Question

If $\left(m,n\right)=1$in Z, prove that $\left(m\right)\cap \left(n\right)$is the ideal$\left(mn\right)$

Short answer

It is proved that$\left(m\right)\cap \left(n\right)$ is an ideal$\left(mn\right)$

Step-by-step solution

Theorem statement 

If a divide the product and the gcd of a and b is 1, then a divides c as well. That is,

If $a|bc$and $\left(a,b\right)=\mathbf{1}$, then$a|c$.

To prove that $\left(m\right)\cap \left(n\right)$ is an ideal $\left(mn\right)$, we need to prove that,

$\left(m\right)\cap \left(n\right)=\left(mn\right)$.

And to prove $\left(m\right)\cap \left(n\right)=\left(mn\right)$, prove that $\left(m\right)\cap \left(n\right)\subseteq \left(mn\right)$and$\left(m\right)\cap \left(n\right)\supseteq \left(mn\right)$

Prove m∩n⊆mn

It is given that$\left(m,n\right)=\mathbf{1}$in Z. Let $x\in \left(m\right)\cap \left(n\right)$, then$x\in \left(m\right)$and $x\in \left(n\right)$so, there exists k and l, such that $x=km$and $x=\ln$respectively. Hence, by the theorem stated in step 1, it can be concluded that n divides km and n divides k, such that$k=k_{0}n$because $\left(m,n\right)=\mathbf{1}$. Then, we get,

$$\begin{gathered} x=k_{0}nm \\ =k_0\left(mn\right)\in \left(mn\right) \end{gathered}$$

Then, we have $\left(m\right)\cap \left(n\right)\subseteq \left(mn\right)......\left(\mathbf{1}\right)$.

Prove m∩n⊇mn

It is given that $\left(m,n\right)=\mathbf{1}$ in z . Then, there exists an integer k, such that $x=k\left(mn\right)$, then, we have:

$$\begin{gathered} x=kmn \\ =\left(kn\right)m\in \left(m\right) \end{gathered}$$

Similarly,

$$\begin{gathered} x=kmn \\ =\left(km\right)n\in \left(n\right) \end{gathered}$$

Because $kn,km\in \mathrm{\mathbb{Z}}$. Which implies that $x\in \left(m\right)\cap \left(n\right)$, then we can conclude that$\left(m\right)\cap \left(n\right)\supseteq \left(mn\right)......\left(\mathbf{2}\right)$

From equation (1) and (2), we have:

$\left(m\right)\cap \left(n\right)=\left(mn\right)$

Hence, it is proved that $\left(m\right)\cap \left(n\right)$is an ideal$\left(mn\right)$