Originally Posted by

**matt.qmar** Hello!

If we take $\displaystyle Z[x]$ to be the ring of polynomials with integer coefficients, and $\displaystyle I$ to be the subset $\displaystyle I = \{a_nx^n + ... + a_0 \in Z[x] : a_i \in 2Z\}$.

So I is the ring of polynomials with even integer coefficients

So far, I have shown that I is an ideal.

I am trying to

(i) find & describe the number of cosets in $\displaystyle Z[x]/I$

(ii) Prove that $\displaystyle I$ is not a principle ideal of $\displaystyle Z[x]$

(iii) Decide if $\displaystyle I$ is a prime ideal.

Well, for (i), I am thinking there is infinitely many cosets? As long as any $\displaystyle x^i$ has coefficient 1, it forms it's own coset... for example

$\displaystyle x + I \neq x^3 + I \neq x^4 + x^3 + I$ etc... in terms of coset equality... That seems to be it to me...

$\displaystyle x+I=x^3+I\Longleftrightarrow x-x^3\in I$ , which is true...Note that an element

in the quotient ring is uniquely determined by its free coefficient's parity...

(ii) I am not quite too sure how to go about this. If we let $\displaystyle I = <p(x)> $for some $\displaystyle p(x) \in Z[x] $and derive a contradiction (show we can't just take multiples of p?)

First show that $\displaystyle I=\langle 2,x\rangle$ , and then suppose $\displaystyle I=\langle f(x)\rangle$ . Check that this

is impossible by noting that both $\displaystyle 2,\,x\in I$ and checking the possible degree of $\displaystyle f(x)$

(iii) I think this will be easy once (i) is figured out... If $\displaystyle Z[x]/I$ is an integral domain, then $\displaystyle I $ is prime. If $\displaystyle Z[x]/I$ is not integral domain, then $\displaystyle I$ is not prime.

Indeed, this follows at once from (i)

Tonio

Any help appreciated! Thanks in advance!