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Let $a$ and $b$ be integers. Since $\mathbb{Z}a+\mathbb{Z}b$ is an ideal of $(\mathbb{Z},+,\cdot)$, $\exists n\geq 0$ s.t. $\mathbb{Z}a+\mathbb{Z}b=\mathbb{Z}n$. Prove that $n=\gcd(a,b)$ (therefore, $\exists f,e\in\mathbb{Z}$ s.t. $ea+fb=\gcd(a,b)$).

I'm having trouble with this one. I know that given a ring $A$ and an ideal $A\alpha_i$, we have that $$\sum_{i=1}^{t}A\alpha_i := \left\{\sum_{i=1}^{t}a_i\alpha_i \biggr\rvert a_1,\cdots, a_t\in A\right\}\quad.$$ This means that $\mathbb{Z}a+\mathbb{Z}b=\{z_1a+z_2b\;|\;z_1,z_2\in\mathbb{Z}\}$. But then $\exists z\in\mathbb{Z}$ s.t. $\mathbb{Z}a+\mathbb{Z}b = \{z(a+b)\;|\;z\in\mathbb{Z}\}$, but $a+b\neq\gcd(a,b)$. What am I missing here?

Anne Bauval
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A. Riba
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    Hint: Bézout!!! – julio_es_sui_glace Jul 02 '23 at 10:29
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    "But then $\exists z\in\mathbb{Z}$ s.t. $\mathbb{Z}a+\mathbb{Z}b = {z(a+b)\mid z\in\mathbb{Z}}$" is nonsense: $\mathbb{Z}a+\mathbb{Z}b = {z(a+b)\mid z\in\mathbb{Z}}$ is equivalent to $\mathbb{Z}a+\mathbb{Z}b =\mathbb{Z}(a+b),$ which does not involve $z$ and is anyway generally false. – Anne Bauval Jul 02 '23 at 10:35

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I did not really read your work.

Since $\Bbb Zn\supseteq \Bbb Za+\Bbb Zb$, we have $n\mid a$ and $n\mid b$. Since every common divisor of $a$ and $b$ divides all the elements of $\Bbb Za+\Bbb Zb$, and $\Bbb Za+\Bbb Zb\supseteq \Bbb Zn$, we have that every common divisor of $a$ and $b$ divides $n$. Therefore $n$ is a greatest common divisor of $a$ and $b$, as per definition. Note that this works in all rings.