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There is a quotient ring given: $$\mathbb Z[i]/(2-i)$$ $\mathbb Z[i] = \{ a + bi: a,b \in \mathbb Z \}$ thus $\mathbb Z[i]$ is a ring. I am to calculate the amount of elements in this ring.
I don't know how to approach to this this task.
How can I rewrite this quotient ring as a set?

Hendrra
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4 Answers4

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Consider the unique ring homomorphism $f:\Bbb Z[x] \to \Bbb Z[i]/(2-i)$ that sends $x$ to $\overline{i}$, more explicitely, we have $f(P(x))=\overline{P(i)}$ for $P(x) \in \Bbb Z[x]$. We have $\Bbb Z[x]/\operatorname{ker}(f) \cong \Bbb Z[i]/(2-i)$ as $f$ is surjective.

It's easy to see that $x^2+1$ and $2-x$ are contained in $\operatorname{ker}(f)$, so we have $(x^2+1,2-x) \subset \operatorname{ker}(f)$.

As $5=(2+x)(2-x)+x^2+1 \in (x^2+1,2-x)$, we see that $(5,2-x) \subset (x^2+1,2-x) \subset \operatorname{ker}(f)$.
Now $\Bbb Z[x]/(5,2-x) \cong \Bbb F_5[x]/(x-2) \cong \Bbb F_5$, so $(5,2-x)$ is a maximal ideal, because $\Bbb F_5$ is a field.
As $\Bbb Z[i]/(2-i)$ is not the zero ring, we have $\operatorname{ker}(f) \neq \Bbb Z[x]$, so we have must have $(5,2-x)=\operatorname{ker}(f)$, thus $\Bbb{Z}[i]/(2-i) \cong \Bbb F_5$, so it has $5$ elements.

Lukas Heger
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  • Why do $x^2 + 1$ and $2 -x$ in ker$(f)$? And by $(x^2 + 1, 2- x)$ you mean a set or an ordered pair? – Hendrra Jan 17 '18 at 01:13
  • @Hendrra $x^2+1$ gets mapped to $i^2+1$ which is $0$ already in $\Bbb Z[i]$. $2-x$ gets mapped to $2-i$ which is zero in the quotient because we mod out the ideal $(2-i)$. By $(x^2+1,2-x)$ I mean the Ideal generated by $x^2+1$ and $2-x$, that is ${a(x^2+1)+b(2-x)\mid a,b \in \Bbb Z[x]}$. – Lukas Heger Jan 17 '18 at 01:19
  • I have edited the post to make it more clear what $f$ does. – Lukas Heger Jan 17 '18 at 01:21
  • Thanks! I think I understand it know. It really helped! – Hendrra Jan 17 '18 at 01:25
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Note that the size of the residue class of $\alpha$ is $|N\alpha|$. In this case $N(2-i)=5$.

Rene Schipperus
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One usually chooses coset representatives to describe the quotient ring.

In the quotient the equation $2-i=0$ holds, so $[a+bi]=[a+2b]$ (here, $[x]$ is the coset containing $x$). Thus, we can restrict ourselves to $\mathbb Z$.

Next, since $2=i$, also $2^2=i^2 \Rightarrow 4=-1 \Rightarrow 5=0$.

Thus, the quotient ring seems to be isomorphic to $\mathbb Z / 5 \mathbb Z$, with five cosets $[0],[1],[2],[3],[4]$. The only thing that is left to do is to prove that these five cosets are not equal (that is, that the quotient ring is not trivial), which I leave as an exercise.

lisyarus
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Note that $[2]$ and $[i]$ are equal in the quotient.

Now let's count.....

$[1]$

$[2] = [i]$

$[3] = [1] + [2] = [1]+[i] = [1+i]$

$[4] = [2][2] = [i] \cdot [i] = [-1]$

$[5] = [4] + [1] = [-1] + [1] = [0]$

$[6] = [5] + [1] = [0] + [1]$

Moreover, for integer $a,b$, our computations show

$$[a+bi] = [a] + [b][i] = [a] + [2b] = [(a + 2b) \mod 5]$$

and it's clear that there are exactly five elements in the quotient.

Badam Baplan
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