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basic question about field theory

I know that if $p(x)$ is an irreducible polynomial in $\mathbb{Q}$, then $$\frac{\mathbb{Q}[x]}{\langle p(x) \rangle}$$ is a field.

But if $p(x)$ is irreducible over $\mathbb{Z}$, then we have $$\frac{\mathbb{Z}[x]}{\langle p(x) \rangle}$$

Now I am wondering that how these two fields differ?

I am extremely confused because the polynomial $x^3+9x+6$ is irreducible over $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ so we have the corresponding quotient fields.

If $\theta$ is a root of $p(x)$ then the element $1+\theta$ as an inverse $\frac{\theta^2-\theta+10}{4}$ in $$\frac{\mathbb{Q}[x]}{\langle p(x) \rangle}$$

but it doesn't lie in $$\frac{\mathbb{Z}[x]}{\langle p(x) \rangle}$$

But both of them are fields. So that element must have inverse in both of them Probably I am missing something but extremely confused.

mr_e_man
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permutation_matrix
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  • If $p\in \Bbb Z[x]$, say $p=x^2-2$, then $\Bbb Z[x]/(p)\cong \Bbb Z[\sqrt{2}]$ is not a field. – Dietrich Burde Nov 08 '22 at 14:43
  • @DietrichBurde, I am saying that $p(x)$ is a polynomial over integers as well. Hence it is definitely an ideal over $\mathbb{Z}[x]$ – permutation_matrix Nov 08 '22 at 14:44
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    On the other hand $\Bbb Q[x]/(x^2-2)\cong \Bbb Q(\sqrt{2})$ is a field, but $\Bbb Z[x]/(x^2-2)$ isn't a field. So your question is in error. For $R/I$ being a field, $I$ should be a maximal ideal. Search for the maximal ideals of $\Bbb Z[x]$ - or see this duplicate. – Dietrich Burde Nov 08 '22 at 14:51

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Irreducibles in UFDs generate prime ideals (and not necessarly maximal ideals), so the quotient in general are integral domains. For your specefic example of $\mathbb{Z}[X]$, maximal ideals have the form $\langle n,p \rangle$, where $n$ is a prime integer and $p$ is a primitive polynomial that is irreducible mod $p$ (u can easily see that the your ideal is not maximal). So when an ideal of $\mathbb{Z}[X]$ is maximal, the quotient the finite field $\mathbb{Z}/p\mathbb{Z}$.

belkacem abderrahmane
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