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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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    I know what $\mathbb Z_q$ is and it is *not* the same as $\mathbb Z/q\mathbb Z$.2012-11-20

3 Answers 3

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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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    It is amazing. Thank you very much.2012-11-22
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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 wrong. An isomorphism doesn't have to lift to the identity map on the level of polynomial rings.2012-11-13