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Edit: I foolishly changed my example right before posting this. The following is what I meant to ask.

On the wikipedia page below, it says "a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units." https://en.wikipedia.org/wiki/Irreducible_element

Then, on the wikipedia page below, it says "an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials." https://en.wikipedia.org/wiki/Irreducible_polynomial

The page then gives an example of a polynomial that is irreducible "over the rationals" (I assume it is implied that "the integers" is referring to the ring of polynomials over the field of rationals, right?)

$p(x) = x^2 + 2 = (x+\sqrt{2})(x-\sqrt{2})$

On the Wofram page below, it says "the irreducible polynomials are examples of irreducible elements."
http://mathworld.wolfram.com/IrreducibleElement.html

If irreducible polynomials are a subset of irreducible elements, how is $x+\sqrt{2}$ a unit? What can it be multiplied by to equal the multiplicative identity?

I think I have it right this time (sorry. long day)

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The following is the original post (for those who already responded and those looking at the corresponding responses)

On the wikipedia page below, it says "a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units." https://en.wikipedia.org/wiki/Irreducible_element

Then, on the wikipedia page below, it says "an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials." https://en.wikipedia.org/wiki/Irreducible_polynomial

The page then gives an example of a polynomial that is reducible "over the integers" (I assume it is implied that "the integers" is referring to some sort of ring, right? due to the claim below)

$p(x) = x^2 + 4x + 4 = (x+2)(x+2)$

On the Wofram page below, it says "the irreducible polynomials are examples of irreducible elements."
http://mathworld.wolfram.com/IrreducibleElement.html

If irreducible polynomials are a subset of irreducible elements, how is $x+2$ a unit? What can it be multiplied by to equal the multiplicative identity?

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    $\begingroup$ Who ever said that $x+2$ was a unit? $\endgroup$ – Lord Shark the Unknown Oct 27 '17 at 2:54
  • $\begingroup$ @walkar I actually had a rationals example at first and changed it last minute to an integer example. So is Wikipedia's example not in the context of a ring? $\endgroup$ – user3146 Oct 27 '17 at 3:06
  • $\begingroup$ @LordSharktheUnknown If it is not, doesn't that contradict Wolfram's claim that "the irreducible polynomials are examples of irreducible elements"? $\endgroup$ – user3146 Oct 27 '17 at 3:07
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    $\begingroup$ Actually a polynomial with integer coefficients factors over the integers if and only if it factors over the rationals since any factorization over the integers gives one in the rationals and one can turn a rational factorization into an integer one by multiplying by the GCD of the coefficients. The point is $x^2+4x+4$ is reducible over the integers and can be factored into irreducibles like $x+2$, notice we similarly want to say $-x-2$ is irreducible since it's only factorization is $(-1)(x+2)$ and $(-1)$ is a unit. $\endgroup$ – walkar Oct 27 '17 at 3:12
  • $\begingroup$ In that sense you should view each single irreducible as a class of "associates," i.e. for $x$ irreducible, $\lambda x$ is irreducible for all units $\lambda$. $\endgroup$ – walkar Oct 27 '17 at 3:15
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Irreducible polynomials in $\mathbb{F}[x]$, for $\mathbb{F}$ a field, are examples of irreducible elements of the polynomial ring $\mathbb{F}[x]$.

The polynomial $x^{2}+2$ is irreducible over the rationals (we consider it to have rational coefficients so that we are working over a field) because it cannot be factored into two rational polynomials that are not units (your only choice is to factor it as something like $(1/2)(2x^{2}+4)$, where $1/2$ is a unit). Notice that $x + \sqrt{2}$ is not a unit, but it can't be used in a factorization of $x^{2}+2$ in $\mathbb{Q}[x]$ because it does not have rational coefficients.

The polynomial $x^{2}+4x+2 = (x+2)(x+2)$ is reducible, not irreducible, precisely because it can be factored as $(x+2)(x+2)$, where $x+2$ is not a unit.

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  • $\begingroup$ Ooohh. I thought the factoring in the example was supposed to be demonstrating how it doesn't fit the definition as I think would make sense and as you actually did in your factoring. Thank you!! Is it me? Or, was that factoring in the example misleading? $\endgroup$ – user3146 Oct 27 '17 at 3:57
  • $\begingroup$ Sorry. I tried to upvote your answer, but I don't have enough points yet. $\endgroup$ – user3146 Oct 27 '17 at 3:58
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    $\begingroup$ I would say the factoring in that example was misleading, it definitely doesn't help illustrate the point they are trying to make (and no worries about the vote). $\endgroup$ – Morgan Rodgers Oct 27 '17 at 6:26
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The point is that $x+2$ is $\textbf{not}$ a unit. This shows you can factor this polynomial into a product of two non-units, so it is $\textbf{not}$ irreducible. This definition just means, $3$ is still irreducible even though you can factor it $(-3)\cdot (-1)$. Since $-1$ is a unit, it doesn't count as a factorization which would make it fail to be irreducible. This also is used for unique factorization even of the integers. Otherwise you could say $6=2 \cdot 3=(-1)(-1)2 \cdot 3$ would be different factorizations. The units don't count as real factors basically. So irreducible in a domain usually means if $a=bc$ then $b$ or $c$ must be a unit, which is how I think about it.

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  • $\begingroup$ Sorry. I spazzed and changed what I was going to post last minute to something pretty dumb. I edited it (quickly as to minimize time wherein someone else can waste time writing an answer to my flawed question). I think I have it straightened out. $\endgroup$ – user3146 Oct 27 '17 at 3:32

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