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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ Possible duplicate of How many elements does $\mathbb Z_7[i]/\langle i+1\rangle$ have? $\endgroup$ – Watson Nov 24 '18 at 16:56

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.