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Will a prime $p^{0}$ be considered a unique prime in prime decomposition of a number? If the answer to the above question is yes then it breaks the uniqueness which the Fundamental Theorem of Arithmetic proposed.

If the answer is no, which I think is true, then why don't we count it.

Isn't it also the power of a prime?

That way I can say $4=2×2$ or $2×2×3^{0}$ and so on.

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    $\begingroup$ See this answer for a discussion. $\endgroup$ – Cameron Buie Apr 15 '15 at 12:17
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    $\begingroup$ decomposition is unique up to units (and order of factors) and $p^0$ is a unit, so you can add/remove how many units you want without breaking essential uniquiness of factorization $\endgroup$ – Danae Kissinger Apr 15 '15 at 12:17
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    $\begingroup$ The question should be in the body, not just in the title. Please edit accordingly. $\endgroup$ – Gerry Myerson Apr 15 '15 at 12:42
  • $\begingroup$ Looked to that Gerry 👍. Thanks Cameron and Danae for clearing my doubt! $\endgroup$ – Niket Parikh Apr 15 '15 at 13:10
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    $\begingroup$ Lol! Sorry guys. Don't know what was going on in my mind that I ended up doing something like this. I will see to that. $\endgroup$ – Niket Parikh Apr 16 '15 at 13:34
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A typical statement of the Fundamental Theorem of Arithmetic is this one from Wikipedia:

Every positive integer $n > 1$ can be represented in exactly one way as $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$ where $p_1 < \dots < p_k$ are primes and the $\alpha_i$ are positive integers.

0 is not a positive integer so powers $\alpha_i = 0$ are not allowed in the representation described here. You are correct that if 0 powers were allowed, the representation would no longer be unique.

It is true that $1 = 3^0$ is a power of prime. The error is in thinking that the FTA asserts that $n$ can be represented uniquely as a product of powers of primes. It claims only that it can be represented uniquely as a product of primes to positive powers.

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    $\begingroup$ I'd upvote if it weren't for the Wikipedia link. May Jar'Edo Wens bless you. $\endgroup$ – Robert Soupe Apr 17 '15 at 3:53
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I'd like to elaborate on the Hardy-Wright standard form which Bob mentioned.

Given a positive, composite number $n$, its prime factorization is $$n = p_1 p_2 \ldots p_k$$ where $k$ is what we now usually call the "big omega" function $\Omega(n)$, prime factors of $n$ counted with multiplicity.

"The primes ... are not necessarily distinct, nor arranged in any particular order," write Hardy and Wright. "If we arrange them in increasing order, associate sets of equal primes into single factors, and change the notation appropriately, we obtain" $$n = {p_1}^{a_1} {p_2}^{a_2} \ldots {p_k}^{a_k}$$ where now $k$ is the "little omega" function $\omega(n)$ which counts distinct prime factors of $n$. Hardy and Wright also give the conditions $a_1 > 0, a_2 > 0, \ldots$ and reiterate algebraically that the primes are now given in ascending order as $p_1 < p_2 < \ldots < p_k$. This is what they call standard form.

On the next page, they state the fundamental theorem of arithmetic this way: "The standard form of $n$ is unique; apart from rearrangement of factors, $n$ can be expressed as a product of primes in one way only."

So then $p^0$ is deliberately excluded. They don't bother to assert that each $a_i$ has to be an integer, but this can be safely assumed. So if you had read Hardy and Wright, I think your doubt would have been addressed before it could even rise to the level of an objection.

As far as I can tell, they never felt the need to declare that $1$ is not a prime number. But as you expand your study of the prime numbers beyond the positive integers, you will need to know the definition of a unit. In $\mathbb Z$, the only units are $1$ and $-1$ (this is true of almost every imaginary quadratic ring as well). Neither rearranging prime factors nor multiplying by units changes uniqueness of factorization.

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    $\begingroup$ Very good answer, but you should clarify that although unique factorization applies to negative integers as well and to many real quadratic rings, only very few imaginary quadratic rings have unique factorization. I can't say much about cubic rings, they haven't been studied with the same zeal as quadratics. $\endgroup$ – Robert Soupe Apr 18 '15 at 2:39
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The use of exponents is merely a convenience, as is sorting the primes into ascending order. Multiplying by units and/or reordering the primes does not create distinct factorizations. So for example $1 \times 1 \times 2 \times 3 \times 2 \times 1 \times 1$ and $(-1)^2 \times 2^2 \times 3$ are not distinct factorizations of 12, though neither is what Hardy and Wright would call "standard form." (By the way, the number 1 has never been prime, because it has always lacked a lot of properties the true primes have; though I concede that maybe it was the fundamental theorem of arithmetic that forced mathematicians to finally accept this truth barely a century ago).

Because exponents in factorizations are just a shorthand for writing a prime a number of times and that number of times is a positive integer, only positive integers are valid exponents for primes in a prime factorization. This means you can do things like $p^5$, $p^{15643424564}$, etc., but not things like $p^{\pi + e}$, $p^\sqrt{2}$, $p^\frac{7}{2}$ etc. (By the way, this also holds when you're operating in a domain in which a prime could be an irrational number: the exponents still need to be purely real, rational positive integers).

You could get all fancy and sophisticated and claim that $p^0$ means writing the prime $p$ zero times and thus giving the "empty product." But the fact is that $p^0 = 1$ and that's a unit, not a prime. Therefore $2^2 \times 3^0 \times 5^0 \times 7^0 \times 11^0 \times \ldots$ does not create a distinct factorization of 4.

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