Irreducibility of polynomials in two variables - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:33:21Z http://mathoverflow.net/feeds/question/14076 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables Irreducibility of polynomials in two variables Hailong Dao 2010-02-03T22:49:54Z 2013-01-15T10:06:09Z <p>Let $k$ be a field. I am interested in sufficient criteria for $f \in k[x,y]$ to be irreducible. An example is Theorem A of this <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa82/aa8237.pdf" rel="nofollow">paper</a>.</p> <p>Does anyone know of similar results in the same vein? How about criterion over fields other than the complex numbers? </p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/14077#14077 Answer by Pete L. Clark for Irreducibility of polynomials in two variables Pete L. Clark 2010-02-03T23:01:07Z 2010-02-03T23:01:07Z <p>I think there must be a lot of work on irreducibility criteria, although I have unfortunately never seen anything like a unified treatment. </p> <p>The first thing that my MathSciNet search turned up was:</p> <p><hr /></p> <p>Ayad, Mohamed Sur les polynômes $f(X,Y)$ tels que $K[f]$ est intégralement fermé dans $K[X,Y]$. (French) [Polynomials $f(X,Y)$ such that $K(f)$ is integrally closed in $K[X,Y]$] Acta Arith. 105 (2002), no. 1, 9--28.</p> <p>MathSciNet review by Maurice Mignotte:</p> <p>As indicated in the title, the author studies polynomials $f(X,Y)$ such that $K[f]$ is integrally closed in $K[X,Y]$ (then $f$ is said to be closed), where $K$ is of characteristic zero. His main result is the following theorem. Let $f(X,Y)$ be a nonconstant polynomial in $K[X,Y]$. Then $K[f]$ is integrally closed in $K[X,Y]$ if and only if there exists $a\in K$ such that $f(X,Y)+a$ is irreducible over the algebraic closure of $K$.</p> <p>Indeed, this article contains many other results and also provides a survey of this question. In particular, several examples of families of closed polynomials are given. The case of positive characteristic is studied briefly at the end of the paper. </p> <p><hr /></p> <p>Explanation for how I found this: I recalled that my colleague Dino Lorenzini has a paper "Reducibility of polynomials in two variables" (J. Algebra, 1993). Upon inspection, this paper didn't seem quite in the spirit of your question -- rather, he gives upper bounds on the number of irreducible components in certain families of polynomials -- but it has several citations on MathSciNet. Ayad's paper is the first. </p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/14080#14080 Answer by David Speyer for Irreducibility of polynomials in two variables David Speyer 2010-02-03T23:25:53Z 2010-10-16T14:31:12Z <p>A trick which works surprisingly often in my experience: If the Newton polytope of $f$ can not be written as a Minkowski sum of two smaller polytopes, then $f$ is irreducible. I think of this as a generalization of Eisenstein's criterion. </p> <p>It is surprisingly easy to test whether a lattice polytope in $\mathbb{R}^2$ can be written as a Minkowski sum of smaller lattice polytopes. Let $P$ be a lattice polytope. For example, the convex hull of $(2,0)$, $(1,1)$ and $(0,0)$. Travel around $\partial P$ and write down the vectors pointing from each lattice point to the next. So, in this case, we would write $(-1,1)$, $(-1,-1)$, $(1,0)$, $(1,0)$. We'll call this sequence $v(P)$.</p> <p>It turns out that $v(A + B)$ is simply the sequences $v(A)$ and $v(B)$, interleaved in a certain manner. So, if $P$ can be written as the Minkowski sum $A+B$, we must be able to partition $v(P)$ into two disjoint sub-sequences, each of which sums to zero. In the above example, this can't be done, so any polynomial of the form $a+bx+c x^2 + d xy$, with $a$, $c$ and $d$ nonzero, must be irreducible.</p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/14081#14081 Answer by Qiaochu Yuan for Irreducibility of polynomials in two variables Qiaochu Yuan 2010-02-03T23:26:51Z 2010-02-03T23:26:51Z <p>If $k$ is algebraically closed, then any two components of the projective closure of $\text{Spec } k[x, y]/(f(x, y))$ intersect by Bezout's theorem, and one can check for the existence of such points by looking at where the partial derivatives simultaneously vanish (the singular points). For example, $f(x, y) = x^2 + 2xy + y^2 - 1$ has projective closure defined by $F(X, Y, Z) = X^2 + 2XY + Y^2 - Z^2$ and the partial derivatives vanish at $(1 : -1 : 0)$, the intersection of the components $X + Y - Z = 0$ and $X + Y + Z = 0$.</p> <p>Generically, $k[x]$ is a UFD, so Eisenstein's criterion applies, although I am not sure how practical this is. </p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/14084#14084 Answer by Graham Leuschke for Irreducibility of polynomials in two variables Graham Leuschke 2010-02-03T23:51:25Z 2010-02-04T00:03:00Z <p>One of my all-time leading candidates for Most Preposterous Theorem Ever:</p> <p>Definition: A polynomial $f(x) \in \mathbb{C}[x]$ is indecomposable if whenever $f(x) = g(h(x))$ for polynomials $g$, $h$, one of $g$ or $h$ is linear.</p> <p>Theorem. Let $f, g$, be nonconstant indecomposable polynomials over $\mathbb C$. Suppose that $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x) = f(ax+b)$ for some $a,b \in \mathbb{C}$, or $$\operatorname{deg} f = \operatorname{deg} g = 7, 11, 13, 15, 21, \text{ or } 31,$$ and each of these possibilities does occur.</p> <p>The proof uses the classification of the finite simple groups [!!!] and is due to Fried [1980, in the proceedings of the 1979 Santa Cruz conference on finite groups], following a the reduction of the problem to a group/Galois-theoretic statement by Cassels [1970]. [W. Feit, <a href="http://books.google.com/books?id=UA-YqrL58dQC&amp;lpg=PP1&amp;ots=2I1C6mSw1t&amp;dq=santa%20cruz%20conference%20finite%20groups&amp;pg=PA176#v=snippet&amp;q=indecomposable&amp;f=false" rel="nofollow">"Some consequences of the classification of finite simple groups,"</a> 1980.]</p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/42351#42351 Answer by viethung for Irreducibility of polynomials in two variables viethung 2010-10-16T01:05:20Z 2010-10-16T01:05:20Z <p>there is a theorem of Ruppert, but I don't know it is useful or not. You can try: <a href="http://pdf%20-%20arXiv%3amath/9808021v1%20%5Bmath.NT%5D%205%20Aug%201998" rel="nofollow">pdf - arXiv:math/9808021v1 [math.NT] 5 Aug 1998</a></p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/46029#46029 Answer by wishcow for Irreducibility of polynomials in two variables wishcow 2010-11-14T10:09:30Z 2010-11-14T10:09:30Z <p>I have a reference for the method mentioned by David Speyer above (sorry couldn't find how to add comment to existing answer):</p> <p>S. Gao, Absolute irreducibility of polynomials via newton polytopes, <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.46.5311" rel="nofollow">link</a></p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/94418#94418 Answer by Hugo Chapdelaine for Irreducibility of polynomials in two variables Hugo Chapdelaine 2012-04-18T15:20:08Z 2012-04-18T15:50:34Z <p>Let $f\in A[x][y]$ where $A$ is a UFD and assume that $f$ is monic i.e that it can be written as $$f=a_0+a_1 y+\ldots a_{n-1}y^{n-1}+y^n,$$ with $a_i\in A[x]$. Assume that there exists an irreducible polynomial $g\in A[x]$ such that $g|a_i$ for $i=0\ldots n-1$ and $g^2\nmid a_0$ then by Eisenstein's criterion $f$ is irreducible over $A[x]$ and therefore irreducible in $A[x,y]$. I've used the fact that $A$ was a UFD only to make sure that $gA[x]$ in $A[x]$ is a prime ideal. (Of course Eisenstein criterion is a very special of the method of the Newton polygon).</p> <p>Using this idea and induction it is easy to see that polynomials like $$(*)\;\;\;\; x_1^{n_1}+x_2^{n_2}+\ldots x_r^{n_r}\in \mathbf{C}[x_1,x_2,\ldots x_r],$$ are irreducible whenever $n_i\geq 1$ and $r\geq 3$, since the polynomial $x_1^{n_1}+x_2^{n_2}$ has always a multiplicity one irreducible divisor.</p> <p>More generally there is the so called Eherenfeucht criterion which says that $$(**) \;\;\;\; f_1(x_1)+f_2(x_2)+\ldots f_r(x_r)\in\mathbf{C}[x_1,\ldots,x_r],$$ is always irreducible if $deg(f_i)\geq 1$ and $r\geq 3$. In the case where $r=2$ it is still irreducible if one has $(deg(f_1),deg(f_2))=1$.</p> <p>Note that the polynomials in $(\star)$ are a very special case of polynomials in $(\star\star)$. A nice proof of this criterion may be found in a paper of Tverberg entitled "A remark on Ehrenfeucht's crieterion for irreducibility of polynomials". Unfortunately, if you have a polynomial with mixed monomials then this criterion does not apply. </p> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/94468#94468 Answer by Julien Puydt for Irreducibility of polynomials in two variables Julien Puydt 2012-04-18T21:38:11Z 2012-04-18T22:25:37Z <p>What follows is more a series of considerations than a practical algorithm, but could still be of interest. The main idea is that it's easier to work with one-variable polynoms, so we trade a bad problem in two variables for several bad problems in one variable.</p> <p>The key point is the following lemma (assume $k$ is of characteristic zero!): if $P(X,Y)$ is a two-variable polynom and there are enough distinct values of $a$ such that $P(X,a)$ is a constant polynomial, then $P(X,Y)$ is a polynom in $Y$ only. The proof is by arguing that $P(X,a)$ being a constant polynom means that $a$ is a root for the polynom (in $Y$) coefficient of $X^n$ for all $n>0$. And that can't happen too often in characteristic zero, unless those coefficients are zero polynoms, hence $P(X,Y)$ is reduced to its constant term as a polynom in $X$, hence is only a polynom in $Y$, as was to be proved.</p> <p>Now, for your question : if you suppose a $P(X,Y)$ isn't irreducible, say factors as $Q(X,Y)R(X,Y)$, but many $P(X,a)$ are irreductible, then that means for each such $a$ either $Q(X,a)$ or $R(X,a)$ is a constant, hence given enough of those $a$, one of $Q$ or $R$ at least is only a polynom in $Y$ by the lemma, say $R$.</p> <p>Then if you manage to fully factor $P(a,Y)=Q(a,Y)R(Y)$ for some $a$ (again dropping to a one-variable polynom), you get a list with $R$ (and divisors of $R$): check each element for divisibility of $P(X,Y)$. If none is good, $P$ is irreducible.</p> <p>EDIT:</p> <ol> <li>In fact, after you have found $R$ is a polynom in $Y$, just consider the gcd of the coefficients of the $X^n$ -- if you get $1$, $P$ is irreducible.</li> <li>The previous considerations mostly prove that a cheating polynom (ie: not irreducible although it appears to be when evaluated along a variable) necessarily has a very precise form, which makes it susceptible to easy factorisation.</li> </ol> http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/118956#118956 Answer by Tim Browning for Irreducibility of polynomials in two variables Tim Browning 2013-01-15T10:06:09Z 2013-01-15T10:06:09Z <p>Quite a useful result can be found in Schmidt's lecture notes on <em>Equations over finite fields</em> (Theorem III.1B in SLNM 536). Suppose that $K$ is any field and let $f(x,y)=c_0y^d+c_1(x)y^{d-1}+ \cdots+c_d(x) \in K[x,y]$, with $c_0 \neq 0$. Let $$\psi(f)=\sup_{1\leq i\leq d}\frac{\deg c_i}{i}.$$ Then $f$ is absolutely irreducible over $K$ provided that that $\psi(f)=m/d$ with $\gcd(m,d)=1$.</p> <p>This shows, for example, that the polynomial $f(x,y)=g(x)-h(y)$ is irreducible when $\deg g$ and $\deg h$ are coprime.</p>