Infinite Field of Characteristic 5 I recently took an exam in which the professor asked to give an example of an infinite field of characteristic 5. 
I had studied this problem, and found examples such as this. 
My answer that I wrote down was $F=\displaystyle\frac{\mathbb{Z}}{5\mathbb{Z}}$. My professor said that was wrong because $F \simeq \mathbb{Z}_5$ which is a finite field. Is he correct? Do I have an argument? How can I prove him wrong?
 A: He is not at all wrong. By mere definition $F=\mathbb{Z}_5$, no? Even if the whole being-a-field/ring thing is screwing you up, you know that, if nothing else, $F$ is a quotient group and $|F|=[\mathbb{Z}:5\mathbb{Z}]=5$.
A: You are wrong, and the professor is correct. Consider the division algorithm: for any $n\in\mathbb{Z}$, there exist $q\in\mathbb{Z}$ and $0\leq r<5$ such that $n=5q+r$. Thus, for any $n\in\mathbb{Z}$, we have
$$n+5\mathbb{Z}=r+5\mathbb{Z}$$
for some $r\in\{0,1,2,3,4\}$. It is also easily checked that these are distinct; thus, $\mathbb{Z}/5\mathbb{Z}$ has exactly 5 elements. In particular, $\mathbb{Z}/5\mathbb{Z}$ is finite. 
A: He is correct, and there is nothing to back up your case I'm afraid. It should be very clear to you that $\mathbb{Z}/5\mathbb{Z}$ is finite. If it isn't, I would respectfully suggest that you go back to review the basics of group theory.
A proper example would have been the field of rational functions with coefficients in $\mathbb{\mathbb{Z}/5\mathbb{Z}}$.
