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Consider the ring $R = \displaystyle\frac {\mathbb Z_2[x]}{\langle x^8-1\rangle}$.

i) Is $R$ a finite ring?

ii) Does $R$ have a zero divisor?

iii) Does $R$ have nilpotent elements?

ii) for zero divisor: $[(x^4+1)(x^2+1)(x+1)](x-1) = (x^8-1) = 0 \mod(x^8-1)$. Is this the right way of showing zero divisor?

And I have no idea how we can solve the other two.

Any help would be appreciated.

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    i) You should prove that $R$ is a vector space over the field $\mathbb{Z}_2$ of dimension $8$.2012-05-05
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    ii) To show $a$ and $b$ are zero-divisors, it's not enough to show that $ab=0$. You also need to show that $a$ and $b$ are not zero themselves.2012-05-05
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    Nilpotency of $f(x)+\langle x^8-1 \rangle\not=0$ is equivalent to existence of positive integer $n$ s.t. $(x^8-1)|f(x)^n$ but $(x^8-1)\nmid f(x)$.2012-05-05
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    @Andrea how do I prove R is a vector space.2012-05-05
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    For finiteness consider when $f(x)\equiv g(x) \pmod{ x^8-1}$ holds for two polynomials $f,g$. This should allow you to choose a suitable polynomial from every equivalence class modulo $x^8-1$.2012-05-05
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    Note that modulo $2$ we have $x^2 -1 \equiv (x-1)^2 \mod 2$ so $x^8-1\equiv (x-1)^8 \mod 2$. That should help you.2012-05-05

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