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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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pigishpig
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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
  • @Patrick, HUH? We write quotients literally all the time in that way. Here we have someone having problems with seeing the finiteness of a $Z/5$: introducing very strange notational complications is hardly helpful. – Mariano Suárez-Álvarez May 27 '23 at 04:57
  • @MarianoSuárez-Álvarez I am writing a comment, not an answer; you're welcome to disagree. I write fractions that way (as in elements of a field or some localized ring), but I avoid writing group or ring quotients (or any kind of algebraic structure quotient) like this at all costs. Even when doing the isomorphism theorems, I prefer $(R/\mathfrak a)/(\mathfrak b/\mathfrak a) \simeq (R/\mathfrak b)$ over $\frac{R/\mathfrak a}{\mathfrak b/\mathfrak a} \simeq \frac R{\mathfrak b}$ all the time, and I can't remember of a book who didn't agree with me. Happy to be proved wrong though! – Patrick Da Silva May 29 '23 at 15:13

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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
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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
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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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Here we have Z/5Z is a set with some elements which are also set. In Z/5Z there are 5 distinct elements(sets). Sets which are belongs to Z/5Z are infinite set but total number of sets(elements) in Z/5Z are finite(5). Thus we conclude Z/5Z is finite set. Obviously Z/5Z is a field with characteristics 5 but this is finite.

Hariom Chauhan
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