# Thread: Prime Ideals

1. ## Prime Ideals

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

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

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

It's easy to show that it's an ideal of $\mathbb{Z}$ 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 $\mathbb{Z}$: if $a,b \in \mathbb{Z}$ and $ab \in P \cap \mathbb{Z},$ then since $ab \in P$ and $P$ is prime in $\mathcal{O}_K,$we'll have $a \in P$ or $b \in P.$ so either $a \in P \cap \mathbb{Z}$ or $b \in P \cap \mathbb{Z}.$

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

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

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

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

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

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

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

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