Commutative ring and ideal

Groups, Rings, Domains, Modules, etc, Galois theory
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Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Commutative ring and ideal

#1

Post by Papapetros Vaggelis »

Prove that the set \(\displaystyle{\mathbb{Z}\,[i\,\sqrt{5}]=\left\{a+i\,b\,\sqrt{5}\in\mathbb{C}: a\,,b\in\mathbb{Z}\right\}}\) is a subring of the ring \(\displaystyle{\mathbb{C}}\) . Consider the two-sided ideal \[\displaystyle{P_{2}=<3+i\,0\,\sqrt{5},1+i\,\sqrt{5}>=\left\{r_1\cdot (3+i\,0\,\sqrt{5})+r_2\cdot (1+i\,\sqrt{5})\in \mathbb{Z}\,[i\,\sqrt{5}]: r_1\,,r_2\in\mathbb{Z}\,[i\,\sqrt{5}]\right\}}\] of this ring. Prove that this ideal is a maximal ideal of the ring \(\displaystyle{\left(\mathbb{Z}\,[i\,\sqrt{5}],+,\cdot\right)}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Commutative ring and ideal

#2

Post by Papapetros Vaggelis »

Obviously, \(\displaystyle{0_{\mathbb{C}}=0+i\,0\,\sqrt{5}\,\,,1_{\mathbb{C}}=1+i\,0\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}}\) .

Let \(\displaystyle{a_{i}+i\,b_{i}\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}\,,a_{i}\,,b_{i}\in\mathbb{Z}\,,i=1,2}\) .

\(\displaystyle{\left(a_{1}+i\,b_{1}\,\sqrt{5}\right)+\left(a_{2}+i\,b_{2}\,\sqrt{5}\right)=\left(a_{1}+a_{2}\right)+i\,\left(b_{1}+b_{2}\right)\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}}\)

\(\displaystyle{\left(a_{1}+i\,b_{1}\,\sqrt{5}\right)\cdot \left(a_{2}+i\,b_{2}\,\sqrt{5}\right)=\left(a_{1}\,a_{2}-5\,b_{1}\,b_{2}\right)+i\,\left(a_{1}\,b_{2}+a_{2}\,b_{1}\right)\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}}\)

since \(\displaystyle{a_{1}+a_{2}\,,b_{1}+b_{2}\,\,,a_{1}\,a_{2}-5\,b_{1}\,b_{2}\,\,,a_{1}\,b_{2}+a_{2}\,b_{1}\in\mathbb{Z}}\) .

Futhermore,

\(\displaystyle{\left(a_{1}+i\,b_{1}\,\sqrt{5}\right)\cdot \left(a_{2}+i\,b_{2}\,\sqrt{5}\right)=\left(a_{2}+i\,b_{2}\,\sqrt{5}\right)\cdot \left(a_{1}+i\,b_{1}\,\sqrt{5}\right)=\left(a_{1}\,a_{2}-5\,b_{1}\,b_{2}\right)+i\,\left(a_{1}\,b_{2}+a_{2}\,b_{1}\right)\,\sqrt{5}}\)

Therefore, the set \(\displaystyle{\mathbb{Z}_{[i\,\sqrt{5}]}}\) is a commutative subring of \(\displaystyle{\mathbb{C}}\) .

In order to prove that the ideal \(\displaystyle{P_{2}}\) is a maximal ideal of the commutative ring \(\displaystyle{\left(\mathbb{Z}_{[i\,\sqrt{5}]},+,\cdot\right)}\) ,

it is sufficient to prove that the commutative ring \(\displaystyle{\left(\mathbb{Z}_{[i\,\sqrt{5}]}/P_{2},+,\cdot\right)}\) is a field.

Let \(\displaystyle{x+P_{2}\in\mathbb{Z}_{[i\,\sqrt{5}]}/P_{2}}\). Then, \(\displaystyle{x=a+i\,b\,\sqrt{5}\,,a\,,b\in\mathbb{Z}}\) and

\(\displaystyle{x+P_{2}=\left(a-b\right)+b\,(1+i\,\sqrt{5})+P_{2}}\). We have that \(\displaystyle{a-b=3\,q+r\,,q\in\mathbb{Z}\,,0\leq r\leq 2}\) and then

\(\displaystyle{\begin{aligned} x+P_{2}&=\left(r+i\,0\,\sqrt{5}\right)+\left(q+i\,0\,\sqrt{5}\right)\,\left(3+i\,0\,\sqrt{5}\right)+\left(b+i\,0\,\sqrt{5}\right)\,\left(1+i\,\sqrt{5}\right)+P_{2}\\&=\left[\left(r+i\,0\,\sqrt{5}\right)+P_{2}\right]+\left[\left(q+i\,0\,\sqrt{5}\right)\,\left(3+i\,0\,\sqrt{5}\right)+\left(b+i\,0\,\sqrt{5}\right)\,\left(1+i\,\sqrt{5}\right)+P_{2}\right]\\&=\left[\left(r+i\,0\,\sqrt{5}\right)+P_{2}\right]+P_{2}\\&=\left[\left(r+i\,0\,\sqrt{5}\right)+P_{2}\right]\end{aligned}}\)

since \(\displaystyle{\left(q+i\,0\,\sqrt{5}\right)\,\left(3+i\,0\,\sqrt{5}\right)+\left(b+i\,0\,\sqrt{5}\right)\,\left(1+i\,\sqrt{5}\right)\in P_{2}}\) .

We define \(\displaystyle{f:\mathbb{Z}_{[i\,\sqrt{5}]}\longrightarrow \mathbb{Z}_{3}\,\,,a+i\,b\,\sqrt{5}\mapsto f(a+i\,b\,\sqrt{5})=\left[a-b\right]_{3}}\)

If \(\displaystyle{a_{i}+i\,b_{i}\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}\,,i=1,2}\), then :

\(\displaystyle{\begin{aligned} f((a_{1}+i\,b_{1}\,\sqrt{5})+(a_{2}+i\,b_{2}\,\sqrt{5}))&=f((a_{1}+a_{2})+i\,(b_{1}+b_{2})\,\sqrt{5})\\&=\left[(a_{1}+a_{2})-(b_{1}+b_{2})\right]_{3}\\&=\left[(a_{1}-b_{1}\right]_{3}+\left[a_{2}-b_{2}\right]_{3}\\&=f(a_{1}+i\,b_{1}\,\sqrt{5})+f(a_{2}+i\,b_{2}\,\sqrt{5})\end{aligned}}\)

\(\displaystyle{\begin{aligned}f((a_{1}+i\,b_{1}\,\sqrt{5})\cdot f(a_{2}+i\,b_{2}\,\sqrt{5})&=f((a_{1}\,a_{2}-5\,b_{1}\,b_{2})+i\,(a_{1}\,b_{2}+a_{2}\,b_{1})\,\sqrt{5})\\&=\left[a_{1}\,a_{2}-5\,b_{1}\,b_{2}-a_{1}\,b_{2}-a_{2}\,b_{1}\right]_{3}\\&=\left[(a_{1}\,a_{2}+b_{1}\,b_{2})-(a_{1}\,b_{2}+a_{2}\,b_{1})-6\,b_{1}\,b_{2}\right]_{3}\\&=\left[a_{1}-b_{1}\right]_{3}\cdot \left[a_{2}-b_{2}\right]_{3}-\left[6\right]_{3}\cdot \left[b_{1}\,b_{2}\right]_{3}\\&=f(a_{1}+i\,b_{1}\,\sqrt{5})\cdot f(a_{2}+i\,b_{2}\,\sqrt{5})\end{aligned}}\)

and also \(\displaystyle{f(1_{\mathbb{Z}_{[i\,\sqrt{5}]}})=f(1+i\,0\,\sqrt{5})=\left[1-0\right]_{3}=\left[1\right]_{3}=1_{\mathbb{Z}_{3}}}\) .

Therefore, the function \(\displaystyle{f}\) is a ring homomorphism.

\(\displaystyle{\begin{aligned} \rm{Ker}(f)&=\left\{a+i\,b\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}: f(a+i\,b\,\sqrt{5})=\left[0\right]_{3}\right\}\\&=\left\{a+i\,b\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}: \left[a-b\right]_{3}=\left[0\right]_{3}\right\}\\&=\left\{a+i\,b\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}: \exists\,k\in\mathbb{Z}: a-b=3\,k\right\}\\&=\left\{a+i\,b\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}: a+i\,b\,\sqrt{5}=3\,k+b\,(1+i\,\sqrt{5})\right\}\\&=P_{2}\end{aligned}}\)

Finally, if \(\displaystyle{\left[a\right]_{3}\in\mathbb{Z}_{3}}\), then \(\displaystyle{a+i\,0\,\sqrt{5}\in\mathbb{Z}_{[i\,\sqrt{5}]}}\) and

\(\displaystyle{f(a+i\,0\,\sqrt{5})=\left[a-0\right]_{3}=\left[a\right]_{3}}\), which means that \(\displaystyle{f}\) is onto \(\displaystyle{\mathbb{Z}_{3}}\) .

According to the 1st Isomorphism Theorem, we have that :

\(\displaystyle{(\mathbb{Z}_{[i\,\sqrt{5}]})/P_{2}\simeq \mathbb{Z}_{3}}\) and \(\displaystyle{\left(\mathbb{Z}_{3},+,\cdot\right)}\) is a field.
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