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I am trying to understand Yuchen's answer to my other question here. The first line is

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

I would like to understand please two things:

$i)$ Can I tell that this is true just by looking at it? It seems weird to me that this is true, because even though the polynomial $x^3+1$ appears on both sides, polynomials seem to behave very differently in different rings.

$ii)$ What is the formal justification for this?

I have tried to prove this isomorphism by finding an explicit map; if we let $I=(2,(x^3+1))$ in $\mathbb{Z}[x]$ and we let $J=((x^3+1))$ in $\mathbb{F}_2[x]$, my best guess was to map $p(x)+I$ to $\overline{p(x)}+J$, where $\overline {p(x)}$ means reducing the coefficients $\pmod2$. However, I think this doesn't work, because our map doesn't depend at all on $(x^3+1)$ being on the left or right sides; I could have different polynomials replacing $x^3+1$ on the left and right sides. So I think the map is most likely not an isomoprhism.

I guess your exercises must come in at some point because I don't think I've used them yet.

Edit for Hurkyl:

I completed your exercises, and I'm still having trouble with your last line. I start like this: $\dfrac{\mathbb{Z}[x]}{(2, x^3+1)} = \dfrac {\mathbb{Z}[x]}{(2)+(x^3+1)} \cong \dfrac {\mathbb{Z}[x]/(2)}{((2)+(x^3+1))/(2)}$.

I know that $\mathbb{Z}[x]/(2) \cong \mathbb{F}_2[x]$. So I must show that $((2)+(x^3+1))/(2) \cong (x^3+1)$, in $\mathbb{F}_2[x]$ and then I think we are done; however, I'm not sure how to do it. Please let me know if I'm right/wrong on this, but I think that if $I \subseteq J$, then $(I+J)/I = J/I$. However, when $I \not \subseteq J$, I think $J/I$ does not make sense under the standard definition.

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That's actually right.

In fact, it's even easier than you describe; you don't even have to reduce the coefficients, you simply need to apply the homomorphism $\mathbb{Z} \to \mathbb{F}_2$. If, for example, you represent elements of $\mathbb{F}_2$ as modular equivalence classes, then you don't have to do any work at all. The image of $7$ is simply $[7]_2$. That's the same element as $[1]_2$, but the point is you don't actually have to bother with the calculation needed to change representations.

This isomorphism is basically just applying the isomorphism theorems. The most useful form for this sort of calculation is

$$ R/(I+J) \cong (R/I) / ((I+J) / I) $$

together with $R[x]/I[x] \cong (R/I)[x]$ and the following facts, which I'll leave as an exercise:

Exercise: Suppose that $\{ j_\alpha \}$ is a set of generators for the ideal $J$ of $R$. Show that $\{ j_\alpha + I \}$ is a set of generators for the ideal $(I+J)/I$ of $R/I$.

Exercise: Suppose that $\{ j_\alpha \}$ is a set of generators for the ideal $J$ of $R$. Show that $\{ j_\alpha \}$ is a set of generators for the ideal $J[x]$ of $R[x]$

With these, we compute

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

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  • $\begingroup$ Remark: Let $\varphi$ be the projection $R \to R/I$. Given $r \in R$, I would normally just write $r$ for $\varphi(r)$ in any situation where it's clear I mean an element of $R/I$. $\endgroup$
    – user14972
    Apr 22 '18 at 17:11
  • $\begingroup$ Hello. I wrote a reply in an edit to the question, would you mind taking a look? $\endgroup$
    – Ovi
    Apr 22 '18 at 21:07
  • $\begingroup$ @Ovi: The trick is that $i+j \equiv j \pmod I$ whenever $i \in I$ and $j \in J$. $\endgroup$
    – user14972
    Apr 22 '18 at 21:10
  • $\begingroup$ @vi: Also, note that $(I+J)/J \cong J/(I \cap J)$. Some people have an easier time thinking about the latter. $\endgroup$
    – user14972
    Apr 22 '18 at 21:12
  • $\begingroup$ Is there a theorem that gets us from $\dfrac {(2)}{(2) \cap (x^3+1)}$ to $(x^3+1)$ in $\mathbb{F}_2$, or at this point we have to construct an explicit isomorphism? $\endgroup$
    – Ovi
    Apr 22 '18 at 21:36
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It is very clear, you do not have to use so many isomorphism theorems. First, you need to know if $A \cong A'$ under map $f$ and $B \cong B'$ also under map $f$. $B$ is an ideal of $A$, then we have $A/B \cong A'/B'$. Then considering a natuanl map $f$ mapping from $\mathbb{Z}$ to $\mathbb{F}_2$. Then $(2,x^3+1)$ reduces to $(x^3+1)$. Using the conclusion we just said, you get $\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1)$

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