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let $\mathbb F$ be a field and let $\mathbb F[x_1, \cdots , x_n]$ be the polynomial ring in the variables $x_1, \cdots x_n$. For $n=1$ there are several irreducible criteria. But if $n>1$ there are methods of determining whether a given polynomial over $\mathbb F$ is irreducible?

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2 Answers 2

up vote 5 down vote accepted

One can consider such a polynomial $P$ as a polynomial $P(x_n)$ in the variable $x_n$ with coefficient in the UFD $R:=\mathbb F[x_1, \dots, x_{n-1}]$. Then $P$ is irreducible if and only if (1) it is primitive (the gcd of its coefficients is $1$) and (2) it is irreducible in $K[x_n]$ where $K=\mathrm{Frac}(R)$. This is called Gauss Lemma.

One also has the Eisenstein criterion : if there exists a prime element $f\in R$, prime to the leading coefficient of $P(x_n)\in R[x_n]$, dividing the other coefficients and such that $f^2$ doesn't divide the constant term. Then $P$ is irreducible if it is moreover primitive.

Edit Another method not reducing to one variable (and only valid for $n\ge 2$) : if $P$ and its partial derivatives generate the unit ideal in $\mathbb F[x_1,\dots, x_n]$, then $P$ is irreducible. The proof uses algebraic geometry. The condition on $P$ and its partial derivatives holds if and only if $P$ and its partial derivatives don't have a common zero with coordinates in an algebraic closure of $\mathbb F$ (this comes from Hilbert Nullstellensatz).

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I hope I did not change your meaning in making a slight edit. Feel free to return to the original. –  Lubin Feb 28 '13 at 23:42
@Lubin: thanks to make the statement clearer. –  user18119 Mar 1 '13 at 8:35
Thanks for the reply. –  zacarias Mar 1 '13 at 10:58

There are different ways to say if a polynomial in F[x] is irreducible:

As in QiL'8 answer:

1/ Eisenstein Criterion (especially useful to determine if a polynomial is irreducible over $\mathbb Q$)

2/ Gauss Lemma

Some other ways:

3/ Rational Roots Theorem (to see if there is a root x in the set of rationals). If such a root exists, then the polynomial is reducible over $\mathbb Q$, thus in $\mathbb Z$

4/ Using ideals : A polynomial f(x) is irreducible over F <==> The ideal generated by f(x) is a maximal ideal.

But of course we can always try the most basic approach by trying to find the root of the polynomial f(x) (if the polynomial is easy enough to work with), and see if the root is in our field F. For example, we see that x^2 + 2 is irreducible over $\mathbb R$ but is reducible over $\mathbb C$, since the roots are 2i and -2i, which are complex numbers.

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