# prime field

• Nov 23rd 2009, 09:42 AM
ux0
prime field
I need to show that the prime field $\mathbb{R}$ and the prime field $\mathbb{C}$ are in $\mathbb{Q}$

could i get help showing one of these (the harder one) then ill try doing the other one... ??
• Nov 23rd 2009, 10:24 AM
Jose27
I assume you mean the prime subfield of $\mathbb{R}$, notice that by definition this is the intersection of all subfields of $\mathbb{R}$ ie. $P(\mathbb{R})= \bigcap_{B\subset \mathbb{R} : \ B \ field } B$. Since $1 \in P(\mathbb{R})$ and it is a field, it contains a copy of $\mathbb{Z}$. Now $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$ so $\mathbb{Q}$ is the smallest field containing $\mathbb{Z}$, so $\mathbb{Q} \subset P(\mathbb{R} )$ and since $P(\mathbb{R}) \cap \mathbb{Q} = P(\mathbb{R})$...
• Nov 23rd 2009, 06:24 PM
ux0
wow i was completely off on the question.. sorry.

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

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

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

iii) Show that the prime field of $\mathbb{C}$ is $\mathbb{Q}$
• Nov 23rd 2009, 09:23 PM
Jose27
Read carefully my first post and you'll notice that it answers all three questions.

As for $\mathbb{Z} [i]$ this is, by definition, the smallest ring that contains both $\mathbb{Z}$ and $i$, and $\mathbb{Z} (i)$ is the aboves fraction field, and since it contains $\mathbb{Z}$ it must contain $\mathbb{Q}$
• Nov 24th 2009, 12:18 AM
Shanks
Yeah, Jose27 is quite right.
• Nov 24th 2009, 07:08 AM
ux0
Quote:

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

Ohhhh so what your saying in words is that

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

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