9
$\begingroup$

When I read "Contemporary Abstract Algebra" by Joseph gallian, under the topic irreducible polynomials, his first example is the polynomial $$2x^2+4=0$$ is reducible over $\mathbb Z$ but irreducible over $ \mathbb Q$. I dont know how is this possible? Since it is of degree 2, we can see the roots of the polynomial, where it lies? if it lies in $\mathbb Z$ then it would lie in $ \mathbb Q$, then how can it be reducible over $\mathbb Z$ when the roots are complex numbers? pls explain

$\endgroup$
13
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ Great explanation....But the exercise you gave "is there a polynomial like that?"? Because there is a theorem "if a polynomial is reducible over Q then it is reducible over Z" and i want to ask you a doubt..what happens to the theorem saying "if the degree of polynomials is of 2 or 3 then it is reducible over F[x] iff it has roots in F." $\endgroup$ Dec 19 '15 at 5:59
  • $\begingroup$ @SamChristopher Re: the first theorem you mention: Well, that sounds like that kills the exercise, doesn't it? There's a reason I phrased it the way I did :). (EDIT: Now, a parallel good exercise: find rings $R\subset S$ and a polynomial $p\in R[x]$ such that $p$ is reducible over $R$ but irreducible over $S$! There's something special about $\mathbb{Z}$ . . .) Re: the second theorem you mentioned: there you're taking $F$ to be a field. The theorem breaks down if $F$ is merely a ring like $\mathbb{Z}$ - look at the proof to see why. $\endgroup$ Dec 19 '15 at 6:11
  • $\begingroup$ Great.....Thank you... $\endgroup$ Dec 19 '15 at 6:47
  • 1
    $\begingroup$ The precise relationship between the two types of irreducibility (over Z and over Q) is solved by Gauss' lemma. More generally, let A be an UFD and K its field of factions. The irreducible elements of A[X] are either irreducible elements of A , or polynomials in A[X] which are irreducible in K[X] and have content 1. For a proof and the definition of "content", see e.g. the chapter on "Polynomials" in Lang's "Algebra" $\endgroup$ Apr 17 '16 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.