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I am trying to understand quotient rings and it would be really helpful if someone could show me a general way of mapping quotient rings to simpler rings.

Consider $\Bbb Z_6[x] / (x^2+5)$. I know with coefficients in $\Bbb Z_6$ we can factor

$(x^2+5) = (x-1)(x-5)$.

My guess is the answer is $\Bbb Z_6 \times \Bbb Z_6$ if we define the homomorphism $\varphi: f(x) \rightarrow (f(1),f(5)) \in \Bbb Z_6 \times \Bbb Z_6$. Am I correctly applying the first isomorphism theorem?

What about $\Bbb Z_4[x] / (x^2+1)$ which isn't reducible in $\Bbb Z_4$. I know the cosets are of the form $ax+b$. I think the quotient is a field since $(x^2+1)$ is irreducible. But I read somewhere that all finite fields have set order that is prime, and so I'm confused.

I do not have a lot of intuition here.

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  • $\begingroup$ I can't understand your question. do you want to find a mapping from $\mathbb Z_6[x]/{(x^2+5)} \rightarrow \mathbb R \times \mathbb R$? $\endgroup$ – Ram Mar 11 '13 at 10:37
  • $\begingroup$ You wanted to mean $\Bbb Z_6\times\Bbb Z_6$ instead of $\Bbb R\times\Bbb R$, no? $\endgroup$ – Berci Mar 11 '13 at 10:38
  • $\begingroup$ The correct theorem is that all finite fields have prime power order, but the converse is not true (e.g. $\mathbb{Z}_4$ is not a field, it has zero divisors). $\endgroup$ – hardmath Mar 11 '13 at 10:41
  • $\begingroup$ @hardmath, I think you meant to say that all the finite fields have prime power order, but it is not true that any ring with prime power order is a field...? $\endgroup$ – DonAntonio Mar 11 '13 at 12:16
  • $\begingroup$ You guess for the answer, @cowchee...the answer to what?! $\endgroup$ – DonAntonio Mar 11 '13 at 12:17
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This is not correct. The map $\mathbb{Z}/6\mathbb{Z}[x]/[x^{2}+5]$ gives you a ring with 36 elements, $ax+b,a,b\in \mathbb{Z}/6\mathbb{Z}$. Using Chinese remainder's theorem, this only gives you $$ \mathbb{Z}/6\mathbb{Z}[x]/[x^{2}+5]=\mathbb{Z}/6\mathbb{Z}[x]/[x-1]\oplus \mathbb{Z}/6\mathbb{Z}[x]/[x-5]\cong \mathbb{Z}/6\mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z} $$

In the second case $(x^{2}+1)$ is not a maximal ideal(thanks for the comment to point it out), and the quotient is just $ax+b$, $a,b\in \mathbb{Z}/4\mathbb{Z}$. But this is not necessarily a field partly because the base ring is not a domain. For example $2\times 2=0$ at here.

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  • $\begingroup$ A maximal ideal? I think that between the pair of them $2$ and $x^2+1$ generate an even bigger one. Even that is promptly topped by tha ideal generated by $2$ and $x+1$! Hint: The quotient of a commutative ring moded out by a maximal ideal is a field :-) $\endgroup$ – Jyrki Lahtonen Mar 11 '13 at 10:54
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We can think of $\mathbb{Z}_6[x]/(x^2 + 5)$ as a simple overring of $\mathbb{Z}_6$ which introduces a "new" root of $x^2 + 5$, and we may represent this as $\mathbb{Z}_6[r]$. As you point out $\pm 1 \in \mathbb{Z}_6$ are already roots, as would be evident if we wrote instead $x^2 + 5 = x^2 - 1$. Ring element $r$ does not belong to $\mathbb{Z}_6$.

Let $r$ be this new root of $x^2 - 1$, and work out the details of the addition and multiplication for this finite ring (with zero divisors, obviously). The addition is somewhat like your idea, but involves two copies of $\mathbb{Z}_6$, i.e $a + br$ where $a,b \in \mathbb{Z}_6$ is a typical element; addition is done componentwise.

For multiplication we need only apply that $r^2 = 1$ (since $r^2 + 5 = 0$ as a result of the quotient ring homomorphism). As a result of expressing $r^2$ in lower order terms, everything in $\mathbb{Z}_6[r]$ is of the form $a + br$.

Here's a way to think about this generally. If we have a ring $R$ and ideal $I$ of $R[x]$ for indeterminate $x$ such that $R \cap I = \{0\}$, then the quotient ring $R[x]/I$ is isomorphic to simple overring $R[r]$ of $R$, where $r$ is the image of $x$ in an "evaluation homomorphism" mapping $R[x]$ to $R[r]$ by $f(x) \mapsto f(r)$. The kernel of this homomorphism is $I$.

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