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How to prove $\Bbb Z[i]/(1+2i)\cong \Bbb Z_5$?

My method is

$$(1+2i)=\big\{a+bi丨a+2b≡0\pmod 5\big\},$$

So any $a+bi$ in $\Bbb Z(i)$,we got

$$a+bi=(b-2a)i+a(1+2i).$$

So $\Bbb Z[i]/(1+2i)=\big\{0,[i],[2i],[3i],[4i]\big\}$.

I know how to prove this ring is isomorphic to $\Bbb Z_5$, but how can I prove that $\Bbb Z[i]/(1+2i)$ equals to $\Bbb Z_5$ directly? Any suggestion ia appreciated.

yLccc
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  • Why are you unhappy with your solution? It looks perfectly reasonable to me. – Arthur Jan 03 '19 at 15:31
  • I can't prove it directly, which makes me sad:( – yLccc Jan 03 '19 at 16:06
  • What does "diectly" mean in this case? – Arthur Jan 03 '19 at 16:17
  • $\mathbb Z[i]/(a+bi)\cong \mathbb Z/N(a+bi)\mathbb Z$ where $(a,b)=1$ – Mustafa Jan 03 '19 at 17:13
  • https://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers – Viktor Vaughn Jan 03 '19 at 17:35
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    $\mathbb Z[i] / (1+2i)$ is not equal to $\mathbb Z_5$. By definition $\mathbb Z_5$ is $\mathbb Z / (5)$ and each element of $\mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $\mathbb Z$. But each element of $\mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $\mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have? – Lee Mosher Jan 03 '19 at 20:31
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    @Mustafa, it is not generally true that $\mathbb Z[i]/(a+bi)\isom \mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $\mathbb Z/p^2$. – paul garrett Jan 04 '19 at 01:11

4 Answers4

7

I think it is easier to do it this way. Let $\varphi:\Bbb Z[x]\to \Bbb Z[i]$ sending $x\mapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $\Bbb Z[i]$ under $\varphi$ is the ideal $J=(x^2+1,x-2)$ of $\Bbb Z[x]$. Thus $\varphi$ induces the isomorphism $$\Bbb{Z}[i]/I\cong \Bbb{Z}[x]/J.$$
Now observe that $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$ This shows that$$\frac{\Bbb Z[i]}{(1+2i)}=\frac{\Bbb{Z}[i]}{(i-2)}\cong \frac{\Bbb Z[x]}{(x^2+1,x-2)}=\frac{\Bbb{Z}[x]}{(5,x-2)}\cong\frac{\Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$ Because under the isomorphism $\Bbb{Z}[x]/(x-2)\cong \Bbb Z$ (which sends $p(x)+{\color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5\Bbb{Z}$ of $\Bbb Z$, we get $$\frac{\Bbb Z[i]}{(1+2i)}\cong\frac{\Bbb Z}{5\Bbb Z},$$ as required.

  • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it – yLccc Jan 03 '19 at 16:55
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    Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{\color{red}{(x-2)}}=5p(x)+{\color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{\color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5\Bbb Z$. –  Jan 03 '19 at 20:00
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    In my answer and in my previous comment, anything that appears in $\color{red}{\mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses). –  Jan 03 '19 at 20:00
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A ring homomorphism $\varphi\colon\mathbb{Z}[i]\to\mathbb{Z}_5$ is uniquely determined by assigning $q=\varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.

With the second choice, you have $\varphi(a+bi)=[a-2b]$. Can you determine $\ker\varphi$?

egreg
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Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)\supset (5)$. Hence $\mathbb Z[i]/(1+2i)\subset \mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)\cong \mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.

To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) $\{0,1,2,3,4\}$. It's easy to see none of them are the same.

0

If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $\mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.

Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.

Mark Bennet
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  • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right? – yLccc Jan 03 '19 at 16:05
  • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $\mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers. – Mark Bennet Jan 03 '19 at 16:09