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Let $R=\mathbb{Z}[i]$ be the ring of Gaussian integers. Let $z=5 + 3i$ and let $I=\left< z \right>$. Let $\phi: \mathbb{Z} \rightarrow R/I$ be the canonical ring homomorphism.

I am trying to show that $\phi$ is surjective. I understand that working in $R/I$ we have $5+3i=0$. I also understand that $\phi(c) = 1 + 1 + ... + 1$ ($c$ times). My problem is that I cannot see how to find a $c$ such that $\phi(c) = 1+i$, for example, because $1+i$ is not a multiple of $5+3i$. And how can I think of $\phi(c)$ when $c$ is negative?

How do I proceed? Any pointers are much appreciated.

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  • $\begingroup$ Also I have previously proven that $(4-i)(1+i) = 5+3i$ but I'm unsure how to use this. $\endgroup$ – TorbenB Jan 2 '16 at 15:21
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Note that it is enough to check that there exists some $c$ such that $\phi(c)=i+I$ since $1$ and $i$ generate $\mathbb{Z}[i]$. Since $5$ and $3$ are coprime, there exist integers $a,b$ such that $5b+3a=-1$. Letting $c=5a-3b$ yields $c-i=(5+3i)(a+bi)$, i.e. $\phi(c)=i+I$.

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  • $\begingroup$ Why is $5b+3a=-1$? Shouldn't the $-1$ be $1$? $\endgroup$ – TorbenB Jan 2 '16 at 15:57
  • $\begingroup$ How does the equality $c-i=(5+3i)(a+bi)$ follow? How are you going from subtracting to a product? $\endgroup$ – TorbenB Jan 2 '16 at 15:58
  • $\begingroup$ @TorbenB since $5$ and $3$ are coprime we can find integers $u,v$ with $5u+3v=1$ (Bezout). Now let $b=-u$ and $a=-v$ and we have $5b+3a=-1$. Finding equality $c-i=(5+3i)(a+bi)$ is just a question of working out the RHS and applying the equalities $5b+3a=-1$ and $c=5a-3b$. $\endgroup$ – drhab Jan 2 '16 at 16:29
  • $\begingroup$ @drhab I see. How do we go from having $c-i = (5+3i)(a+bi)$ to $\phi(c) = i+I$? $\endgroup$ – TorbenB Jan 2 '16 at 17:04
  • $\begingroup$ @TorbenB $\phi(c)=c+I=i+I$. The second equality because $c-i\in I$. $\endgroup$ – drhab Jan 3 '16 at 13:33

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