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Igor Rivin
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This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over)?

This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over?

This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over)?

Source Link
Igor Rivin
  • 96.4k
  • 11
  • 153
  • 366

irreducibility of discriminant

This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over?