Feel lost in the scheme of the reducibility of polynomials over $\Bbb Z$ or $\Bbb Q$

Question

I'm studying abstract algebra. I have learned Eisenstein criterion, rational root test, Guass's lemma and so on. However, I can't see the scheme and the whole picture that these things plays the roles in. I think I should better directly ask here. I don't know where to start asking, so I list some questions below. Hope you can see what I'm stucking in. Need to be given the direction.

• Is the ultimate target here is that we want to tell whether a polynomial can be factored over $\Bbb Z$, or over $\Bbb Q$, or over $\Bbb R$, or over $\Bbb C$?
• Which do we really care about in the factorization? Integer coefficients polynomials, or rational coefficients? Or both?
• Is it a standard produre that whenever dealing with the (ir)reducibility problem of a polynomial of rational coefficients, we turn it into integer coefficient first? Why? Or is it vice versa?
• What is the importance or usage of Gauss's lemma?
• How about reducibility over $\Bbb R$ or $\Bbb C$ when there're no many tools can be used as in the $\Bbb Q[x]$?

Answer 1

Warning: my bullet points are not isomorphic to yours!

• If a polynomial factors over a smaller field, then it also factors in the bigger ones above (for general rings, this is more tricky: $ax^2+a$ factors over $\mathbb{Z}$ as $a(x^2+1)$ for all $a\neq 0,\pm1$ (since $a$ is not a unit), but in $\mathbb{R}$ it is irreducible, because $a$ is a unit). Also, if a polynomial has a zero in a smaller ring, then it also has (the same) zero in the bigger rings above.

• Being irreducible doesn't coincide with having no zeros, in general. For example, $(x^2+1)(x^2+1)$ is reducible over anything, but it doesn't have a zero in $\mathbb{R}$. Conversely, $x$ is always irreducible but it always has a zero. Nevertheless, the two things are related: for instance, if $F$ is a field and $P(x) \in F[x]$ is a polynomial with degree greater than one that has a zero in $F$, then $P$ is reducible (this comes from division with remainder of polynomials).

• Integers are also rational numbers. So integer coefficient polynomials are also rational coefficient polynomials. But it is also true that a polynomial with rational coefficients is just a rational multiple of a polynomial with integer coefficients (just clear the denominators), and their zeros coincide (and the zeros are usually the most interesting thing). Irreducibility over $\mathbb{Q}$ implies irreducibility over $\mathbb{Z}$, if the polynomial is primitive. One does not immediately see why the converse should be true, but it is, by Gauss lemma! So irreducibility over $\mathbb{Z}$ and $\mathbb{Q}$ is the same for primitive polynomials. Why do we consider only primitive polynomials? For the reason mentioned above: if you can divide the polynomial by some integer $a\neq 0,\pm 1$, then it is reducible over $\mathbb{Z}$, but not necessarily over $\mathbb{Q}$. But since roots don't depend on a division by a constant factor $a$, note that Gauss lemma implies that a polynomial with integer coefficients and degree greater than one has an integer root if and only if it has a rational root! That's amazing!

• What I find very helpful is to always see any (complex) polynomial of degree $n$ as $u(x-\alpha_1)\dots(x-\alpha_n)$, where $u$ is some complex number and the $\alpha_i$ are the roots (not necessarily distinct). This is always possible, since $\mathbb{C}$ is algebraically close (maybe you have not learnt it yet, but it is true and there are many very beautiful proofs of this fact).

• It is easier to check irreducibility in $\mathbb{Z}$ than in $\mathbb{Q}$, that's why you often do so.

• As I already said, polynomials over $\mathbb{C}$ always have a very easy form. In $\mathbb{R}$ you have that irreducible polynomials always have degree at most $2$. This comes from the fact that if $\alpha$ is a complex root of some real polynomial, then also $\overline{\alpha}$ is (why?). So let $P$ be any real polynomial that has at least a non-real root. Then it is divisible by $(x-\alpha)(x-\overline{\alpha})=x^2 + |\alpha|^2-2xRe(\alpha)$, which is a real polynomial. In particular, if it has degree at least $3$, it is not irreducible.

Many things will become clearer when you will do some field theory: there irreducibility of polynomials becomes very important. Also at this stage, doing some exercises involving irreducibility criteria will certainly clarify things.

Answer 2

What we really care about depends, as always, on our viewpoint, i.e., on the problems we want to solve. In number theory, polynomials with integer and rational coefficients are very interesting, and finding roots is a famous, old question going back to "Diophantos". To solve Diophantine equations, like $x^n+y^n=z^n$, we need at least some algebraic number theory, e.g., the ring of integers of the cyclotomic field $\mathbb{Q}(\zeta_n)$. For Dedekind rings, and commutative rings in general, we need the basics from commutative algebra, e.g., irreducible and prime elements, respectively ideals. So one reason to study this, comes from this motivation. Cyclotomic fields are very important examples of number fields, and cyclotomic polynomials very important examples of irreducible polynomials, you were asking about. The name of Gauss appearing in this context, is directing towards number theory, I would say:)

Answer 3

Here is one story: not the only one, but a very common one.

Often we are interested in the factorisation of a polynomial $f\in\mathbb{Q}[X]$.

Multiplying by any $c\in\mathbb{Q}$ doesn't change anything, so we can clear the fractions and suppose $f\in\mathbb{Z}[X]$, although we are still interested in factorisations in $\mathbb{Q}[X]$.

Gauss's Lemma tells us how factorisations in $\mathbb{Q}[X]$ relate to factorisations in $\mathbb{Z}[X]$, so our focus moves to such factorisations.

Now Eisenstein's Criterion (and the ideas behind it) relate the factorisations of $f\in\mathbb{Z}[X]$ in $\mathbb{Z}[X]$ with the factorisations of $\bar{f}\in\mathbb{Z}_p[X]$ in $\mathbb{Z}_p[X]$, a potentially "easier" problem.