Does ring homomorphism from $F_p$ to $\mathbb{Q}$ exist?

353 Views Asked by At

I was asked this question by my classmates , Let $p$ be a prime, I want to know whether there exists ring homomorphism between $\mathbb{Q}$ and $F_p$ .

  1. Can we construct ring homomorphisms from $F_p$ to $\mathbb{Q}$?

  2. Can we construct ring homomorphisms from $\mathbb{Q}$ to $F_p$?

I know the fact that if there exists a ring homomorphism $f : R \to S$ then the characteristic of $S$ divides the characteristic of $R$. And the characteristic of $\mathbb{Q}$ is $0$, and that that of $F_p$ is $p$. therefore [2]there are no homorphisms

Is this correct ? And how can I solve [1]?

I did not come up with example of this homomorphism but cannot proof there exists no homomorphism either.

1

There are 1 best solutions below

0
On BEST ANSWER

I assume that your ring homomorphism need to preserve $1$, which is a common convention.

It is correct that for $f:R \to S$ a ring homomorphism the characteristic of $S$ divides that of $R$.

Yet note that $p \mid 0$ while $0 \nmid p$. Whence you can exclude, by this argument, the existence of a ring homomorphism in the first point. Not, the second one.

For the second point, assume there is ring homomorphism $f$. We know that $f(1)=1$. Consider what happens to $1/p$. Let's put $f(1/p)=y$.

Then we have, in $F_p$:

$$1=f(1) = f(p(1/p))= p f(1/p)=0$$ a contradiction.

Another way to argue, if you know that result, is that a ring homomorphism of fields is always injective, but it is possible/likely that you are not yet aware of this result.