Is this polynomial irreducible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:03:20Z http://mathoverflow.net/feeds/question/72147 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72147/is-this-polynomial-irreducible Is this polynomial irreducible? zroslav 2011-08-05T05:26:22Z 2011-08-05T15:10:05Z <p>I'm just curious about the polynomial $\det (x_k^iy_k^j)_{0\leq i\leq d-1, 0\leq j\leq e-1, 1\leq k\leq de}$ (determinant of $de\times de$-matrix, $x_k$, $y_k$ are all independent variables). Is anything known about its factorization on irreducible polynomials? </p> <p>The question is based only on my own interest.</p> http://mathoverflow.net/questions/72147/is-this-polynomial-irreducible/72184#72184 Answer by Jason Starr for Is this polynomial irreducible? Jason Starr 2011-08-05T14:50:53Z 2011-08-05T15:10:05Z <p>Geometrically, your question is the following. Let $v_{d-1}:\mathbb{P}^1 \to \mathbb{P}^{d-1}$ and $v_{e-1}:\mathbb{P}^1 \to \mathbb{P}^{e-1}$ be the Veronese embeddings. Let $s_{d-1,e-1}:\mathbb{P}^{d-1}\times \mathbb{P}^{e-1} \to \mathbb{P}^{de-1}$ be the Segre embedding. Consider the composition $c$, i.e., $s_{d-1,e-1}\circ (v_{d-1},v_{e-1}):\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^{de-1}$. The composition is "linearly nondegenerate", i.e., for a general choice of $de$ points $(x_k,y_k)$ of the domain $\mathbb{P}^1\times \mathbb{P}^1$, the image points in $\mathbb{P}^{de-1}$ span the target. And you are now considering the divisor $\Delta$ in $(\mathbb{P}^1\times \mathbb{P}^1)^{de}$ where the image points are linearly degenerate. You want to know if this divisor is reducible.</p> <p>I claim $\Delta$ is irreducible. Choose some index $k=1,\dots,de$ and consider the projection $\pi_k:(\mathbb{P}^1\times \mathbb{P}^1)^{de} \to (\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$ which forgets the $k^{\text{th}}$ component. Consider the restriction $\pi_k:\Delta \to (\mathbb{P}^1 \times \mathbb{P}^1)^{de-1}$. Consider a generic point in the target $(\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$. This parameterizes $de-1$ points in $\mathbb{P}^1\times \mathbb{P}^1$ whose images in $\mathbb{P}^{de-1}$ span a generic hyperplane $H$. The intersection of $H$ with $c(\mathbb{P}^1\times \mathbb{P}^1)$ is a generic curve $\Gamma$ of bidegree $(d-1,e-1)$ in $\mathbb{P}^1\times \mathbb{P}^1$, which is irreducible. For the $k^{\text{th}}$ point of $\mathbb{P}^1\times \mathbb{P}^1$, the total collection of $de$ points are linearly independent in $\mathbb{P}^{de-1}$ unless the $k^{\text{th}}$ point maps into $H$, i.e., unless the $k^{\text{th}}$ point is in the irreducible curve $\Gamma$.</p> <p>What this proves is that there is a unique irreducible component of $\Delta$ which dominates $(\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$. And for any component which does not dominate, the fiber dimension over its image must be precisely $2$ and the image must be a divisor, i.e., for some codimension $1$ subset of $(\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$, every choice of $k^{\text{th}}$ points makes the total collection of $de$ points linearly degenerate. Since $c(\mathbb{P}^1\times \mathbb{P}^1)$ spans $\mathbb{P}^{de-1}$, the image is contained in no hyperplane. So the only possibility is that the collection of $de-1$ points is itself linearly dependent. But, we expect $de-1$ points in $\mathbb{P}^{de-1}$ to be linearly independent in codimension $2$, not in codimension $1$. Of course the "expected codimension" may be wrong, but since this situation is so homogeneous, I bet it is easy to prove the expected codimension equals the actual codimension. I will think about it a bit more and post soon.</p> <p><b>Edited.</b> Okay, by the above, we have only to prove that for every effective Cartier divisor $D$ in $(\mathbb{P}^1\times \mathbb{P}^1)^{de}$, i.e., $D$ is an irreducible component of $\Delta$, there exists an index $k=1,\dots,de$, such that for the projection $\pi_k:(\mathbb{P}^1\times \mathbb{P}^1)^{de} \to (\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$, $\pi_k:D \to (\mathbb{P}^1\times \mathbb{P}^1)^{de-1}$ is dominant, i.e., the intersection of $D$ with a general fiber $F_k$ is nonzero Now the Picard group of $(\mathbb{P}^1\times \mathbb{P}^1)^{de}$ is $\mathbb{Z}^{2de}$, i.e., every invertible sheaf is of the form $\text{pr}_1^*\mathcal{O}(a_1,b_1)\otimes \dots \otimes \text{pr}_{de}^*\mathcal{O}(a_{de},b_{de})$ for some choice of integers $a_l,b_l$. If the divisor $D$ is effective, then its corresponding invertible sheaf has every $a_l,b_l$ nonnegative. And the fiber $F_k$ is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$. The restriction of the invertible sheaf to $F_k$ is isomorphic to $\mathcal{O}(a_k,b_k)$. In particular, the intersection of $D$ with $F_k$ can be empty only if $a_k$ and $b_k$ equal $0$. But if this holds for <b>every</b> choice of $k$, then the invertible sheaf os just $\mathcal{O}$ which forces $D$ to be empty. </p>