$\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,then the number of elements in $\mathbb F$ is prime

Question

Assume $\mathbb F$ is a field and $\phi:\mathbb Z \to \mathbb F$ is a surjective ring homomorphism,prove that the number of elements in $\mathbb F$ should be a prime number.

Based on my knowledge

• $\phi(a+b)=\phi(a)+\phi(b)$ for all $a,b \in \mathbb Z$
• $\phi(ab)=\phi(a)\phi(b)$ for all $a,b \in \mathbb Z$
• $\phi(1)=\phi(1_{\mathbb F})$

Moreover surjectivity of $\phi$ implies that for every element $x \in \mathbb F$ there is some $a \in \mathbb Z:\phi(a)=x$

But I don't understand how this is related to the number of elements in $\mathbb F$ and how it should be a prime number,maybe the surjectivity of such a mapping implies that $\mathbb F$ is a finite field and hence the order of $\mathbb F$ should be of the form $p^n$ ,where $n$ is a natural number and $p$ is a prime number,but even if it happens then $\text{ord}(\mathbb F) $ is not necessarily a prime number.

Answer 1

Hint: By the homomorphism theorem $\phi$ induces an isomorphism $\mathbb Z/\operatorname{ker}(\phi)\to \mathbb F$. What are the possible ideals $I\subseteq \mathbb Z$, quotients $\mathbb Z/I$ and when is such a quotient a field?

Answer 2

For each integer $n$, one has $\phi(n) = \phi(n1) = n\phi(1) = n1_F$. Since $\phi$ is onto, $F = \{n1_F\mid n\in\Bbb Z\}$.

If $F$ has characteristic 0, it contains the rational numbers and so the mapping cannot be onto.

Thus $F$ must have characteristic $p>0$ and $p$ is prime. In $F$, $n1_F=0$ holds only if $n$ is prime. Otherwise, it would contain zero divisors.

Answer 3

Hint: Since $\phi$ is a ring homomorphism, $\phi(n)=n\phi(1)$. Since the field is finite (otherwise $\phi$ is injective -- what goes wrong in this case?), there must be some integer $m$ for which $\phi(m)=0$, and thus $\mathbb F$ consists exactly of $$\{0,\phi(1),\dots,(m-1)\phi(1)\}.$$ What does this mean about $m$, and thus the size of the field?