Question

Let P be as in Exercise 17. Prove that ${\mathit{P}}^{\mathbf{2}}$ is the principal ideal (2).

Short answer

It is proved that $P^2$is the principal ideal (2).

Step-by-step solution

Product of Ideals

The product $\mathit{I}\mathit{J}$ of ideals I and J is the set of all sums of elements in the form of role="math" localid="1654760959652" $\mathit{a}\mathit{b}$, with$\mathit{a}\mathbf{\in }\mathit{I}$ and$\mathit{b}\mathbf{\in }\mathit{J}$ that is$\mathit{I}\mathit{J}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathbf{a}}_{\mathbf{n}}{\mathbf{b}}_{\mathbf{n}}\mathbf{|}\mathbf{n}\mathbf{\ge }\mathbf{1}\mathbf{,}{\mathbf{a}}_{\mathbf{k}}\mathbf{\in }\mathbf{I}\mathbf{,}{\mathbf{b}}_{\mathbf{k}}\mathbf{\in }\mathbf{J}\mathbf{\}}$.

Prove that P2 is the principal ideal (2)

The prime ideal in $Z\left[\sqrt{-5}\right]$is expressed as the product of prime ideals. The irreducible factorization of elements is $6=2\cdot 3$. But $\left(2\right)$is not a prime ideal because the product $(1+\sqrt{-5})(1-\sqrt{-5})=6$is in $\left(2\right)$, but neither of the factors is in $\left(2\right)$.

Let $P$ be the ideal in $\mathrm{Z}\left[\sqrt{-5}\right]$generated by $2$ and $1+\sqrt{-5}$.

$P=\{2a+(1+\sqrt{-5})b|a,b\in Z\left[\sqrt{-5}\right]\}$

Then $P$ is an ideal shows that $r+s\sqrt{-5}\in P$if and only if $r$and $s$are both even or both odd. This implies that the only cosets in $\mathrm{Z}\left[\sqrt{-5}\right]|\mathrm{P}$are $0+p$and $1+p$. Similarly for $2,1-\sqrt{-5}$.

Let $Q$be the ideal generated, so that

$\mathrm{Q}=\{2\mathrm{a}+(1+\sqrt{-5})\mathrm{b}|\mathrm{a},\mathrm{b}\in \mathrm{Z}\left[\sqrt{-5}\right]\}$

This can also be written as $r-s\sqrt{-5}\in P$. The only distinct cosets in $\mathrm{Z}\left[\sqrt{-5}\right]|\mathrm{P}$are $0+P,1+P$.

It follows the quotient ring $Z\left[\sqrt{-5}\right]|P$is isomorphic and $P=-Q$. But $\left(P\right)=\left(Q\right)=\left(Q\right)$.

Therefore,$P^2$ is the principal ideal $\left(2\right)$.