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Let $P(\mathbb{F})$ be the set of all polynomials of one variable over a field $\mathbb{F}$. It is known that such a set, with usual polynomials operations, is a Euclidean domain, that is, a domain that has the Euclidean division algorithm. So we have the following

Theorem. Let $\deg: P(\mathbb{F})\setminus 0 \to\mathbb{N}$ the degree function. Then $(P(\mathbb{F}), \deg)$ is a Euclidean domain, that is:

a) $P(\mathbb{F})$ is a domain;

b)For all $f,g\in P(\mathbb{F}), g\ne 0$, exist unique polynomials $t,r\in P(\mathbb{F})$ such that $$ f(x) = g(x)q(x) + r(x), \text{ with }\deg(r)<\deg(g) \text{ or }r = 0. $$

What I want to know: In such a domain, what is meant by a irreducible or prime polynomial? Moreover, if $p, m\in P(\mathbb{F})$ are such that $p\vert m$ and $p$ is irreducible (or prime) and $m$ is monic (coefficient of the higher degree term is $1$), why we have that $\deg\left(\dfrac mp\right)<\deg(m)$, that is, why $p$ can't have zero degree?

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    $\begingroup$ Are you sure you don't mean irreducible? rather than irreductible? $\endgroup$ – jgon Aug 26 '17 at 22:13
  • $\begingroup$ I don't know for sure, I'm using a text from my teacher that is incomplete and dont have such definition, besides uses it in propositions. But the text is not in english (neither I'm all proficient in that language). $\endgroup$ – AnalyticHarmony Aug 26 '17 at 22:14
  • $\begingroup$ @jgon irréductible is French for irreducible. $\endgroup$ – André 3000 Aug 27 '17 at 16:05
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I'll talk about the definitions in general domains, and it should be clear how they apply to the polynomial rings you're interested in.

In general, in any domain, $R$, a non-unit element, $p$, is said to be prime if $p\mid ab$ implies $p\mid a$ or $p\mid b$. A non-unit element, $r$, is said to be irreducible if $r=ab$ implies that one of $a$ or $b$ is a unit.

All primes, $p$, are irreducible.

Proof: If $p=ab$, then $p \mid ab$, so $p\mid a$ or $p\mid b$, without loss of generality, we may assume $p\mid a$. Then $a=pv$ for some $v\in R$, and $p=pvb$. Since $R$ is a domain, we may cancel to get $1=vb$, so $b$ is a unit. Hence $p$ is irreducible.

On the other hand, it isn't always the case that irreducible elements are prime. This is however a necessary condition for a ring to be a unique factorization domain, and hence it is in fact true in every Euclidean domain. For a proof, see the answers here, one of which gives a direct proof from the Euclidean property.

To sum up the answer to your first question, in a Euclidean domain, an element is prime ($p\mid ab \implies p\mid a\text{ or } p\mid b$) if and only if it is irreducible ($s=ab \implies a$ or $b$ is a unit).

As for your second question, note that polynomials over a field with degree zero are units, and hence aren't considered irreducible. Thus if $p\mid m$, with $p$ prime/irreducible, then $\deg(m/p) = \deg(m) - \deg(p) < \deg m$. (I've assumed you meant $m$ rather than $f$ in the fraction in your question.)

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    $\begingroup$ What you wrote applies only to domains, not general commutative rings. In non domains basic notions such as irreducible, associate etc, bifurcate into a few inequivalent notions, e.g. see here. $\endgroup$ – Bill Dubuque Aug 26 '17 at 22:48
  • $\begingroup$ @BillDubuque Ah yes, that's a fair point, that was careless. Edited. Primes still ought to be the right notion, but looking at the answer you linked to, it appears the definition I gave corresponds to the notion of strong associate, rather than associate. I'll record the definition of irreducible given in one of the papers in the linked answer for future curious people, because it's quite nice, and nicely generalizes the idea that it is weaker than prime. $s$ is irreducible in a commutative ring if $s=ab$ implies $s\mid a$ or $s\mid b$. $\endgroup$ – jgon Aug 26 '17 at 22:56
  • $\begingroup$ Though I should record your note on that answer, that the terminology isn't entirely standardized. $\endgroup$ – jgon Aug 26 '17 at 23:01
  • $\begingroup$ I often use that definition of irreducible since it makes the implication $\rm prime\Rightarrow irreducible$ completely obvious, e.g. see this answer. $\endgroup$ – Bill Dubuque Aug 26 '17 at 23:20
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    $\begingroup$ Ah, sorry, yes, that is the unit element of the ring. A unit element, or more succinctly, a unit, is an invertible element, i.e. a $u$ such that there exists $v$ such that $uv=1$. $\endgroup$ – jgon Aug 26 '17 at 23:44
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In $P(\mathbb{F})$, an irreducible polynomial $f$ is a a polynomial OF POSITIVE DEGREE that cannot be written as the product of two polynomials of smaller degree. This is the same as being prime (a prime in a ring is an element $p$ such that if $p | ab$ then $p | a$ or $p | b$) and this is a consequence of $P(\mathbb{F})$ being a unique factorization domain, (because it is a Euclidean domain). The note that Hungerford's Abstract Algebra gives on this matter is:

You could just as well all such a polynomial "prime," but "irreducible" is the customary term with polynomials.

As for your next question, I think you are then asking why an irreducible polynomial cannot be constant since you are assuming that $p$ is irreducible, then asking why it cannot be constant. Irreducible polynomials are nonconstant by definition. The reason is that in $P(\mathbb{F})$, ever constant polynomial (except $0$) is a unit, they have inverses since $\mathbb{F}$ is a field. So, when you talk about a polynomial being irreducible or prime, one runs into the same sort of reason that $1$ is not prime, because the notion of irreducibility is only a useful concept when you ignore the constants since you can always multiple or divide by a constant and nothing really changes much.

I am not sure what exactly is going on with $f$ and $m$, but I hoped I answered your questions.

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  • $\begingroup$ I would argue for being prime being stronger just because if you exclude non-constant polynomials/ scalars like wikipedia does for irreducible you might find $3x^2+2x+1$ is irreducible on the integers, but only a subset of these don't have a factor of 2 ( namely only when x is even, is the whole thing odd, in the x is even case it reduces to: $12y^2 + 4y + 1$) $\endgroup$ – user451844 Aug 27 '17 at 0:41
  • $\begingroup$ I would note that $2$ dividing a polynomial does not mean that it divides the values of the polynomial when you plug in a value for $x$ (though the converse is true: if $2$ divides the polynomial, then whenever you plug in values for $x$ you get a multiple of $2$). The definition of a polynomial $a(x)$ dividing a polynomial $b(x)$ is that there is a polynomial $q(x)$ such that $a(x) = q(x)b(x)$, where the multiplication is done by using the rules of algebra. Note that $x$ in a polynomial is an indeterminate, not an integer/value of our field: $\endgroup$ – user357980 Aug 27 '17 at 4:20
  • $\begingroup$ See: math.stackexchange.com/questions/495465/… for a description of an indeterminate. $\endgroup$ – user357980 Aug 27 '17 at 4:20
  • $\begingroup$ I would also comment that what I said about polynomial division was for polynomials over a field $\mathbb{F}$. $\mathbb{Z}$ is not a field, but due to Gauss's lemma being irreducible over $\mathbb{Q}$ is the same as being irreducible over $\mathbb{Z}$, so irreducible is the same as being prime. $\endgroup$ – user357980 Aug 27 '17 at 4:23
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$\underline {\text{Irreducible polynomial}} \text{: A non-constant polynomial that can't be factored within a given field or ring}$

so in the integers: $x^2+1$ is irreducible but $x^2-1$ isn't, this is because the latter factors as $(x+1)\cdot (x-1)$ both of which have coefficients that are integer.

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    $\begingroup$ okay you got me, though quoting wikipedia it seems to also apply in rings: "In mathematics, an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials. The property of irreducibility depends on the field or ring to which the coefficients are considered to belong." $\endgroup$ – user451844 Aug 26 '17 at 22:26

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