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The problem says "Find the generator for the ideal of $\mathbb{Z}[i]$ generated by $3 + 4i$ and $4 + 7i$."

I don't understand the question. It asks us to find the generator of the ideal, but then it tells us what the ideal is generated by.

Here is an attempted solution using the tip from kjo:

We are looking for the GCD of $3 + 4i$ and $4 + 7i$ in $\mathbb{Z}[i]$.

We will find it using a Guassian version of the Euclidean Algorithm.

$||3 + 4i|| = 25$ and $||4 + 7i|| = 65$

The GCD of these norms is 5. Any common divisor of our numbers $3 + 4i$ and $4 + 7i$ must divide $5$.

$5 = (2 + i)(2 - i)$

Calculate: $\frac{3 + 4i}{2 + i} = \frac{(3 + 4i)(2 - i)}{(2 + i)(2 - i)} = \frac{10 + 5i}{5} = 2 + i$

So $2 + i$ divides $3 + 4i$.

Calculate: $\frac{4 + 7i}{2 + i} = \frac{(4 + 7i)(2 - i)}{(2 + i)(2 - i)} = \frac{15 + 10i}{5} = 3 + 2i$

So $2 + i$ is the GCD of $3 + 4i$ and $4 + 7i$ so it is the generator for the ideal.

EDIT: I think I got it. I made some silly mistakes on my solution before. Thanks for the help everyone!

willkurada
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1 Answers1

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Hint $\ $ Eliminate to get the coef of $\,i\,$ to be $= 1\!:\ $ $ 2(3\!+\!4i)-(4\!+\!7i) = 2+i =:\alpha \in I.\,$

$$\begin{eqnarray}\alpha\mid \alpha\alpha' = \color{#0a0}5,\ \ {\rm so}\ \ {\rm mod}\,\ 2\!+\!i\!:\,\ \color{#c00}{i\equiv -2} &\,\Rightarrow\,& 3\!+\!4\color{#c00}i\equiv 3\!+\!4(\color{#c00}{-2})\equiv\ \ \ {-}\color{#0a0}5\ \equiv\ 0\\ &\Rightarrow& 4\!+\!7\color{#c00}i\equiv 4\!+\!7(\color{#c00}{-2})\equiv -2(\color{#0a0}5)\equiv 0 \end{eqnarray}\qquad\ \ $$

Therefore $\ I = (i+2)\ $ (why?). $ $ Alternatively, apply the Euclidean algorithm.

Bill Dubuque
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  • I attempted a solution of the problem using the Euclidean algorithm and I got a different answer. Could you tell me where I went wrong? – willkurada Mar 16 '14 at 22:05
  • Nevermind. I made a silly mistake. Thanks for your help :) – willkurada Mar 16 '14 at 22:39
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    Great! The Euclidean algorithm is the best way to proceed in general. For special small problems, other ways might be faster (and more elementary), e.g. as above. – Bill Dubuque Mar 16 '14 at 22:42