Question

If is greatest common divisor of a and b in $\mathbf{\square }$, show that $\left(a\right)\mathbf{+}\left(b\right)\mathbf{=}\left(d\right)$. (the sum of ideals is defined in exercise 20.)

Short answer

It is proved that$\left(a\right)+\left(b\right)=\left(d\right)$

Step-by-step solution

Prove a+b⊂d

Let$x\in \left(a\right)+\left(b\right)$

Then, there exist$\underline{s}$$y\in \left(a\right)$ and$z\in \left(b\right)$ such that$x=y+z$

Furthermore, $\underline{s}\sout{S}$ince $y\in \left(a\right)$, there exists$k\in Z$ such that$y=ka$

Similarly, since $z\in \left(b\right)$, there exists $I\in Z$such that$z=Ib$

Now,$x=ka+Ib$

By the definition of$\_$ the greatest common divisor of a and b, d divides both a and b.

Since d divides a, there exists an integer n such that$a=nd$

Similarly, d divides b, there exists an integer m such that$b=md$

Therefore,

$$\begin{gathered} x=k\left(nd\right)+I\left(md\right) \\ =\left(kn\right)d+\left(IM\right)d \\ =\left(kn+Im\right)d\in \left(d\right) \end{gathered}$$

Because, role="math" localid="1653674555696" $kn+Im\in Z$.

Since x was taken arbitrarily,$\left(a\right)+\left(b\right)\underline{\subset \left(d\right)}$

Prove d⊂a+b

Let $x\in \left(d\right)$. Then$x=kd$ for some$k\in Z$

Using Theorem 1.2, which $\sout{is}$ says that $\mathbf{!}\sout{\mathbf{L}}$et a and b be integers, not both 0, and let d be their greatest common divisor. Then$\mathbf{,}$ there existrole="math" localid="1653674847448" $\underline{\mathbf{s}}$integer and v such that $d=au+bv$

Therefore, there exists integers u and v $\mathit{\_}$such that $d=au+bv$.

So,

$$\begin{gathered} x=kd=k\left(au+bv\right)=kau+kbv \\ =\left(ku\right)a+\left(kv\right)b\in \left(a\right)+\left(b\right) \end{gathered}$$

Because we have$\left(ku\right)a\in \left(a\right)$ and$\left(ku\right)b\in \left(b\right)$

Since x was taken arbitrarily,

$\left(d\right)\underline{\subset \left(a\right)+\left(b\right)}$

Finally, since$\left(a\right)+\left(b\right)\underline{\subset \left(d\right)}$ and$\left(d\right)\underline{\subset \left(a\right)+\left(b\right)}$

We can conclude that role="math" localid="1653675153587" $\left(a\right)+\left(b\right)=\left(d\right)$.