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I am self studying algebra, and have come across this exercise:

Show that $\mathbb{Z}[i]/(2+3i)$ has 13 elements and is a field.

My proof: Since Z[i] is a euclidian domain with euclidian norm function $f(c+id)= c^2+d^2$, for any element $a+bi \in Z[i]$, we can "divide" it by $2+3i$ to get $a+bi= q (2+3i) + r$ where either $r=0$ or $r=c+di$ for some integers $c,d$ not both $0$ with $c^2 + d^2=f(c+di) < f(2+3i)= 13$. But then $a+bi$ belongs to the coset $ r + (2+3i) $. So every element in the quotient ring belongs to some $r + (2+3i)$ for $r=0$ or $r=c+di$ with $c^2 + d^2<13$. Now we can count all the possible pairs of integers satisfying this condition, ie $(0,0),(0,1),(0,2),(0,3), ..., (3,1)$ to get 13 and so there are 13 elements in the quotient ring.

Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance!

user26857
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PaulDavis
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  • If the product of two (non-units) Gaussian integers were $2+3i$ then their norms would b (non-units) factors of $13$, the norm of $2+3i$. Therefore, $2+3i$ is a Gaussian prime. – orole Jan 19 '18 at 04:22
  • Thanks orole I have not heard of gaussian primes though and what that would imply? – PaulDavis Jan 19 '18 at 04:24
  • That implies that the quotient is a domain. The division thing you can do by doing Euclid's algorithms. If $a$ is not divisible by $2+3i$ (they are relatively prime). Apply Euclid's algorithm to get $m,n$, such that $am+n(2+3i)=1$. Therefore, in the quotient, the class of $a$ has as inverse the class of $m$. – orole Jan 19 '18 at 04:30
  • Thank you orole, also is my proof for the 13 elements correct? – PaulDavis Jan 19 '18 at 05:07
  • Yes. Maybe insert the explanation of why any two of the representatives you chose don't belong to the same class. You have shown that their classes represent all the classes. Now show that your list doesn't have repetitions. – orole Jan 19 '18 at 05:10
  • Oh yes that is right, I will try to show there are no repetition thanks. – PaulDavis Jan 19 '18 at 05:18

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I probably posted this before, but here it goes. You can determine the size of $\mathbb Z[i]/(a+bi)$ using the Smith normal form by considering it as a quotient of $\mathbb Z\oplus \mathbb Z$ by the subgroup generated by $(a,b)$ and $(-b,a)$. This has corresponding matrix $\begin{pmatrix} a&b\\ -b &a\end{pmatrix}$ whose determinant is $a^2+b^2$. This gives that the quotient has this size, which in your case is $2^2+3^2 = 4+9=13$. The fact the quotient is a field follows immediately since there is only one ring of prime size $13$, which is the field $\mathbb Z/13$.

Pedro
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$\mathbb Z[i]/(2+3i) \cong \mathbb Z/N(2+3i)\mathbb Z =\mathbb Z/13 \mathbb Z$

or $\mathbb Z[i]/(2+3i) \cong \mathbb Z[x]/(x^2+1,2+3x)\cong \mathbb Z[x]/(13,2+3x)$ where $(x^2+1,2+3x)=(13,2+3x)$, so $\mathbb Z[x]/(13,2+3x)\cong \mathbb Z_{13}[x]/(2+3x)\cong \mathbb Z_{13} $

Mustafa
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You can do this, but counting one-by-one the elements of the quotient ring would not be practical if you replaced $2+3i$ by say $2323419+3803616i$.

Note that $(2+3i)(2-3i)=13$ so that $(2+3i)\subseteq (13)$ as ideals. Therefore $R/(2+3i)$ is a quotient of the ring $R/(13)$ where $R=\Bbb Z[i]$. But $|R/(13)|=13^2$ since $a+bi+(13)=c+di+(13)$ iff $a\equiv c$ and $b\equiv d\pmod{13}$. So the size of $R/(2+3i)$ is a factor of $13^2$, so is $1$, $13$ or $13^2$.

$|R/(2+3i)|$ cannot be $1$ since that would entail $(2+3i)\mid 1$ in $R$, which is false. $|R/(2+3i)|$ cannot be $13^2$ since that would entail $2+3i\in( 13)$ which is false. So $|R/(2+3i)|=13$.

As $13$ is prime, $R/(2+3i)$ equals its own prime subring, which is a field.

Angina Seng
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  • Thanks, here you are using Lagrange's theorem by looking at the quotient rings as groups? Could you explain more the reasoning behind why it can't be 13^2 and also the final point, what you mean by "it equals its own prime subring, which is a field"? – PaulDavis Jan 19 '18 at 05:53
  • @PaulDavis Each unital ring has a subring consisting of the integer multiples of the multiplicative identity: this is the prime subring. It is isomorphic to $\Bbb Z$ or to $\Bbb Z/n\Bbb Z$ for some $n$. When $n$ is prime, as here, it is a field. – Angina Seng Jan 19 '18 at 05:56