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We all know that numbers can be prime. Can the concept of primality be generalised to functions?

If so, how? What are examples of prime functions and under what circumstances is a function prime?

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    If you have a ring of functions, you can talk about prime ideals (or principal and prime ideals). – 2017-02-23
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    Absolutely. Prime ideal generalizes to general rings.https://en.wikipedia.org/wiki/Prime_ideal – 2017-02-23
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    Are [irreducible polynomials](https://en.wikipedia.org/wiki/Irreducible_polynomial) sort of in the direction you want? – 2017-02-23
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    @BrianTung With the great advantage that it can be efficiently verfied whether a polynomial in $\mathbb Z[x]$ is squarefree. This is not possible in $\mathbb Z$. I remember a joke, when a student said "It would be useful to have a derivate on positive integers" :) – 2017-02-23
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    @Peter Do you mean $\mathbb Q[x]$? $\mathbb Z$ sits in $\mathbb Z[x]$ and its primes are still prime. – 2017-02-24
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    No, I mean, there is no efficient way to determine in general whether a positive integer is squarefree. But it is possible to verify efficiently whether a polynomial over the integers is squarefree. – 2017-02-24
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    What are Zx and Qx? Rings of polynomials in Q and Z? – 2017-02-25
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    @Peter Integers are polynomials over the integers and they're square free if and only if they're square free as polynomials over $\mathbb Z$. They're units as polynomials over $\mathbb Q$ so in that situation they're always square free. But if you're restricting yourself only to $\mathbb Z[x]$ determining whether it is square free is strictly harder. – 2017-02-25
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    @Peter For example, $4x+4$ is not square free in $\mathbb Z[x]$ despite being of degree $1$. But it's square free in $\mathbb Q[x]$. – 2017-02-25
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    @Peter Alternatively you could restrict yourself to monic polynomials, in which case the easy method works. – 2017-02-25
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    @MattSamuel What are Zx and Qx? Rings of polynomials in Q and Z? – 2017-02-25
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    @Robert yes they are – 2017-02-25
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    @MichaelBurr Perhaps it's a ring of functions I'm interested in. Can there be some concept of primality of a function, which prevents it ever being cyclic? i.e. we might imagine $f^n(x)=x\implies f(x)$ factors $f^n(x)$. For example if $f^n(x)=x$ for some $x$, can that preclude $f^n(x)=x$ for any other $x$? It's possible I'm thinking of an identity relation rather than primality. – 2017-02-25
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    Consider $K[x]$ where $K$ is a field. In this case, all ideals are principal ($K[x]$ is a Euclidean domain). Moreover, the prime ideals correspond to irreducible polynomials. – 2017-02-25
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    @MichaelBurr I'm shocked (and pleased) you've mentioned Euclidean domains because I'm sure this is profoundly important to the problem I'm working on. What does "all ideals are principal" mean? – 2017-02-25

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What kind of functions? Functions on the naturals? Complex numbers? Strings? Or functions on functions? Secondly, the notion of "primality" usually only makes sense once you have defined multiplication on your chosen functions. Things are going to be very different depending on your choices.

For example if $F$ is the ring of functions on $\mathbb{R}$ with pointwise addition and multiplication, then the prime elements of $F$ are precisely those elements in $F$ which have exactly one zero. Why? Those with no zero are units, while those with more than one zero are clearly not prime. (Prove it!)

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    Can you give an example of an element in $F$ with a zero in it? Because it would seem to me that $f=x+0$ is a function on R with a zero in it, but I'm not sure this is what you mean. – 2017-02-27
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    @RobertFrost: I'm using standard terminology; see [Wikipedia](https://en.wikipedia.org/wiki/Zero_of_a_function). – 2017-02-28
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    Thanks. I was looking for the most general answer; which appears to be that primality is some notion of "irreducibility" of a function. But the **specific** case I want to apply it to, to understand better, is I want to understand when is it true that $\nexists n,m:f^n(x)=f^m(x)$. – 2017-02-28
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    @RobertFrost: Irreducibility and primality are two **different notions**, and any proper textbook on abstract algebra will tell you about them. In particular, in any ring prime elements are irreducible, but irreducible elements may not be prime in some rings. You need to learn at least some basic abstract algebra if you wish to grasp these concepts. – 2017-02-28
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    I haven't fully thought through what you put yet. Good things come to those who wait. – 2017-02-28
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    Do I understand correctly, this notion of primality you describe relates to functions being multiplied together? Or are they prime under some other form of combination? Because $f(x)=x^2$ is prime as it has one zero, and $f(x)=x$ is prime also, but $f=x^3$ is a product of the two, so not prime, but that also has exactly one zero suggesting it is. OR are we talking about primality under combination? – 2017-03-01
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    @RobertFrost: As I had stated in my answer, the notion of "primality" usually only makes sense **once you have defined multiplication** on your chosen functions. Also, I specified **pointwise multiplication**, but for primality we must count zeroes with their multiplicity. $( x \mapsto x^2 )$ has two zeroes and is not prime, because it is the pointwise product of $( x \mapsto x )$ and itself. In general, if $p,q$ are primes then $pq$ is not a prime, otherwise since $pq \mid pq$ by symmetry we can assume that $pq \mid p$ and hence $qk = 1$ for some $k$, contradicting $q$ not being a unit. – 2017-03-02
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    @RobertFrost: And the lemma I just showed can be used to easily prove the claim I had left as an exercise in my post. – 2017-03-02