1
$\begingroup$

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.

  • 1
    What is $x$? Do you mean $\mathbb Z_2[x]/\left$?2012-11-13
  • 0
    Also, isn't $2=0$ and $3=1$ in $\mathbb Z_2$?2012-11-13
  • 2
    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?2012-11-13
  • 0
    You might consider, assuming such an isomorphism exists, the image of x under such a map.2012-11-13
  • 0
    The author might intend $\mathbb{Z}_2$ to be the localization at $(2)$, which would explain the 2 and 3. But that's just a stab in the dark.2012-11-13
  • 1
    Isn't the map which is constant on $Z/2Z$, and sending $x$ to $x-1$ an isomorphism between these two rings?2012-11-13
  • 0
    @rschwieb Originally the notation $\mathbb{Z}_2/\langle x^2 - 2 \rangle$ was used, so I made the best guess I could. The author will have to clarify.2012-11-13
  • 1
    @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 :)2012-11-13
  • 0
    Sory, I edited the problem more accurately2012-11-14
  • 0
    @rschwieb Lesson learned.2012-11-14
  • 0
    @Firmino We are still waiting for you to confirm what you mean by $\mathbb{Z}_2$.2012-11-14
  • 0
    I think everyone know definition $\mathbb{Z}_2$, so I don't need confirm what its mean.2012-11-15
  • 0
    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

1

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.

1

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".

  • 0
    Can you tell me more?. Why $x^2+\langle x^2 \rangle =0$?. What we will do next step?2012-11-14
  • 0
    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!2012-11-21
  • 0
    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.2012-11-21
  • 0
    @Firmino: Correct.2012-11-21
  • 0
    It is amazing. Thank you very much.2012-11-22
-3

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

  • 1
    This is very difficult to follow.2012-11-13
  • 1
    This is wrong. An isomorphism doesn't have to lift to the identity map on the level of polynomial rings.2012-11-13