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How can we determine whether the quotient rings $R/I$ where where $R=\mathbb Z[x], \mathbb Z_2[x], \mathbb Z_3[x]$ and $I=(x^2+1), (x^2+x+1)$ are fields, principal ideal domains, unique factorization domains or integral domain? (Of course they are all integral domains, but I am looking for the strictest condition.) Thank you.

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Are you quite sure that they’re all integral domains? What do you get when you simplify $(x+1)^2$ in $\Bbb{Z}_2[x]$? What about $(x+2)^2$ in $\Bbb{Z}_3[x]$? –  Brian M. Scott Mar 4 '12 at 12:41
Thank you, @Brian . –  Lost Mar 4 '12 at 13:26
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2 Answers

Note that $R/I$ is a field if and only if $I$ is maximal and $R/I$ is an integral domain if and only if $I$ is prime.

Also note that if $F$ is a field then $(p(x))$ is maximal in $F[x]$ if and only if $p(x)$ is irreducible over $F$.

Note that $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are fields. Now use that $f(x) \in F[x]$ with $\operatorname{deg}{(f)} \in \{2,3\}$ is reducible if and only if it has a root in $F$.

First let $p(x) = x^2 + 1$ and $R = \mathbb{Z}_3 [x]$. Note that $p(0) = 1 \neq 0$, $p(1) = 2 \neq 0$ and $p(2) = 2 \neq 0$. Hence $p(x)$ is irreducible and therefore $I = (p(x))$ is maximal and therefore $R/I$ is a field in this case. Every field is also an integral domain, a PID and a UFD, so you're done.

Now let $p(x) = x^2 + x + 1$ and $R = \mathbb{Z}_2[x]$. Is $p(x)$ irreducible over $R$?

Hope this helps. I'll add some more details later.

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Thank you, Matt. –  Lost Mar 4 '12 at 13:23
+1 $\\\\\\\\\\\\\\\\$ –  user21436 Mar 4 '12 at 13:33
Thanks, @KannappanSampath : ) –  Matt N. Mar 4 '12 at 13:59
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Her are some (disorganized) ramblings.

First of all: they are not all integral domains. For example, $x^2+1$ is reducible modulo 2, namely as $(x+1)^2$, so $\mathbb{Z}_2[x]/(x^2+1)$ is not an integral domain.

Similarly, $x^2+x+1=x^2+x+1-3x=(x-1)^2$ mod 3, so $\mathbb{Z}_3[x]/(x^2+x+1)$ is not an integral domain either.

For the other case, note that since $\mathbb{Z}_2$ and $\mathbb{Z}_2$ are fields, $\mathbb{Z}_2[x]$ and $\mathbb{Z}_3[x]$ are PID's. Hence so are their quotient rings.

Note also that $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$.

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Thank you, Fredrik –  Lost Mar 4 '12 at 13:23
Incidentally, why is $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$? –  Lost Mar 4 '12 at 13:25
@Lost: Every polynomial in $\Bbb{Z}[x]$ is congruent mod $(x^2+1)$ to a polynomial of the form $m+nx$, and $x^2\equiv -1 \pmod{(x^2+1)}$, so $m+ni$ corresponds to the equivalence class of $m+nx \pmod {(x^2+1)}$. –  Brian M. Scott Mar 4 '12 at 13:32
@Lost: Think of modding out by a polynomial in $\mathbb{Z}[x]$ as a way to add new elements to $\mathbb{Z}$. If you for example want an element $a$ such that $a^2=-1$, then note that $a$ must satisfy the polynomial $x^2+1=0$. So you see that the image of $x$ in $\mathbb{Z}[x]/(x^2+1)$ acts precisely that way. –  Fredrik Meyer Mar 4 '12 at 20:28
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