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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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    $\begingroup$ $\mathbb{Z}/5\mathbb{Z}$ 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. $\endgroup$ – Arturo Magidin Dec 1 '11 at 5:30
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    $\begingroup$ $\mathbb{Z}/5\mathbb{Z}$ only has $5$ elements, which, regrettably, is finite. Sorry mate. $\endgroup$ – Elchanan Solomon Dec 1 '11 at 5:31
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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Dec 1 '11 at 6:09
  • $\begingroup$ See also math.stackexchange.com/q/58424/18880 $\endgroup$ – Marc van Leeuwen Dec 1 '11 at 14:37
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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 $|F|=[\mathbb{Z}:5\mathbb{Z}]=5$.

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  • $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Dec 1 '11 at 6:07
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    $\begingroup$ @Patrick: Your nitpicking is out of place here. "The" cyclic group with $5$ elements, $C_5$, is an isomorphism class of groups, each having four automorphisms, which groups are not endowed with a ring or field structure (though they can be so equipped). For those for which $\mathbb{Z}_5$ does not mean the $5$-adic integers, it is a shorthand for $\mathbb{Z}/5\mathbb{Z}$, so (for once) I think equality may be asserted. $\endgroup$ – Marc van Leeuwen Dec 1 '11 at 7:27
  • $\begingroup$ @Marc van Leeuwen : Fine then. If I don't see things your way that doesn't mean I am "out of place". You don't need to be aggressive because of a comment... calm down man. $\endgroup$ – Patrick Da Silva Dec 1 '11 at 17:19
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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.

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

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  • $\begingroup$ 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). $\endgroup$ – Marc van Leeuwen Dec 1 '11 at 7:40

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