Facts about quotient rings - example

Question

I have three quotient rings:

$R_1 = \frac{\mathbb{Q}[x]}{(x^2 -1)}$
$R_2 = \frac{\mathbb{Q}[x]}{(x^2 +1)}$
$R_3 = \frac{\mathbb{Q}[x]}{((x -1)^2)}$

I am trying to decide whether these are integral domains, fields, if they have any elements $u$ s.t $u^2 = 0$ and if any of them are isomorphic. So far I have:

For the first one: This polynomial is not irreducible, it splits into $x+1$ and $x-1$ which are coprime, so by the chinese remainder theorem we can decompose it and clearly see that it has no such elements $u$. Also it's not an i.d. as the polynomial isn't prime

For the second one, this is a field (the poly. is irreducible), so there are no zero divisors.

For the third one I am having difficulty, we cannot use the chinese remainder theorem because the factors of $(x-1)^2$ are not coprime. It's not a field, but I am having trouble showing whether it is an i.d or if there is such a $u$.

Also for showing whether these rings are isomorphic, is it true that the three properties I am testing for are preserved by isomorphisms? Hence none of them should be isomorphic?

Thanks

So if I divide $(x^2-1)$ into $(x-1)^2$ the coset is $2(x-1) + (x-1)^2$, I'm not quite sure how that helps? — Wooster, Jun 25 '14 at 14:25
In a domain, if $,z^2 = 0,$ then what can you infer about such a nilpotent element $,z,?\ $ Recall that in a domain, by hypothesis, $\ yz = 0,\Rightarrow, y= 0,$ or $,z = 0.\ \ $ — Bill Dubuque, Jun 25 '14 at 14:38
It must be zero because an integral domain has no non-zero integral divisors? — Wooster, Jun 25 '14 at 14:40
@BillDubuque But R3 is not necessarily an integral domain? In fact it's not because the polynomial is not irreducible? — Wooster, Jun 25 '14 at 18:39
@BillDubuque Ah but I am trying to say whether there are any elements $u$ s.t $u^2 = 0$, even if $R_3$ is not an integral domain, that isn't automatically true is it? — Wooster, Jun 25 '14 at 18:47
Oh, I thought you knew that from the multiple hints above. Recall that in $,\Bbb Q[x]/(f(x)),$ the image (coset) $,\bar x,$ of $,x,$ is a (generic) root of $,f(x).,$ Apply this to $,R_3.\ \ $ — Bill Dubuque, Jun 25 '14 at 18:54

score 0 · Answer 1 · answered Aug 12 '14 at 05:04

$R_1 = \frac{\mathbb{Q}[x]}{(x^2 -1)}$
- is not an ID since $(x+1 + (x^2 -1)) \cdot (x-1 + (x^2 -1) = 0 + (x^2 -1)$
$R_2 = \frac{\mathbb{Q}[x]}{(x^2 +1)}$
- is a field: Since $x^2+1$ is irreducible over $\mathbb{Q}$, the ideal $(x^2+1)$ is maximal in $\mathbb{Q}[x]$, thus $R_2$ is a field.
$R_3 = \frac{\mathbb{Q}[x]}{((x -1)^2)}$
- is not an ID since $(x-1 + (x-1)^2) \cdot (x-1 + (x-1)^2) = 0 + (x-1)^2$

Use the fact that every field is an integral domain to fill in the gaps.

Facts about quotient rings - example

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