Prove that $(3, 2 + \sqrt{-5})$ is prime ideal in $\mathbb Z[\sqrt{-5}]$. I want to show that $(3, 2 + \sqrt{-5})$ is maximal directly from definition, ie. show that by adding any other element to it we get the entire ring. Any hints on how to do this? I already know that the fact can be shown by using different isomorphisms.
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Consider $(3,2+\sqrt{-5},a+b\sqrt{-5})$. You have $$(3,2+\sqrt{-5},a+b\sqrt{-5})=(3,2+\sqrt{-5},a+b\sqrt{-5}-b(2+\sqrt{-5}))=\\=(3,2+\sqrt{-5},a-2b)=\begin{cases}(3,2+\sqrt{-5})&\text{if }3\mid a-2b\text{ in }\Bbb Z\\ (1)&\text{if }3\nmid a-2b\text{ in }\Bbb Z\end{cases}$$
The first case occurs when $a+b\sqrt{-5}\in(3,2+\sqrt{-5})$.
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This can be viewed as a special case of reduction modulo Hermite normal form - see my answer. – Bill Dubuque Feb 08 '17 at 17:42
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Hint $ $ Write $\, w = \sqrt{-5}\ $ and employ $\rm\color{#0a0}{modular}$ reduction as in Euclid's GCD algorithm
$$\begin{align} (w\!+\!2,\,3,\, a\!+\!bw) &=\, (w\!+\!2,\,3,\: a\!+\!\color{#c00}w\,b\color{#0a0}{\bmod}{w\! +\! 2})\\ &=\, (w\!+\!2,\,3,\, a\!\color{#c00}{-2}\,b)\ \ {\rm by}\ \ \color{#c00}w\equiv\color{#c00}{-2}\!\!\!\pmod{\!w\!+\!2}\\ &=\, (w\!+\!2,\,3)\ \ \,{\rm if}\ \ (3,\,a\!-\!2b)=3\ \ {\rm else}\\ &=\, (1)\qquad\quad \ \, {\rm if}\ \ (3,\,a\!-\!2b)=1 \end{align} $$
Hence $\ a\!+\!bw\not\in I = (3,w\!+\!2)\,\Rightarrow\, (I,a\!+\!bw) = (1),\ $ so $\,I\,$ is maximal
Remark $\ $ The above modular reduction is essentially a special case of normal form reduction modulo a Hermite normal form basis - a powerful method which proves very useful for computation. See this answer for more.
Bill Dubuque
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Thanks for your answer. i just want to confirm, the reason that $(a - 2b)$ can only be either $(3)$ or $(1)$ is because $3$ is prime in $Z[\sqrt{-5}]$? Also, could you briefly expand on how we go from $(3, w + 2, a - 2b)$ to $(3, w + 2, (a - 2b, 3))$? I kind of get it but I want to make sure. – b_pcakes Feb 08 '17 at 21:51
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@b_pcakes $ $ For integers $,m,n,$ we have $, m,n\in I\iff \gcd(m,n)\in I,$ because by Bezout $,\gcd(m,n) = im+jn,$ for some $,i,j\in \Bbb Z.,$ So any integers occurring as generators in an ideal can be replaced by their gcd. In particular the generators $,3,n,$ can be replaced by $,\gcd(3,n) = 1$ or $,3,,$ because $3$ is prime (so irreducible) in $\Bbb Z.,$ Above is special case $,n=a-2b.\ \ $ – Bill Dubuque Feb 08 '17 at 22:18
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$\mathbb Z[\sqrt{-5}]\cong \mathbb Z[x]/(x^2+5)$, then $\mathbb Z[\sqrt{-5}]/(3,2+\sqrt{-5})\cong \mathbb Z[x]/(x^2+5,3,2+x)\cong \mathbb Z_3[x]/(x^2+5,2+x)=\mathbb Z_3[x]/(x^2+2,x+2)$
In other hand we have in $\mathbb Z_3$ that $x^2+2=(x+2)(x+1)$, so $x^2+2\in(x+2)$ and $(x^2+2,x+2)=(x+2)$, so $\mathbb Z_3[x]/(x^2+2,x+2)=\mathbb Z_3[x]/(x+2)\cong \mathbb Z_3[-2]=\mathbb Z_3[1]=\mathbb Z_3$, then $\mathbb Z[\sqrt{-5}]/(3,2+\sqrt{-5})\cong \mathbb Z_3$ and $(3,2+\sqrt{-5})$ is maximal ideal.
Mustafa
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In the second paragraph it is simpler to use Euclid'a algorithm $$(x!+!2,,f(x)), =, (x!+!2,,f(x)\bmod x+2), =, (x!+!2,f(-2))$$ – Bill Dubuque Feb 08 '17 at 16:52
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