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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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$\mathbb Z[\sqrt{-5}]\cong \mathbb Z[x]/(x^2+5)$, 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.

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    In the second paragraph it is simpler to use Euclid'a algorithm $$(x\!+\!2,\,f(x))\, =\, (x\!+\!2,\,f(x)\bmod x+2)\, =\, (x\!+\!2,f(-2))$$ – 2017-02-08
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    Thank you for comment and '+1' for answer. – 2017-02-08
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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. – 2017-02-08
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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.

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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. – 2017-02-08
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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.\ \ $ – 2017-02-08