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Is the polynomial $(2x + 2)$ irreducible in $\mathbb{Z}[x]$?

I know that the units in $\mathbb{Z}[x]$ are $\pm$1.

Please check my proof below:

Suppose $f$ = $(2x + 2)$ $\in$ $\mathbb{Z}$[x] is a degree 1-polynomial. Then $f=1(2x + 2)$ where $g = 1$ and $h = (2x + 2)$ for $g, h$ $\in$ $\mathbb{Z}$[$x$]. Since $g$ is a unit, we have $f =\text{unit}\cdot$degree-$1$ polynomial. Hence $f$ is irreducible in $\mathbb{Z}[x]$ .

yagayeet
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    I get that $2x+2=2(x+1)$. That looks like a non-trivial factorisation to me. – Angina Seng Nov 21 '19 at 07:26
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    The units in $\Bbb Z[x]$ are $1$ and $-1$. – Angina Seng Nov 21 '19 at 07:30
  • @LordSharktheUnknown Then can I instead say, Suppose f = $(2x + 2)$ $\in$ $\mathbb{Z}$[$x$] is a degree 1-polynomial. Then f = $1(2x + 2)$ where g = 1 and h = $(2x + 2)$ for g, h $\in$ $\mathbb{Z}$. Since g is a unit, we have f = unit$\cdot$degree-1 polynomial. Hence f is irreducible in $\mathbb{Z}$. – yagayeet Nov 21 '19 at 07:32
  • "Since $g$ is a unit, we have $f$ = unit⋅degree-1 polynomial. Hence $f$ is irreducible in $\Bbb Z$." The second sentence doesn't follow here; not all degree-one polynomials are irreducible over $\Bbb Z[x]$. In particular, $2x+2$ is a counterexample. What you're saying is true over $F[x]$ for $F$ a field though! – Henry Swanson Nov 21 '19 at 07:47
  • @HenrySwanson I'm not sure why $(2x + 2)$ is reducible over $\mathbb{Z}$[$x$]. Can you point me in the right direction please? Thanks! – yagayeet Nov 21 '19 at 07:54
  • $x=2(x + 1)$ and both $2$ and $x + 1$ are polynomials (none of them invertible, and hence no unit) – Felix Nov 21 '19 at 08:00
  • Something is irreducible if it can be factored into two factors, both of which are non-units. Since neither $2$ nor $x+1$ is a unit in $\Bbb Z$, $2x+2$ is not irreducible. – Henry Swanson Nov 21 '19 at 08:01
  • How would this change if instead of $\mathbb{Z}$[$x$] it was $\mathbb{Q}$[$x$]? I was looking online that Einstein's Irreducibility Criterion might suffice but since $(2x + 2)$ is a degree-1 polynomial would we even need Einstein's Irreducibility Criterion? Thanks in advance! – yagayeet Nov 21 '19 at 08:09
  • The conditions for Eisenstein's (not Einstein's) Criterion are not met by $2x+2$. – DanielWainfleet Nov 21 '19 at 09:04
  • @DanielWainfleet Can I use Gauss' Lemma here? Since $(2x + 2)$ is reducible in $\mathbb{Z}$[$x$], $(2x + 2)$ must be reducible in $\mathbb{Q}$[$x$]. – yagayeet Nov 21 '19 at 09:27
  • Any $f\in \Bbb Z[x]$ is reducible in $\Bbb Z[x]$ if $f=g\cdot h$ for some non-zero $g,h\in \Bbb Z[x]$ such that neither $g$ nor $h$ has a multiplicative inverse in $\Bbb Z[x]$.... E.g. $g=2$ and $h=x+1.$ – DanielWainfleet Nov 21 '19 at 09:31
  • In Gauss' Lemma the definition of "reducible in $\Bbb Z[x]$" is different. – DanielWainfleet Nov 21 '19 at 11:03

2 Answers2

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Your proof is incorrect - you display one factorisation in which one of the factors is a unit. However to be irreducible you have to consider all possible factorisations.

As in the comments the factorisation $2x+2=2(x+1)$ is a factorisation into non-units in $\mathbb Z[x]$ which shows that the polynomial is reducible.

The situation in $\mathbb Q[x]$ is different, because $2$ is a unit in this context.

You don't need Eisenstein here - since the product of polynomials has degree the sum of the degrees, and a linear polynomial has degree $1$ the only possible factors have degree $0$ (constants) or $1$. If there is a factorisation in the $\mathbb Z$ context, one of the factors must be a constant, and this will be a product of primes. The prime factors of the constant must clearly be factors of each of the coefficients of the polynomial, and any prime which divides the coefficients leads to a factorisation.

In the $\mathbb Q$ context every constant other than $0$ is a unit, so a non-trivial factorisation into irreducibles must involve two factors of degree greater than $0$.

Mark Bennet
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  • Thank you for the detailed response. So in $\mathbb{Q}$[$x$], implementing what you said, $(2x + 2)$ = $2(x + 1)$ which is a factorization of a degree-0 (2) polynomial with a degree-1 $(x + 1)$ polynomial. Hence, $(2x + 2)$ is not irreducible in $\mathbb{Q}$[$x$] as $2(x + 1)$ is factored as a degree-0 polynomial $\cdot$ a degree-1 polynomial. – yagayeet Nov 21 '19 at 08:48
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    @yagayeet In $\mathbb Q[x]$ we have $2\times \frac 12=1$ so that $2$ is a unit. In this context $2x+2$ is an associate of $x+1$ and both are irreducible. – Mark Bennet Nov 21 '19 at 11:23
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$(2x+2)$ is reducible in $\mathbb{Z}[x]$.

To see why let's start with the definition.

An element $f$ in a commutative ring $R$ is called irreducible if there does not exist non-units(Or non-invertible elements) $x$ and $y$ such that $f = xy$.

Thus an element is reducible if it can be written as the product of two non units.

We can write $2x+2$ as $2 * (x+1)$, both of which are non-units in $\mathbb{Z}[x]$.(The only units in $\mathbb{Z}[x]$ are $1$ and $-1$). Thus $2x+2$ is reducible.

Lior Ben-David
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