a_{k-1}x^{k-1}+...+a_0)=(a_k+I)x^k+(a_{k-1}+I)x^{k-1}+...+a_0+I$

With this definition of $\displaystyle \phi$, it's "easy" to show $\displaystyle \phi$ is a homomorphism with image (R/I)[x] and kernel I[x].
• Feb 8th 2013, 07:09 PM
Deveno
Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2
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 +...+ (an+I)xn in (R/I)[x], we clearly have for

g(x) = a0 + a1x +...+ anxn, φ(g(x)) = f(x), which shows that φ is surjective.
• Feb 8th 2013, 08:42 PM
Bernhard
Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2
Thanks Deveno!