How to check if a polynomial is irreducible over the rationals

[MathematicalPhysicist]

Homework Statement

I have for example the polynomial: $7x^6-35x^4+21x-1$ and I want to check if it's irreducible or not in $\mathbb{Q}[x]$.

Relevant Equations

none

I first checked for rational roots for this polynomial. The options are $x=\pm 1/7$, both don't nullify the polynomial thus this polynomial doesn't have rational roots.

Now, if it's reducible the only possible factors are: $(ax^3+bx^2+cx+d)(Ax^3+Bx^2+Cx+D)=7x^6-35x^4+21x-1$ or a product of a quadratic polynomial and a quartic polynomial. I need to equate coefficients of similar powers of $x$ and then if there's a solution or not.
If for both cases there isn't a solution then it's irreducible.

Is there a more easy approach for this problem?

 

[fresh_42]

Is there a more easy approach for this problem?

Yes. Assume $7x^6-35x^4+21x=1$ for a reduced quotient $x=p/q.$ Then $q\equiv 0 \mod 7.$ Hence $x=p/7k$ and
\begin{align*}
7x^6-35x^4+21x=1 &\Longrightarrow 7p^6-35p^4q^2+21pq^5=q^6=7^6k^6\\
&\Longrightarrow p^6-5p^4q^2+3pq^5-7^5k^6=0\\
& \ldots
\end{align*}
which is a contradiction.

To change from $\mathbb{Q}$ to $\mathbb{Z}$ is always a good idea in those cases since we gain modular arithmetic in general and Eisenstein in particular.

 

[MathematicalPhysicist]

Yes. Assume $7x^6-35x^4+21x=1$ for a reduced quotient $x=p/q.$ Then $q\equiv 0 \mod 7.$ Hence $x=p/7k$ and
\begin{align*}
7x^6-35x^4+21x=1 &\Longrightarrow 7p^6-35p^4q^2+21pq^5=q^6=7^6k^6\\
&\Longrightarrow p^6-5p^4q^2+3pq^5-7^5k^6=0\\
& \ldots
\end{align*}
which is a contradiction.

To change from $\mathbb{Q}$ to $\mathbb{Z}$ is always a good idea in those cases since we gain modular arithmetic in general and Eisenstein in particular.

Wait a minute, didn't you just check for a rational root for this polynomial?
This doesn't mean that the polynomial can't be a product of two irreducible polynomials, it just just mean it doesn't have a factor of degree 1 in $\mathbb{Q}[x]$.

What am I missing here?

BTW I thought of using Eisenstein criterion at first, but it only works for monic polynomials if I am not mistaken here.

 

[fresh_42]

Wait a minute, didn't you just check for a rational root for this polynomial?
This doesn't mean that the polynomial can't be a product of two irreducible polynomials, it just just mean it doesn't have a factor of degree 1 in $\mathbb{Q}[x]$.

What am I missing here?

BTW I thought of using Eisenstein criterion at first, but it only works for monic polynomials if I am not mistaken here.

Yes, sorry. I only bothered zeros. There still could be quadratic or cubic factors. Looking at it over the integers, however, is still a good idea. I looked it up and IIRC then monic isn't necessary for Eisenstein, only another condition on $a_0$.

 

[MathematicalPhysicist]

Yes, it seems you are correct. In wiki you can use it also for non-monic polynomial.
Ok, Thanks!

 

[MathematicalPhysicist]

@fresh_42 when I look at the criterion in this wiki entry:
https://en.wikipedia.org/wiki/Eisenstein's_criterion
It seems for my case of $7x^6-35x^4+21x-1$, there's only one prime that divides $7,35,21$ which is seven, but it doesn't divide $1$, and according to wiki, I need to find a prime number that divides 35,21 and 1 and that its square doesn't divide 1.
But the fact that such a number doesn't exist doesn't mean that the polynomial is reducible, this condition only goes in one direction.

 

[fresh_42]

@fresh_42 when I look at the criterion in this wiki entry:
https://en.wikipedia.org/wiki/Eisenstein's_criterion
It seems for my case of $7x^6-35x^4+21x-1$, there's only one prime that divides $7,35,21$ which is seven, but it doesn't divide $1$, and according to wiki, I need to find a prime number that divides 35,21 and 1 and that its square doesn't divide 1.
But the fact that such a number doesn't exist doesn't mean that the polynomial is reducible, this condition only goes in one direction.

Yes, I haven't found a better method either. I could only show (if I made no mistake) that if it splits over $\mathbb{Q}$ then it already splits over $\mathbb{Z}.$ The consideration of zeros rules out the combination of degrees $(1,5),$ but $(2,4),(3,3)$ are still possible.

 

[MathematicalPhysicist]

@fresh_42 it seems there's a theorem that states the following:"A polynomial, $Q$ is irreducible over $\mathbb{Q}$ iff the polynomial that is made of reversal of the coefficients $Q$ is irreducible over $\mathbb{Q}$. In that case I only need to check that $7$ indeed divides 7,35,21 but $7^2$ doesn't divide 7.

 

[fresh_42]

@fresh_42 it seems there's a theorem that states the following:"A polynomial, $Q$ is irreducible over $\mathbb{Q}$ iff the polynomial that is made of reversal of the coefficients $Q$ is irreducible over $\mathbb{Q}$. In that case I only need to check that $7$ indeed divides 7,35,21 but $7^2$ doesn't divide 7.

Sounds funny. Do you have a reference?

 

[MathematicalPhysicist]

Sounds funny. Do you have a reference?

Well I have a reference but it's not written in English.
It's mentioned in an exercise that they covered it in the recitation.
I guess they changed variables $y=1/x$ and argued that the same polynomial should be irreducible in $x$.

 

[fresh_42]

Well I have a reference but it's not written in English.
It's mentioned in an exercise that they covered it in the recitation.
I guess they changed variables $y=1/x$ and argued that the same polynomial should be irreducible in $x$.

I have seen a similar trick in my book: If $f(x)$ is reducible, so is $f(x+1)$ ... but this didn't help here.

 

[SammyS]

Well I have a reference but it's not written in English.
It's mentioned in an exercise that they covered it in the recitation.
I guess they changed variables $y=1/x$ and argued that the same polynomial should be irreducible in $x$.

Right.

Multiply through by $-y^6 ~ ,$ where $y=\dfrac{1}{x}$.

Gives you the monic polynomial, $y^6-21y^5 +35y^2 -7$ .

Perhaps you already realized that.

 

[mathwonk]

The "reverse Eisenstein" criterion is on page 25-26 of these class notes, referenced there to a proof in Van der Waerden. The point however is as Fresh42 notes, just to use integers and mod 7 integers. I.e. if there were a rational factorization then there would also be an integral factorization of that polynomial, and then mod 7, both factors would be units. Hence all higher coefficients of both factors are divisible by 7, hence the lead coefficient of the given polynomial would be divisible by 49, but it isn't.
https://www.math.uga.edu/sites/default/files/inline-files/8000b.pdf

i discovered this reverse Eisenstein result myself while teaching the class, and subsequently found the much more general Dumas result in Van der Waerden. Reverse Eisenstein is true essentially just because polynomial multiplication is a symmetric operation, so the mirror image of the statement must also be true, and can be proved by the mirror image of any proof of usual Eisenstein. Again as Fresh said, the simplest proof is reduction mod p. I.e. as exercise, take any proof of Eisentein and reverse it yourself to prove reverse Eisenstein.

 

[WWGD]

The "reverse Eisenstein" criterion is on page 25-26 of these class notes, referenced there to a proof in Van der Waerden. The point however is as Fresh42 notes, just to use integers and mod 7 integers. I.e. if there were a rational factorization then there would also be an integral factorization of that polynomial, and then mod 7, both factors would be units. Hence all higher coefficients of both factors are divisible by 7, hence the lead coefficient of the given polynomial would be divisible by 49, but it isn't.
https://www.math.uga.edu/sites/default/files/inline-files/8000b.pdf

i discovered this reverse Eisenstein result myself while teaching the class, and subsequently found the much more general Dumas result in Van der Waerden. Reverse Eisenstein is true essentially just because polynomial multiplication is a symmetric operation, so the mirror image of the statement must also be true, and can be proved by the mirror image of any proof of usual Eisenstein. Again as Fresh said, the simplest proof is reduction mod p. I.e. as exercise, take any proof of Eisentein and reverse it yourself to prove reverse Eisenstein.

Just curious: is there a notion of reducibility for multivariate polynomials?

 

[fresh_42]

Just curious: is there a notion of reducibility for multivariate polynomials?

A multivariate polynomial generates an ideal in its polynomial ring. The coordinate ring (quotient) describes an algebraic variety of which we can count the irreducible components. It is the starting point of algebraic geometry. It is less about the factorization of the polynomial than it is about the properties of this ideal.

The Buchberger algorithm deals with a generalization of long division for multivariate polynomials.

 

[mathwonk]

The notion of reducibility for polynomials of several variables is the same as for polynomials of one variable, i.e. whether or not it is a product of two other polynomials. it is even harder to determine irreducibility howvever. A polynomial of n variables determines an algebraic set of dimension n-1 in affine n space, and conversely every algebraic set of dimension n-1 in n space is determined by a single polynomial. Algebraic varieties of lower dimension in n space must be determined as common zeroes of more than one polynomial, and as Fresh says, it is usual to consider the ideal of all polynomials vanishing on the set. Then the set is irreducible in the sense of not being a union of two or more algebraic sets if this ideal is a "prime" ideal. I.e. the algebraic set is irreducible if whenever a product of two polynomials vanishes on it, then already one of the factors vanishes on it. (In these geometric situations, it is simplest to use coefficients from an algebraically closed field. Then an irreducible algebraic set in 1-space, for example, is a single point, just as an irreducible polynomial is linear, of form X-a.))

Thus in this setting, the theory of factoring individual polynomials of several variables into irrreducible factors, is identical to the theory of decomposing algebraic sets of codimension one into irreducible components, also of codimension one. Since an algebraic set of higher codimension is defined by more than one polynomial, deciding whether a given polynomial vanishes on such a set reduces to asking whether the polynomial is a linear combination (with polynomial coefficients), of generators for the ideal of that set. This problem is addressed by a several variable version of the division algorithm, such as Buchsberger's algorithm, but I am not expert on this. One reference, for undergraduates, is Ideals, Varieties, and Algorithms, by Cox, Little, and O'Shea. The underlying idea can be glimpsed already by studying a proof of Hilbert's theorem that if every ideal in a ring R has a finite number of generators, then the same is true for the polynomial ring in one variable over R. Recall that the classic division theorem for polynomials in one variable begins by dividing the lead coefficient of the dividend by the lead cofficient of the divisor. Hence writing a polynomial as a linear combination of a given family of polynomials begins by writing the leading coefficient as a linear combination of their coefficients. Then presumably one tries to proceed by induction on the number of variables.

if you look at the class notes I linked in post #14, you will see two examples on p. 26 of polynomials in two variables shown to be irreducible by considering them as polynomials in one variable, with cofficients which are themselves polynomials in the other variable.

 

[WWGD]

A multivariate polynomial generates an ideal in its polynomial ring. The coordinate ring (quotient) describes an algebraic variety of which we can count the irreducible components. It is the starting point of algebraic geometry. It is less about the factorization of the polynomial than it is about the properties of this ideal.

The Buchberger algorithm deals with a generalization of long division for multivariate polynomials.

Sounds a bit like Nustellensatz, but I haven't looked into it for a long while now.

 

[fresh_42]

Sounds a bit like Nustellensatz, but I haven't looked into it for a long while now.

Yes, the Nullstellensatz belongs to the context.

 

[mathwonk]

given a finite set of polynomials over an algebraically closed field, the nullstellensatz gives a geometric criterion for whether the constant 1 can be written as a linear combination of them, namely this is so if and only if the polynomials have no common zero. The Buchberger algorithm gives an algebraic procedure to decide whether there is such a linear combination, and if so, to actually find one.