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$I=\langle x^2,2x,4\rangle$ is an ideal of $\Bbb Z[x]$.

Prove that $I$ is not a principal ideal and find the size of $\Bbb Z[x]/I$.

Using the theorem that ideals are principal iff the generator is irreducible, I think the first part is obvious since $x^2$ is reduced to x(x). I'm slightly thrown by having multiple elements in the generator in this case, so if that complicates things I'd love a heads up.

Mostly I'm confused about finding the size of $\Bbb Z[x]/I$. Noting that $\pi:Z[x]\rightarrow Z[x]/I$ is the canonical homomorphism and it maps $f(x)\mapsto f(x)+I$...but I have no clue how to find the size of this...

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There is no such theorem as the one you state. The definition of "principal" is "having a single generator" --- the reducibility of the generator is irrelevant. – Gerry Myerson Nov 20 '12 at 23:14
up vote 2 down vote accepted

To show it's not principal, you need to show there is no polynomial $f$ such that $I$ is the set of all multiples of $f$. Probably the easiest way to do this is to show that there is no polynomial $f$ such that $x^2$ and $4$ are both multiples of $f$, other than $f=\pm1$, and then show that $\pm1$ isn't in $I$.

For the second part, you have to think about representatives of cosets of $I$ in ${\bf Z}[x]$. Since $x^2$ is in $I$, all coset representatives can be taken to be linear. Since $2x$ is in $I$, the linear term of a coset representative can be taken to be $0$ or $1$. That should get you started.

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For the second part I have a less elementary approach. Take $m=(2,X)$ the ideal of $R=\mathbb{Z}[X]$ generated by $2$ and $X$. Then $R/m\cong\mathbb{Z}_2$, so $m$ is maximal and $I=m^2$. We have an exact sequence of $R$-modules: $0\to m/m^2\to R/m^2\to R/m\to 0$. Note that $m/m^2$ is an $R/m$-vectorspace of dimension $2$ and therefore it has $4$ elements. Now we get that $R/I$ has $8$ elements.

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