Describe the ring $R = \mathbb Z_4[x]/((x^2+1)\mathbb Z_4[x])$ by

  1. listing all the cosets (for example by using coset representatives)
  2. describing the relations that hold between the elements in this ring, that is, describe the relations that hold between these cosets.
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    $\begingroup$ I see no question. $\endgroup$ – Najib Idrissi Mar 30 '12 at 12:51
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    $\begingroup$ Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. $\endgroup$ – Arturo Magidin Mar 30 '12 at 16:38
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    $\begingroup$ Can't you just ask Dr Smoktunowicz if you are stuck? $\endgroup$ – user28109 Apr 1 '12 at 22:55
  • $\begingroup$ @AFellowStudent If only he could pronounce her name. $\endgroup$ – Alex Youcis Apr 1 '12 at 23:04
  • $\begingroup$ @AFellowStudent: Who is Dr Smoktunowicz? $\endgroup$ – spohreis Apr 1 '12 at 23:12

You need to use the division algorithm here; you can use it since $x^2 + 1$ is monic in the ring $\mathbb{Z}_4[x]$. For every polynomial $f \in \mathbb{Z}_4[x]$, by the division algorithm we can write it as

$$f = (x^2 + 1)q(x) + r(x)$$

where the degree of $r$ is bigger than or equal to zero, less than 2. You can now see that the cosets in the quotient are of the form

$$(\text{linear polynomial}) + I$$

where $I$ is the ideal generated by $x^2 + 1$. Now the linear polynomial can be written as $ax + b$ for $a,b \in \Bbb{Z}_4$. But then recall that $x^2 + 1 = 0$ in the quotient, so that we get ring a new ring (the quotient ring) where multiplication between cosets $A + I$ and $B + I$ is defined by $$(A + I)(B+ I)= (AB) + I$$ and where we have the relation $x^2 + 1 = 0$.

  • $\begingroup$ "evaluating the left and right at $x=i$"? The complex number $i$? Anyway, if you are looking for a square root of $-1$ in $\mathbb{Z}/4\mathbb{Z}$, it doesn't exist. $\endgroup$ – M Turgeon Mar 30 '12 at 12:59
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    $\begingroup$ @MTurgeon I was totally mistaken, I have removed that from my answer now. $\endgroup$ – user38268 Mar 30 '12 at 13:12

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