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Find the cardinality of the quotient ring, $$\mathbb Z_5[i]/\langle 1+i\rangle$$

My Attempt:

Since, $$\mathbb Z_5[i]\cong Z[x]/\langle x^2+1,5\rangle$$

Hence, $$\mathbb Z_5[x]/\langle 1+i\rangle\cong \mathbb Z[x]/\langle x^2+1,x+1,5\rangle$$

In the quotient ring we have the following,

$x^2+1=0,x+1=0$ and $5=0$, i.e., $x=-1\;\; \implies 2=(-1)^2+1=0$

And, $5=0 \implies 1=0$. Thus, $\mathbb Z_5[i]/\langle 1+i\rangle\;\;=\{0+\langle 1+i\rangle\}$.

Is the above reasoning correct?

Initially, I had come across the following argument,$$1=(3+2i)(1+i)\implies 1\in \langle 1+i\rangle$$

And hence the quotient ring has just one element namely the zero element. This argument is quite good but it is not always easy to make such guesses. So I wanted to use the traditional method to find the cardinality.

Moreover, is there any different approach to the problem?

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    $\begingroup$ What is $i$ in this context? $\endgroup$ – Bernard Jan 25 '17 at 17:39
  • $\begingroup$ @Bernard Nothing specific has been mentioned in the question. So I guess it has the usual meaning. $\endgroup$ – Naive Jan 25 '17 at 17:41
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    $\begingroup$ You can't mix a finite field like $\mathbf Z/5\mathbf Z$ with complex numbers so easily – all the more so as $-1$ has a square root in $\mathbf Z/5\mathbf Z$. $\endgroup$ – Bernard Jan 25 '17 at 17:45
  • $\begingroup$ @Bernard As far as I remember in such problems we usually say, $x $ corresponds to $i$, and follow a similar procedure as to what I have done. Am I worng at this? $\endgroup$ – Naive Jan 25 '17 at 17:50
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    $\begingroup$ Possible duplicate of How many elements does $\mathbb Z_7[i]/\langle i+1\rangle$ have? $\endgroup$ – Watson Nov 24 '18 at 16:56
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I might have analyzed it this way:

$\mathbb Z_5[x]/\langle x^2+1\rangle$ has two maximal ideals, $(x+2)$ and $(x-2)$ since $x^2+1$ is reducible mod $5$. If $1+x$ was not a unit in this ring, then it would be contained in one of these two ideals. But obviously $x+2-(x+1)=1$ and $x+1-(x-2)=3$ are both units, so $x+1$ is not contained in either maximal ideal, and is a unit. So the quotient is zero.

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  • $\begingroup$ This is a good one! Thank you:) $\endgroup$ – Naive Jan 26 '17 at 7:24

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