# Prime Ideals

#### 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!

#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle P \cap \mathbb{Z} \neq \{0\}.$$ (Nod)

#### 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?

#### chiph588@

MHF Hall of Honor
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 $$\displaystyle a,b \in \mathbb{Z}$$.

Boysilver

#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle \mathbb{Z}.$$

Boysilver

#### Jim63

a less trivial fact is this: if $$\displaystyle P \neq \{0\},$$ then $$\displaystyle P \cap \mathbb{Z} \neq \{0\}.$$ (Nod)
Why is this true?

#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle \mathcal{O}_K.$$

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