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?

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    $\mathbb{Z}/5\mathbb{Z}$ has five elements and only five elements (each element is a set which contains infinitely many integers, but *this* set has only five things in it), so it is not an infinite field. You cannot prove him wrong, because he is not wrong. – Arturo Magidin Dec 01 '11 at 05:30
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    $\mathbb{Z}/5\mathbb{Z}$ only has $5$ elements, which, regrettably, is finite. Sorry mate. – Elchanan Solomon Dec 01 '11 at 05:31
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    I just wanted to point out that writing $\frac{\mathbb Z}{5\mathbb Z}$ is not exactly a good idea in terms of notation because this is not a fraction. The standard notation for quotients is the forward slash, thus $\mathbb Z / 5\mathbb Z$ is what you should use from now on to denote quotients. – Patrick Da Silva Dec 01 '11 at 06:09
  • See also http://math.stackexchange.com/q/58424/18880 – Marc van Leeuwen Dec 01 '11 at 14:37

3 Answers3


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 $|F|=[\mathbb{Z}:5\mathbb{Z}]=5$.

Alex Youcis
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  • By definition $\mathbb Z_5$ is the cyclic group with $5$ elements. $\mathbb Z / 5\mathbb Z$ is just isomorphic to it =) it is impressive how most of us think of isomorphic things as "the same things". I am actually liking your answer. – Patrick Da Silva Dec 01 '11 at 06:07

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.

Zev Chonoles
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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 $\mathbb{\mathbb{Z}/5\mathbb{Z}}$.

Bruno Joyal
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  • This example is indeed correct. It requires understanding well the difference between polynomials and polynomial functions though. To my amazement I found that Lang, Linear Algebra, *defines* polynomials (over *any* field) as polynomial functions, and starts stating a patently false theorem that their coefficients can be recovered (which is conveniently only proved for the reals and the complexes). – Marc van Leeuwen Dec 01 '11 at 07:40