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I'm interested in $V(\operatorname {det}-1) \subset \mathbb{A}^{n^2}$ with the determinant seen as a polynomial. I know that $\det$ is irreducible. But I want to show that $\det-1$ is irreducible. In a paper, it says that if $\det-1=fg$ is a non trivial factorization so the top homogeneous components of $f$ and $g$ gives a non trivial factorization for $\det$. I don't understand why.

Cordialy, doeup

Tina
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doeup
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1 Answers1

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Observe that $\det$ is a homogeneous polynomial in $n$ variables of the same degree. Then $\det-1$ has two homogeneous components, one in homogeneous degree $n$ and one in homogeneous degree $0$. If $f,g$ are any two polynomials and $\det-1=fg$, the homogeneous components of $fg$ of degree $n$ must be $\det$, the $0$th one must be $-1$, and all others should vanish.

Pedro
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