# $(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$

today I have a problem.

Let $R_1=\mathbb{Z}_2[x] /\langle x^2 -2\rangle$ and $R_2=\mathbb{Z}_2[x] /\langle x^2 -3\rangle$

prove or disprove $R_1$ and $R_2$ are isomorphic.

I felt confuse because $x^2 =2$ and $x^2=3$ have solution in $\mathbb{Z}/2\mathbb{Z}$

I don't know what to do.

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What is $x$? Do you mean $\mathbb Z_2[x]/\left<x^2-2\right>$? –  Thomas Andrews Nov 13 '12 at 15:58
Also, isn't $2=0$ and $3=1$ in $\mathbb Z_2$? –  Thomas Andrews Nov 13 '12 at 15:59
Finally, why are you sure they are not isomorphic? Is the problem to prove that they are not, or is the problem to prove or disprove they are isomorphic? –  Thomas Andrews Nov 13 '12 at 16:00
Isn't the map which is constant on $Z/2Z$, and sending $x$ to $x-1$ an isomorphism between these two rings? –  the L Nov 13 '12 at 17:40
@ThomTyrrell I saw that, and actually I opposed the edit on the grounds we'd need the author's input. Really it's best to be conservative about edits like that. Now instead of being merely unclear the question may be in fact incorrect. It's OK though, things'll get straightened out :) –  rschwieb Nov 13 '12 at 18:54

Both sets have four elements, namely the equivalence classes represented by all polynomials over $\mathbb{Z}_2$ of up to first order. Now make multiplication tables for both sets. If they are essentially the same (up to relabelling entries and reordering rows and columns), the rings are isomorphic. Otherwise they aren't.

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Hint: The coset $\alpha=x+\langle x^2-2\rangle$ satisfies the equation $\alpha^2=0$ in the ring $R_1$. The coset $\beta=x+1+\langle x^2-3\rangle$ satisfies the equation $\beta^2=0$ in the ring $R_2$, because $$\beta^2=(x+1)^2+\langle x^2-3\rangle=x^2+2x+1+\langle x^2-3\rangle=x^2-3+\langle x^2-3\rangle=0.$$

Extend "linearly".

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Can you tell me more?. Why $x^2+\langle x^2 \rangle =0$?. What we will do next step? –  Muniain Nov 14 '12 at 10:03
The polynomials $x^2$ and $0$ belong to the same coset of the ideal $\langle x^2\rangle$ of $\mathbb{Z}_2[x]$. The coset $0+\langle x^2\rangle$ is the zero element of the quotient ring $R_1$. This calculations suggests (but does not guarantee) that there might exist an isomorphism $f:R_1\to R_2$ such that $$f(\alpha)=\beta.$$ Next I suggest that you show that this actually works! –  Jyrki Lahtonen Nov 21 '12 at 8:06
I think isomorphism: $$\varphi : f(x)+\langle x^2 \rangle \mapsto f(x+1)+\langle (x+1)^2 \rangle$$. This is 1-1 mapping and homomorphism. –  Muniain Nov 21 '12 at 12:38
@Firmino: Correct. –  Jyrki Lahtonen Nov 21 '12 at 13:12
It is amazing. Thank you very much. –  Muniain Nov 22 '12 at 5:35

I'm not good at editing.I'm sorry:(
First of all, 2 in Z2 is O, because 2 mod 2=0 (the remainder of the division 2 / 2 is 0), then 3 in Z2 is 1 (3/2= 1 , remainder=1).
=> X^2 = X ^2 - 1
-1 is 2-1=1 in Z2

You should know that -m in Zn is n-m.
=> X^2= X^2 + 1 | +1 X^2 + 1=x^2+2 (2 is 0 in Z2)=>

=> X^2 +1= X^2 | + X ^2

X^2+X^2+1=X^2+ X^2

X^2 (1+1) +1=X^2(1+1)

1+1 is 0 =>

0+1=0


1=0 False.
Your ecuation doesn't have a solution. x ^2=2 means x^2=0, with the only solution x=0, and x^2=3 it means x^2=1, x=1. We have a contradiction: x=1=0 False

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This is very difficult to follow. –  Cameron Buie Nov 13 '12 at 17:16
This is wrong. An isomorphism doesn't have to lift to the identity map on the level of polynomial rings. –  the L Nov 13 '12 at 17:39