2
$\begingroup$

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
    What is $i$ in this context? – 2017-01-25
  • 0
    @Bernard Nothing specific has been mentioned in the question. So I guess it has the usual meaning. – 2017-01-25
  • 1
    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 $\mathbf Z/5\mathbf Z$. – 2017-01-25
  • 0
    @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? – 2017-01-25
  • 0
    @rschwieb: I think I've fixed the comment. For your last comment: $-1$ has two roots mod. $5$ ($\pm2)$, whence the factorisation: $x^2+1=(x-2)(x+2)$. – 2017-01-25
  • 0
    @Bernard You mean $(x-2)(x+2)=(x-2)\cap(x+2)$, not $(x-2, x+2)$, because the latter one is the whole ring. In fact (with that modification and the poster's agreement) it would be a good solution to post. – 2017-01-25
  • 0
    @Bernard I get the explanation you just gave, but can you please tell me where have I gone wrong? – 2017-01-25
  • 0
    @rschwieb: Oh! yes: I'll fix it. Thanks for pointing it. – 2017-01-25
  • 0
    @Naive It's not the same as the solution you gave. Currently you are both saying true things. – 2017-01-25
  • 0
    The problem is that we can consider, as you do, the quotient ring $Z[x]/(x^2+1,5)$. But it is isomorphic to $\mathbf Z/5\mathbf Z[x]/(x^2+1)=\mathbf Z/5\mathbf Z[x]/((x-2)(x+2))\simeq\mathbf Z/5\mathbf Z[x]/(x-2)\times\mathbf Z/5\mathbf Z[x]/(x+2) \simeq \mathbf Z\,/\;5\,\mathbf Z \times\mathbf Z/5\mathbf Z$ and this one isn't an integral domain. – 2017-01-25
  • 0
    @Bernard It's true: but for what step of the argument would that be a problem? I think if you go the extra step to map $1+i$ into that ring, you'll find its a unit of the product ring. – 2017-01-25
  • 0
    @Naive: What I find confusional in your approach is that if you consider $i $ denotes a square root of $-1$ in $\mathbf Z/5\mathbf Z$, it is one of $2$ or $-2$, hence $\mathbf Z/5\mathbf Z[i]/(1+i)$ is nothing else than $\mathbf Z/5\mathbf Z$. If you consider it denotes an external element, you obtain *in fine* $(\mathbf Z/5\mathbf Z)^2$. – 2017-01-25
  • 0
    @Bernard $i=\sqrt{-1}$ here. – 2017-01-25
  • 0
    @Sure, but is it the one in $\mathbf C$ or in $\mathbf Z/5\mathbf Z$? In my opinion, the problem is not well posed. – 2017-01-25
  • 0
    @Bernad $i \in \Bbb C$ is what must be considered i guess, otherwise there is nothing much in the problem as you said. – 2017-01-25
  • 0
    @Bernard I understand why my isomorphism was incorrect.Thank you for enlightening me. But can I say, $Z_5[i]= Z[x]/\langle x^2+1,5\rangle$, so $Z_5[x]/\langle 1+i\rangle= Z[x]/\langle x^2+1,x+1,5\rangle$ and proceed with my argument. Or is there something incorrect with that also.? – 2017-01-26
  • 0
    @Naive: It seems correct to me. You can simpllify what follows observing the ideal $(x^2+1,x+1,)$ contains $2=x^2+1+(1-x)(x+1)$, so it contains both $2$ and $5$, hence it contains $1$. – 2017-01-26
  • 0
    @Bernard Thank you for your valuable inputs. – 2017-01-26
  • 2
    Possible duplicate of [How many elements does $\mathbb Z_7[i]/\langle i+1\rangle$ have?](https://math.stackexchange.com/questions/1100352/how-many-elements-does-mathbb-z-7i-langle-i1-rangle-have) – 2018-11-24

1 Answers 1

1

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.

  • 0
    This is a good one! Thank you:) – 2017-01-26