## Is this polynomial irreducible?

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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?

The question is based only on my own interest.

-

1
Uhm, a matrix with three indices? – Federico Poloni Aug 5 2011 at 7:11

2
I think, rows are indexed by pairs (i,j) and columns by k – Fedor Petrov Aug 5 2011 at 7:16

1
Please change either question or title. This polynomial is not irreducible for sure, it is divisible by x_1. – Fedor Petrov Aug 5 2011 at 7:22
@Fedor: Done! Thanks! – zroslav Aug 5 2011 at 12:34

## 1 Answer

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.

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 $\Gamma$.

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.

Edited. 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 every choice of $k$, then the invertible sheaf os just $\mathcal{O}$ which forces $D$ to be empty.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9058181047439575, "perplexity_flag": "head"}