# Thread: prime field

1. ## prime field

I need to show that the prime field $\displaystyle \mathbb{R}$ and the prime field $\displaystyle \mathbb{C}$ are in $\displaystyle \mathbb{Q}$

could i get help showing one of these (the harder one) then ill try doing the other one... ??

2. I assume you mean the prime subfield of $\displaystyle \mathbb{R}$, notice that by definition this is the intersection of all subfields of $\displaystyle \mathbb{R}$ ie. $\displaystyle P(\mathbb{R})= \bigcap_{B\subset \mathbb{R} : \ B \ field } B$. Since $\displaystyle 1 \in P(\mathbb{R})$ and it is a field, it contains a copy of $\displaystyle \mathbb{Z}$. Now $\displaystyle \mathbb{Q}$ is the field of fractions of $\displaystyle \mathbb{Z}$ so $\displaystyle \mathbb{Q}$ is the smallest field containing $\displaystyle \mathbb{Z}$, so $\displaystyle \mathbb{Q} \subset P(\mathbb{R} )$ and since $\displaystyle P(\mathbb{R}) \cap \mathbb{Q} = P(\mathbb{R})$...

3. wow i was completely off on the question.. sorry.

i) Show that every subfield of $\displaystyle \mathbb{C}$ contains $\displaystyle \mathbb{Q}$

I thought $\displaystyle \mathbb{Z}[i]$ was a subfield of $\displaystyle \mathbb{C}$ but there is no $\displaystyle \mathbb{Q}$ subfield of $\displaystyle \mathbb{Z}[i]$... but wouldn't that disprove this statement... $\displaystyle (\mathbb{Z}[i]$ units are 1,-1,i,-i )

ii) Show that the prime field of $\displaystyle \mathbb{R}$ is $\displaystyle \mathbb{Q}$

iii) Show that the prime field of $\displaystyle \mathbb{C}$ is $\displaystyle \mathbb{Q}$

4. Read carefully my first post and you'll notice that it answers all three questions.

As for $\displaystyle \mathbb{Z} [i]$ this is, by definition, the smallest ring that contains both $\displaystyle \mathbb{Z}$ and $\displaystyle i$, and $\displaystyle \mathbb{Z} (i)$ is the aboves fraction field, and since it contains $\displaystyle \mathbb{Z}$ it must contain $\displaystyle \mathbb{Q}$

5. Yeah, Jose27 is quite right.

6. Originally Posted by Jose27
I assume you mean the prime subfield of $\displaystyle \mathbb{R}$, notice that by definition this is the intersection of all subfields of $\displaystyle \mathbb{R}$ ie. $\displaystyle P(\mathbb{R})= \bigcap_{B\subset \mathbb{R} : \ B \ field } B$. Since $\displaystyle 1 \in P(\mathbb{R})$ and it is a field, it contains a copy of $\displaystyle \mathbb{Z}$. Now $\displaystyle \mathbb{Q}$ is the field of fractions of $\displaystyle \mathbb{Z}$ so $\displaystyle \mathbb{Q}$ is the smallest field containing $\displaystyle \mathbb{Z}$, so $\displaystyle \mathbb{Q} \subset P(\mathbb{R} )$ and since $\displaystyle P(\mathbb{R}) \cap \mathbb{Q} = P(\mathbb{R})$...

Ohhhh so what your saying in words is that

Because the intersection of the prime subfiled of $\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{Q}$ is just the prime subfield of $\displaystyle \mathbb{R}$ , we can conclude that $\displaystyle \mathbb{Q} = P(R)$ ,

and the proof is the same thing for $\displaystyle \mathbb{C}$ to show the prime subfiled of $\displaystyle \mathbb{C}$ is $\displaystyle \mathbb{Q}$ ??? (basically just change the R's to C's)