Irreducible polynomials over $\mathbb Q$ and $\mathbb Z $

Solution 1:

It's not the roots, it's the "$2$"!

A polynomial is irreducible over a ring if it cannot be written as a product of two non-invertible polynomials. In $\mathbb{Z}$, "$2$" is noninvertible, so $(x^2+2)2$ is an appropriately "nontrivial" factorization.

Meanwhile, over in $\mathbb{Q}$, the polynomial "$2$" is invertible, since ${1\over 2}$ is rational (proof: exercise :P). So the factoriztion $(x^2+2)2$ is "trivial" in the context of $\mathbb{Q}$, since we can always extract a factor of $2$ from any polynomial.

EDIT: Think of it this way: saying that a polynomial is irreducible over a ring means it has no "nontrivial" factorizations. Now, when we make the ring bigger (e.g. pass from $\mathbb{Z}$ to $\mathbb{Q}$) two things happen:

• More factorizations become possible.

• More factorizations become trivial.

So even though your first instinct might be "polynomials will only go from "irreducible" to "reducible" as the ring gets bigger," actually the opposite can happen!

In fact, here's a good exercise:

Can you find a polynomial $p\in\mathbb{Z}[x]$ which is irreducible over $\mathbb{Z}$ but reducible over $\mathbb{Q}$?

Note that the definition of reducibility over a field may sound different:

For $F$ a field, a polynomial $p\in F[x]$ is irreducible if $p$ cannot be written as the product of two nonconstant polynomials.

But this is actually equivalent to the definition I gave above, in case we're over a field: the noninvertible elements of $F[x]$ are precisely the nonconstant polynomials!