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Is the polynomial $$ x_0^2y_0+x_0x_1y_1+x_1^2y_2+x_1x_2y_3+x_2^2y_4+x_0x_2y_5 \in \mathbb K[x_i, y_j] $$ reducible over an algebrically closed field $\mathbb K$?

I've noticed that the polynomial is bi-homogeneous of degree $(2,1)$ and I think that the question can be stated in the following more general way: fix an integer $d \ge 1$ and consider the space of homogeneous polynomials of degree $d$ in $x_0,x_1,x_2$. This is a vector space (over $\mathbb K$) of dimension $N$ (it doesn't matter, anyway I think $N=\binom{d+2}{2}$). Order somehow the polynomials $M_i(x_0,x_1,x_2)$ of a base of this space and then consider $$ p(x_i,y_j) = \sum_{i=0}^{N-1} M_i(x_0,x_1,x_2)y_i $$

Is $p$ reducible? I do not know how to begin... Do you know any useful tricks to prove reducibility/irreducibility in this case? Thanks.

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1 Answer 1

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We claim that your polynomial is irreducible at least when the characteristic is not $2$. To see this consider your polynomial as a quadratic polynomial in $x_0$ with coefficients in the UFD $R:= k[x_1,x_2,y_0,y_1,\ldots,y_5]$. Your polynomial is reducible iff it admits a root in $R$. So to show it does not admit a root in $R$, it is enough to show that the discriminant $\Delta$ is not a perfect square in $R$. I compute the discriminant to be

$$\begin{eqnarray*} \Delta &=& (x_1y_1 + x_2y_5)^2 - 4y_0(x_1^2 y_2 + x_1x_2y_3 + x_2^2y_4) \\ &=& (y_1^2 - 4y_0y_2)x_1^2 + (\text{lower terms}).\end{eqnarray*}$$

Now if the discriminant is a perfect square I can write $\Delta = (ax_1 + b)^2$ where $a,b$ are polynomials in $k[x_2,y_0,\ldots,y_5]$. Comparing coefficients, we get that

$$a^2 = (y_1^2 - 4y_0y_2).$$

But this results in a contradiction because the R.H.S. is irreducible by Eisenstein with the prime element $y_0 \in k[y_0,y_2]$. Thus your original polynomial must be irreducible.

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