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I just took my final exam for abstract algebra and have this problem stuck in my head.

Prove that $2$ is irreducible in $\mathbb{Z}[\sqrt{-3}]$ but not prime.

My Solution: Proving that it is irreducible was easy. However, to prove that it is prime, I said as follows. Consider $\mathbb{Z}[\sqrt{-3}]/(2)$. Since this is not an integral domain (since $4 \in \mathbb{Z}[\sqrt{-3}]/(2)$ and $2 \in \mathbb{Z}[\sqrt{-3}]/(2)$ and $4 \times 2 = 0$ in $\mathbb{Z}[\sqrt{-3}]/(2)$), it follows that $(2)$ is not a prime ideal. Since this is not a prime ideal, $2$ is not prime.

Is this a valid solution or is there a big conceptual flaw?

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  • $\begingroup$ both $4$ and $2$ are $0$ in that quotient. Are you familiar with the norm? $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 8:02
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    $\begingroup$ BTW, the idea is fine, you just need to pick elements that are not $0$ in the quotient. $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 8:03
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$$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+3,2)$$ But $(X^2+3,2)=(X^2+1,2)$, so $$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+1,2)\simeq(\mathbb Z/2\mathbb Z)[X]/(X^2+1)=(\mathbb Z/2\mathbb Z)[X]/(X+1)^2$$ which is not an integral domain, so $2$ is not prime.

(If you don't want to identify the quotient $\mathbb{Z}[\sqrt{-3}]/(2)$, only to show it's not an integral domain, then try $(1+\sqrt{-3})(1-\sqrt{-3})$.)

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Hint: $2\mid 4=(1+\sqrt{-3})(1-\sqrt{-3})$

So, using your method, can you find two non-zero elements in the quotient that multiply to make $0$? Your solution fails as whilst it's true that $4\times 2=0$ in the quotient, both $4$ and $2$ are zero in the quotient!

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