# Finding all the ideals of $\mathbb{Z} [x]/(2, (x^3+1))$?

This is from Dummit and Foote (Section 9.2):

7. Determine all the ideals of the ring $\mathbb{Z} [x]/(2, (x^3+1))$.

This is my attempt to understand what's going on:

My plan is to find a nice list of cosets of $(2, x^3+1)$; hopefully once I do, I will be able to tell what kind of nice ring it is isomorphic to.

I wrote out the definition: $(2, (x^3+1)) = \{2p_1(x) + p_2(x) (x^3+1) |p_1(x), p_2(x) \in \mathbb{Z} [x]\}$. However, I cannot really get a grip on what this ideal looks like and what it's cosets are. So $2p_1(x)$ gives us all polynomials with even coefficients; $p_2(x) (x^3+1)$ gives us $some$ of the polynomials with some odd coefficients. All the polynomials from this term will have degree $\ge 3$. They will have a root at $x = -1$, and they also have $(x^2-x+1)$ as a factor.

How should I proceed?

Since $$\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1) ,$$

and $$x^3+1=(x+1)(x^2+x+1)$$

in $\mathbb{F}_2[x]$, and $(x^2+x+1)$ is irreducible polynomial over $\mathbb{F}_2$.

In addition, every ideal in $\mathbb{F}_2[x] /(x^3+1)$ corresponds to an ideal $I$ of $\mathbb{F}_2[x]$ which contains $(x^3+1)$. That is $$(x^3+1)\subset I .$$ Since $\mathbb{F}_2[x]$ is a principal ideal domain, we can write $I=(f(x))$, where $f(x)\in\mathbb{F}_2[x]$. Thus we have $$(x^3+1)=(x+1)(x^2+x+1)\subset I=(f(x)) ,$$

which means $$f(x)|(x+1)(x^2+x+1)$$ in $\mathbb{F}_2[x]$.

So $f(x)=1, (x+1), (x^2+x+1)$ or $(x^3+1)$ since $\mathbb{F}_2[x]$ is unique factorization domain. And by corresponding theorem, ideals in $\mathbb{F}_2[x]/(x^3+1)$ are $(x+1), (x^2+x+1), (0), (1)$.

• +1 Nice, this is more or less how I would've written it. Technically, I think you're missing one ideal, namely the whole ring. A mild rephrasing of this answer: by the Chinese remainder theorem, $\mathbb{F}_{2}[X]/(X^{3}+1) \cong \mathbb{F}_{2}[X]/(X-1) \times \mathbb{F}_{2}[X](X^{2}+X+1)$. Both $\mathbb{F}_{2}[X]/(X-1)$ and $\mathbb{F}_{2}[X](X^{2}+X+1)$ are fields, each having two ideals, and so there are precisely four ideals of the product. One can pull these ideals back under the CRT isomorphism to determine them explicitly, which gives the ideals found in your answer. Apr 20 '18 at 21:07
• @AlexWertheim I was thinking CRT as well. It's at least easy to count this way Apr 20 '18 at 22:32
• Hello Yuchen, thank you for your answer. Only now I have time to look at it in detail; if I can ask, how do we know that $\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1)$? I have tried to prove this by finding an explicit isomorphism; if we let $I = (2, (x^3+1))$ in $\mathbb{Z}[x]$ and we let $J = ((x^3+1))$ in $\mathbb{F}_2[x]$, I tried mapping $p(x)+I$ to $\overline{p(x)}+J$, where $\overline{p(x)}$ means reducing the coefficients $\pmod 2$. However, I think this doesn't work because our map doesn't depend at all on $(x^3+1)$; I could have differing such polynomials in $I$...
– Ovi
Apr 22 '18 at 16:31
• ... and $J$, and it would still work; so I think that map must not be an isomorphism.
– Ovi
Apr 22 '18 at 16:31
• I have also posted these comments as a question here in case you want to answer it: math.stackexchange.com/questions/2748999/…
– Ovi
Apr 22 '18 at 16:52