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Prove that in ring $\mathbb{Z}, \forall a,b \in \mathbb{Z} $, $$(a,b) = (d),$$ where $ d = \mathrm{gcd} (a,b)$. Also, generalize this for all finite generated ideals in $\mathbb{Z}$.

I don't even know how to start. I am not sure if I got the definition of generated ideal. When I have $(a)$, what does that mean?

Syzygy Yuan
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user560461
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    Your book should at some point tell you exactly what something like $(a,b)$ means. Read that. – Arthur Aug 06 '19 at 10:01

1 Answers1

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Hint:

  • $x\in(a,b)\iff$ integers $n,m$ exist with $na+mb=x$
  • $x\in(d)\iff$ integer $k$ exists with $kd=x$
drhab
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