# I Field of zero characteristics

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1. Jun 14, 2018

### Danijel

I am interested in the following theorem:
Every field of zero characteristics has a prime subfield isomorphic to ℚ.
I am following the usual proof, where we identify every p∈ℚ as a/b , a∈ℤ,beℕ, and define h:ℚ→P as h(a/b)=(a*1)(b*1)-1 (where a*1=1+1+1... a times) I have worked out the multiplicative rule for this homomorphism, but I am not sure how to prove the additive. I am also interested in what does the notation
(a*1)(b*1)-1 exactly mean. Thanks.

2. Jun 14, 2018

### Staff: Mentor

All we know about the prime field is its characteristic and $0,1 \in \mathbb{P}$. $a\in \mathbb{P}$ is not sure. However $\underbrace{1+\ldots +1}_{a \text{ times}}=a \,\cdot \, 1$ for $a \in \mathbb{N}$ is, and likewise for negative $a$. That's why $a$ times $1$ is used instead of $a$. We only have isomorphic images of $a$ in $\mathbb{P}$.

For the homomorphism, I guess $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$ will be needed.

3. Jun 14, 2018

### mathwonk

this probably won't help much until you have some commutative algebra but it gives the outline. First of all since the field F is an abelian group, there is exactly one additive homomorphism from the integers Z into F for each choice of the image of 1 from Z, so choose that image to be the unit element 1 in P. Now you have the unique additive homomorphism from Z to P that fresh described sending each positive integer n to 1+...+1 (n times). By definition of the multiplication in Z this is also a multiplicative map hence a ring map. Then since each non zero integer goes to an invertible element of F, (because F has characteristic zero), there is a unique extension of the previously defined ring homomorphism Z-->F to a ring homomorphism Q --> F. Then since Q is a field, this ring map is necessarily injective, hence defines an isomorphism onto the smallest subfield of F, i.e. the prime field.