Is the polynomial x 0 2 y 0 + x 0 x 1 y 1 + x 1 2 y 2 + x 1 x 2 y 3 + x 2 2 y 4 + x 0 x 2 y 5 ∈ K [ x i , y j ]reducible over an algebrically closed field K ?

Question

Is the polynomial      x    0    2        y    0    +      x    0        x    1        y    1    +      x    1    2        y    2    +      x    1        x    2        y    3    +      x    2    2        y    4    +      x    0        x    2        y    5    ∈      K    [      x    i    ,      y    j    ]reducible over an algebrically closed field       K   ?

Bernard Scott · Accepted Answer

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    ,  …  ,      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   Δ is not a perfect square in R. I compute the discriminant to be                    Δ                            =                    (                  x          1                          y          1                +                  x          2                          y          5                          )          2                −        4                  y          0                (                  x          1          2                          y          2                +                  x          1                          x          2                          y          3                +                  x          2          2                          y          4                )                                          =                    (                  y          1          2                −        4                  y          0                          y          2                )                  x          1          2                +        (        lower terms        )        .            Now if the discriminant is a perfect square I can write   Δ  =  (  a      x    1    +  b      )    2   where  where a,b are polynomials in  k  [      x    2    ,      y    0    ,  …  ,      y    5    ]. Comparing coefficients, we get that      a    2    =  (      y    1    2    −  4      y    0        y    2    )  .But this results in a contradiction because the R.H.S. is irreducible by Eisenstein with the prime element       y    0    ∈  k  [      y    0    ,      y    2    ]. Thus your original polynomial must be irreducible.

Is the polynomial x^2_0y_0+x_0x_1y_1+x^2_1y^2+x_1x_2y_3+x^2_2y_4+x_0x_2y_5 in K[x_i,y_j]

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