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How does $\Bbb Z[X]/(2, X^2+1)$ look like? I've got a problem that asks a proof for the fact that this ring has $4$ elements, yet it's not isomorphic with $\Bbb Z_2 \times \Bbb Z_2$, but I can't even understand how this set looks like.

Watson
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  • Can you show that it's isomorphic to $ \mathbf Z/2 \mathbf Z[X]/(X^2 + 1) $? – Ege Erdil Jan 14 '17 at 15:52
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    $$\mathbf{Z}[X]/(p,X^2+1)\simeq \mathbb{F}_p[X]/(X^2+1)$$ – Watson Jan 14 '17 at 15:58
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    I don't know if this could be helpful: http://math.stackexchange.com/questions/1253187, http://math.stackexchange.com/questions/1145015, http://math.stackexchange.com/questions/1826188 – Watson Jan 14 '17 at 15:59

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Note that in this ring $x^2=-1$ and $2=0$. so the elements are

$$0,1,x,1+x$$

If we start with $x$ we have $x^2=-1=1$. Note that in $\mathbb{Z}_2\times \mathbb{Z}_2$ every element satisfies $y^2=y$.

Rene Schipperus
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  • That I understand, but how does one say how the set looks like? I mean you told me its elements, but I don't understand how you knew that those were it –  Jan 14 '17 at 16:40
  • You have all polynomials $a+bx+cx^2+\cdots$. All even coefficients are zero. And $x^2=1$ so $x^3=x$, $x^4=1$ etc so all higher powers collapse. You are left with the four elements. – Rene Schipperus Jan 14 '17 at 16:47