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
2026-04-06 07:05:06.1775459106
Irreducible polynomials over $\mathbb Q$ and $\mathbb Z $
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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:
Note that the definition of reducibility over a field may sound different:
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!