difference between irreducible element and irreducible polynomial 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?
 A: 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.
A: 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.
