Is $4x+2$ irreducible over $\Bbb{Q}[x]$ and over $\Bbb{Z}[x]$

Question

I need to show if $4x+2$ is irreducible over $\Bbb{Q}[x]$ and over $\Bbb{Z}[x]$.

I claim its irreducible over $\Bbb{Q}$ because

$$4x+2=2(2x+1)$$

where $2$ is a unit in $\Bbb{Q}$.

But I don't know a method for showing its reducible in $\Bbb{Z}[x]$.

Do I suppose

$$4x+2=ab; \space \text{$a,b \in \Bbb{Z}[x]$}$$

Then the degree of ab would have to equal 1 which means the degree of their sums needs to equal 1 so one has to have degree $0$ and the other degree $1$. Then where would I do from there? Though I read somewhere that irreducible over rrationals implies irreducible over integers. is this true????

And in general (without use of Eeinsteins criteria) how do you determine wether or not a polynomial is or is not reducible in either $\Bbb{Q}[x]$ or $\Bbb{Z}[x]$, can one use the fact that the product of degrees is the sum of the products? I don't think my exam will have any polynomials with higher powers but I'd like to know how to generalize this to higher degree polynomials. Thanks in advance. I am preparing to a PhD qualifying exam and he briefly covered polynomial rings with us.

Answer 1

You've got your two methods pretty much entirely backwards.

First off, writing $4x+2 = 2(2x+1)$ doesn't prove that it is irreducible in $\mathbb{Q}[x]$. You've shown that one particular decomposition requires a unit, but that doesn't mean that every decomposition requires a unit (Unless you already have reason to believe that $2x+1$ is irreducible).

However, $4x+2 = 2(2x+1)$ is exactly the method for showing that $4x+2$ is reducible in $\mathbb{Z}[x]$. You've expressed it as a product of two factors. $2$ isn't a unit in $\mathbb{Z}[x]$. $2x+1$ isn't a unit in $\mathbb{Z}[x]$. Therefore this give a reduction into two different factors.

At the same time, your degree argument that you mention under $\mathbb{Z}[X]$ is exactly how you go about proving that it's irreducible in $\mathbb{Q}[X]$. If $4x+2=ab$ then one of the factors has degree one and the other has degree zero. The factor with degree zero is a non-zero constant, so it has an inverse in $\mathbb{Q}[X]$ - it is a unit. Therefore given any expressing as factors at least one of the factors is a unit: i.e. it is irreducible.

It's concerning that you managed to switch around the two methods like this. Suggests that you read them without understanding what was going on in them.

Answer 2

I recall : a non-zero polynomial $P\in A[X]$ where $A$ is a ring is irreducible if $P$ is not invertible and if $P=QR$ with $Q,R\in A[X]$ implies that $Q$ or $R$ is a inversible.

What you wrote in $\mathbb{Q}[x]$ does not work because $2$ is a unit in $\mathbb{Q}[x]$. In fact any polynomial of degree $1$ in $\mathbb{Q}[x]$ is irreducible because if it is the product of two polynomials, one of them is constant because of the degree, so is a unit because $\mathbb{Q}$ is a field, so $4x+2$ is irreducible in $\mathbb{Q}[x]$.

In $\mathbb{Z}[x]$, the units are the constant polynomials $-1$ and $1$ so the product $4x+2=2(2x+1)$ is a product of two non-invertible polynomials therefore $4x+2$ is not irreducible in $\mathbb{Z}[x]$.

If a polynomial $P\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Z}[x]$ then it is irreducible in $\mathbb{Q}[x]$. Indeed, if $P=QR$ with $Q,R\in\mathbb{Q}[x]$ then by a lemma of Gauss, there exists $r\in\mathbb{Q}^*$ such that $rQ,\frac{1}{r}R\in\mathbb{Z}[x]$. Since $P=(rQ)(R/r)$ and $P$ is irreducible in $\mathbb{Z}[x]$ then $rQ$ or $R/r$ is $\pm 1$ which means that $Q$ or $R$ is a constant polynomial, therefore $P$ is irreducible in $\mathbb{Q}[x]$. The converse is not true, we've just seen that $4x+2$ is a counter-example.

As for the general case, Eisenstein criterion is a very useful tool to find irreducible polynomials in $\mathbb{Q}[x]$. A trick for polynomials of degree $3$ is to prove that they don't have any rational root, a potential factor should be of degree $1$ and thus should have a rational root. I should mention that if a polynomial in $\mathbb{Z}[x]$ is such that its coefficients have no common factors, then by the same Gauss lemma that I used, it is sufficient to prove that it is irreducible in $\mathbb{Q}[x]$ to prove that it is irreducible in $\mathbb{Z}[x]$.