Prime ideal in $\mathbb Z[\sqrt{10}]$

Question

I am trying to solve this exercise:

Prove that $\langle 2,\sqrt{10} \rangle$ is a prime ideal in $\mathbb Z[\sqrt{10}]$.

I could do the following:

I pick an element of the form $zw \in \langle 2,\sqrt{10} \rangle$.

If $z=a+b\sqrt{10},w=c+\sqrt{10}$ and I define for any element $z$ in the ring $\overline{z}=a-b\sqrt{10}$, then $zw\overline{zw}=(a^2-10b^2)(c^2-10d^2)$.

If $zw$ is in the ideal, then $$zw=(x_1+\sqrt{10}x_2)2+(y_1+y_2\sqrt{10})\sqrt{10}$$

So $$(a^2-10b^2)(c^2-10d^2)=(2x_1+10y_2)^2-10(2x_2+y_1)^2$$

Since the member of the right is even, so is $(a^2+10b^2)(c^2+10d^2)$, this implies $2$ divides one of the two factors. Suppose $2$ divides $(a^2+10b^2)$, then $2$ divides $a$. So we have $a+b\sqrt{10}=2a'+b\sqrt{10}$, this means $z$ is contained in the ideal. It follows $\langle 2,\sqrt 10 \rangle$ is prime.

I would like to check if my proof is correct, also, if someone has an alternative solution, he or she is welcome to share it.

Answer 1

Recall $I$ prime ideal in ring $R \iff R/I$ is a integral domain.

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$$\frac{\mathbb{Z}[\sqrt{10}]}{\langle 2,\sqrt{10} \rangle} \cong \mathbb{Z}/\langle 2 \rangle$$

Answer 2

If you know something about the norm of an ideal, the following would be a straightforward argument.

Check that $$\langle 2, \sqrt{10} \rangle^2 = \langle 2\rangle$$ so the norm of the ideal $\langle 2, \sqrt{10} \rangle$ is $2$. In particular, $\Bbb{Z}[\sqrt{10}]/\langle 2, \sqrt{10} \rangle = \Bbb{F}_2$, and this implies that it is a prime ideal.

Answer 3

What you're using when you define the conjugate $\bar{z} = a - b\sqrt{10}$ and then multiply by it is that the norm $|a+b\sqrt{10}| = a^2 - b^2 10$ is well-defined and multiplicative. Your proof looks good. The other solutions are also good but if you're practicing this stuff or doing it as homework then I like your way of doing it.