MHB Rings of Fractions and Fields of Fractions

Click For Summary
Rings of Fractions and Fields of Fractions are explored in the context of understanding the prime subfield of a field F. The exercise requires proving that F contains a unique smallest subfield F_0, which is isomorphic to either the rational numbers (ℚ) or the finite field (ℤ/pℤ) for some prime p. A hint suggests that if F has zero characteristic, F_0 can be constructed using fractions of integers, while if F has characteristic p, F_0 is formed from multiples of a prime. The discussion reflects a need for clarity on these concepts and how to approach the proof. Understanding these foundational elements is crucial for further studies in abstract algebra.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am seeking to understand Rings of Fractions and Fields of Fractions - and hence am reading Dummit and Foote Section 7.5

Exercise 3 in Section 7.5 reads as follows:

Let F be a field. Prove the F contains a unique smallest subfield F_0 and that F_0 is isomorphic to either \mathbb{Q} or \mathbb{Z/pZ} for some prime p. (Note: F_0 is called prime subfield of F.)

I am somewhat overwhelmed with this exercise and need help to get started. Can anyone help with this exercise.

Peter
 
Physics news on Phys.org
Peter said:
Let F be a field. Prove the F contains a unique smallest subfield F_0 and that F_0 is isomorphic to either \mathbb{Q} or \mathbb{Z/pZ} for some prime p. (Note: F_0 is called prime subfield of F.)

Hints: If $$\mathbb{F}$$ has zero characteristic, then $F_0=\{m\cdot 1/n\cdot 1:m\in\mathbb{Z},n\in\mathbb{N^*}\}$ is a subfield of $\mathbb{F}$ isomorphic to $\mathbb{Q}$. If $$\mathbb{F}$$ has characteristic $p$ then, $F_0=\{m\cdot 1:m\in\mathbb{N}\}$ is a subfield of $\mathbb{F}$ somorphic to $\mathbb{Z}/(p)$.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
854
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K