1. ## Prime Ideals

If $\displaystyle K$ is a number field with ring of integers $\displaystyle \mathcal{O}_K$, then prove that if $\displaystyle P$ is a prime ideal of $\displaystyle \mathcal{O}_K$ then $\displaystyle P \cap \mathbb{Z}$ is a prime ideal of $\displaystyle \mathbb{Z}$.

It's easy to show that it's an ideal of $\displaystyle \mathbb{Z}$ but I'm struggling to show that it must be prime; any help would be very useful!

2. Originally Posted by Boysilver
If $\displaystyle K$ is a number field with ring of integers $\displaystyle \mathcal{O}_K$, then prove that if $\displaystyle P$ is a prime ideal of $\displaystyle \mathcal{O}_K$ then $\displaystyle P \cap \mathbb{Z}$ is a prime ideal of $\displaystyle \mathbb{Z}$.

It's easy to show that it's an ideal of $\displaystyle \mathbb{Z}$ but I'm struggling to show that it must be prime; any help would be very useful!
it's even easier to show that it's prime in $\displaystyle \mathbb{Z}$: if $\displaystyle a,b \in \mathbb{Z}$ and $\displaystyle ab \in P \cap \mathbb{Z},$ then since $\displaystyle ab \in P$ and $\displaystyle P$ is prime in $\displaystyle \mathcal{O}_K,$we'll have $\displaystyle a \in P$ or $\displaystyle b \in P.$ so either $\displaystyle a \in P \cap \mathbb{Z}$ or $\displaystyle b \in P \cap \mathbb{Z}.$

note that $\displaystyle P \cap \mathbb{Z} \neq \mathbb{Z}$ because $\displaystyle 1 \notin P.$ a less trivial fact is this: if $\displaystyle P \neq \{0\},$ then $\displaystyle P \cap \mathbb{Z} \neq \{0\}.$

3. No doubt you're right, but I guess my issue is that I don't see how $\displaystyle a \in P$ or $\displaystyle b \in P$ implies $\displaystyle a \in P \cap \mathbb{Z}$ or $\displaystyle b \in P \cap \mathbb{Z}$. Just because $\displaystyle ab \in \mathbb{Z}$ this doesn't mean both $\displaystyle a,b \in \mathbb{Z}$. Am I missing something obvious?

4. Originally Posted by Boysilver
No doubt you're right, but I guess my issue is that I don't see how $\displaystyle a \in P$ or $\displaystyle b \in P$ implies $\displaystyle a \in P \cap \mathbb{Z}$ or $\displaystyle b \in P \cap \mathbb{Z}$. Just because $\displaystyle ab \in \mathbb{Z}$ this doesn't mean both $\displaystyle a,b \in \mathbb{Z}$. Am I missing something obvious?
Since our ideal is over $\displaystyle \mathbb{Z}$ we assume $\displaystyle a,b \in \mathbb{Z}$.

5. Originally Posted by Boysilver
No doubt you're right, but I guess my issue is that I don't see how $\displaystyle a \in P$ or $\displaystyle b \in P$ implies $\displaystyle a \in P \cap \mathbb{Z}$ or $\displaystyle b \in P \cap \mathbb{Z}$. Just because $\displaystyle ab \in \mathbb{Z}$ this doesn't mean both $\displaystyle a,b \in \mathbb{Z}$. Am I missing something obvious?
don't forget that $\displaystyle a$ and $\displaystyle b$ are in $\displaystyle \mathbb{Z}$ because you want to prove that $\displaystyle P \cap \mathbb{Z}$ is a prime ideal of $\displaystyle \mathbb{Z}.$

6. Originally Posted by NonCommAlg
a less trivial fact is this: if $\displaystyle P \neq \{0\},$ then $\displaystyle P \cap \mathbb{Z} \neq \{0\}.$
Why is this true?

7. Originally Posted by Jim63
Why is this true?
let $\displaystyle 0 \neq a \in P$ and suppose that $\displaystyle f(x)=x^n + c_1x^{n-1} + \cdots + c_n$ is the minimal polynomial of $\displaystyle f(x)$ over $\displaystyle \mathbb{Q}.$ it's a known fact that, since $\displaystyle a \in \mathcal{O}_K,$ we have $\displaystyle c_j \in \mathbb{Z},$ for all $\displaystyle j.$

now if $\displaystyle n=1,$ then since $\displaystyle 0=f(a)=a+c_1,$ we have $\displaystyle a=-c_1 \in \mathbb{Z}$ and so $\displaystyle 0 \neq a \in P \cap \mathbb{Z}.$ if $\displaystyle n \geq 2,$ then $\displaystyle c_n \neq 0$ because of minimality of $\displaystyle f(x).$ thus $\displaystyle f(a)=0$ gives us

$\displaystyle c_n=-(a^n + c_1a^{n-1} + \cdots + c_{n-1}a) \in P$ and so $\displaystyle 0 \neq c_n \in P \cap \mathbb{Z}.$

Note: in the proof we didn't need $\displaystyle P$ to be a prime ideal. so the above result holds for any non-zero ideal $\displaystyle P$ of $\displaystyle \mathcal{O}_K.$