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Let $R=\mathbb{Z}[i]/(3+i)$. Why is this a ring and show that $|R|=10$. I'm not allowed to use isomorphisms for the second bit.

For the first bit I said $(3+i)$ is an ideal, $\mathbb{Z}[i]$ is a ring, and $\mathbb{Z}[i]/(3+i)$ is a quotient group so it is a ring.

Second bit no idea.

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    $\begingroup$ Think of this as polynomials with integer coefficients which then get evaluated at the complex number i. Why is i^3 equal to -i equal to 3 in this ring? $\endgroup$ – Tac-Tics Mar 6 '16 at 0:47
  • $\begingroup$ We have $i^3=-i$ but why $=3$? $\endgroup$ – Polp Mar 6 '16 at 0:53
  • $\begingroup$ What is the equivalence class of $3+i$? $\endgroup$ – Zach Effman Mar 6 '16 at 1:00
  • $\begingroup$ $R=(3+i)+r $where $r$ is in $\mathbb{Z}[i]$. If $3=-i$ then $a+bi=a-b(3)=a-3b$? $\endgroup$ – Polp Mar 6 '16 at 1:06
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    $\begingroup$ When you quotient out by something you are effectively treating that "thing" as "zero". In other words, $\mathbb{Z}[i]/(3+i)$ is the structure you get by taking $\mathbb{Z}[i]$ and imposing $3+i=0$. $\endgroup$ – Julien Godawatta Mar 6 '16 at 1:15
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I like to think of a quotient thing as we keep everything but modify the equality on the given structure.
So, things like (some ring)$/(a-b)$ translate to [$a=b$ in the new structure].

Now, the original structure is $\Bbb Z[i]$, the Gaussian integers.

Then we modify the equality so that it respects the structure ($+,-,\cdot$), by posing $3=-i$.

In the new structure therefore every Gaussian integer $a+bi$ will be equal to the integer $a-3b$.

We get the integers $0,1,2,\dots,9$, and, as $3^2=(-i)^2$, we have $10=0$.

Now the only thing left to prove is that these ten numbers are still different in the quotient ring.

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  • $\begingroup$ Btw, observe that $(3-i)\overline{(3-i)}=3^2+1^2=10$. $\endgroup$ – Berci Mar 6 '16 at 1:51

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