How to calculate the cardinality of $\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17})$ and $\mathbb{Z}[\sqrt{-17}]/(\sqrt{-17})$? thanks for taking the time to look at my problems.
I was trying to calculate the norm of $(3, 1 + \sqrt{-17})$ and $(\sqrt{-17})$.
The second one is 17 because of the norm of the element $\sqrt{-17}$, but how does this follow from $|\mathbb{Z}[\sqrt{-17}]/(\sqrt{-17})|$?
I tried to calculate $|\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17})|$ and concluded that $\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17}) \cong \mathbb{Z}/3\mathbb{Z}$ such that $|\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17})| = 3$. Is this correct?
Thanks in advance!
 A: Your computations are correct.
Since $-17\equiv3\bmod{4}$ our ring of integers is $\mathbb{Z}[\sqrt{-17}]$, so we may factor the ideal $(3)$ in $\mathbb{Z}[\sqrt{-17}]$ by factoring 
$$x^2 + 17 \equiv x^2 - 1 \equiv (x+1)(x+2) \bmod{3}.$$  
This yields the ideal $(3,1+\sqrt{-17})$, and since 3 splits the norm of this ideal is 3.  
To see this more directly, we can use the ring isomorphism theorems.  We have $(x^2+17) \subseteq (3,1+x) \subseteq \mathbb{Z}[x]$ from above, hence 
$$\mathbb{Z}[\sqrt{-17}]/(3,1+\sqrt{-17}) \cong
   \mathbb{Z}[x]/(3,1+x) \cong
   \mathbb{Z}/3$$
As for $(\sqrt{-17})$, the same argument works.
A: Given $a+b\sqrt{-17}$, you can subtract $b(1+\sqrt{-17})$ to get a rational integer, then subtract an appropriate multiple of 3 to get 0, 1, or 2. So the quotient ring has at most 3 elements, indeed, has number of elements a divisor of 3, so it now suffices to show it's not 1. 
If it's 1, then 1 is in the ideal, $1=3(a+b\sqrt{-17})+(1+\sqrt{-17})(c+d\sqrt{-17})$. Multiply everything out, and equate rational terms and equate irrational terms, to get the equations, $1=3a+c-17d$, $0=3b+c+d$. Subtraction yields $1=3(a-b-6d)$, a clear impossibility, so it's 3. 
A: You already seem know that the norm of a prinicipal ideal is the norm of its generator. Hence $|\mathbb Z[\sqrt{-17}]/(3)|=9$.
We have $(3) \subsetneq (3,1+\sqrt{-17}) \subsetneq (1)$, hence $\mathbb Z[\sqrt{-17}]/(3,1+\sqrt{-17})$ is a non-trivial quotient of $\mathbb Z[\sqrt{-17}]/(3)$. A non-trivial quotient of a group with $9$ elements must have $3$ elements, hence we obtain
$$|\mathbb Z[\sqrt{-17}]/(3,1+\sqrt{-17})|=3$$
without having calculated the quotient (Which would be of course an easy task, too, as shown in the other answers).
