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Let $R$ be a unique factorization domain, let $X=(x_{ij})_{1\leq i\leq n,1\leq j\leq n}$, be a family of independent indeterminates, and let $R_{nn}$ be the polynomial ring $R[x_{11},x_{12},...,x_{21},x_{22},...,x_{nn}]$. Let $D_{n}\in R_{nn}$ be the determinant of the $n\times n $ matrix $X$, and let $D_{n-1}\in R_{nn}$ be the cofactor of $x_{nn}$, i.e. the determinant of the matrix obtained from the above one by deleting the $n$th row and column.

The question is: (1) if $n>1$, then in $R_{nn}$, $D_{n-1}$ doesn't divide $D_{n}$; (2) $D_{n}$ generates a prime ideal in $R_{nn}$.

user53800
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1 Answers1

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Both statements are immediate consequences of the fact that the determinant $D_n$ is an irreducible polynomial, as is proved in this question.

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Marc van Leeuwen
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