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Let $D$ be the set of all gaussian integers of the form $m+ni$ where $m,n \in Z$. Carry out the construction of the quotient field $Q$ for this integral domain. Show that this quotient field is isomorphic to the set of all complex numbers of the form $a + bi$ where $a,b$ are rational numbers.

Well I know how to show that the quotient field is isomorphic, I just don't know how to construct such a quotient field.

In the case of integers, we have the quotient field $[a,b] = [c,d]$ if and only if $ab = bc$, so I am guessing maybe the quotient field for the gaussian integers should follow that, $[m+ni,r+si] = [m_1 + n_1i,r_1+s_1i]$ if and only if $(m+ni)(r_1+s_1i)=(r+si)(m_1+n_1i)$, is that right?

starblue
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alkabary
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    Your definition of the quotient field as pairs $[a,b]$ subject to a certain equivalence relation has nothing to do with $a$ and $b$ being integers. It works the same if they are Gaussian integers, or elements of any integral domain at all. – KCd Nov 25 '14 at 05:03
  • so the quotient field that i constructed is right , meaning the equivalent classes of $[m+ni,r+si]$ ? – alkabary Nov 25 '14 at 05:08
  • Yes. And the term is equivalence classes, not equivalent classes. – KCd Nov 25 '14 at 05:22

1 Answers1

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Let $K$ be the field you want to construct and let $\Bbb Q(i)$ the field of complex numbers of the form $r+si$ with both $r$ and $s$ in $\Bbb Q$.

First thing you need to convince yourseld that $K\subset{\Bbb Q}(i)$. If you take a fraction $\frac{a+bi}{c+di}$ with $a,b,c,d\in\Bbb Z$ you can rewrite it as $$ \frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i $$ (why?) and you're done.

On the other hand, if $\frac mn+\frac pqi\in\Bbb Q(i)$, we can rewrite this as $$ \frac{mq+npi}{nq} $$ where both numerator and denominator are Gaussian integers. Thus you also have $\Bbb Q(i)\subset K$ and you're finished.

Andrea Mori
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