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Let $\left\{ p_{k}\in\mathbb{R}\left[ x_{1},\ldots,x_{n-1}\right] :k=1,\ldots,n\right\} $ be a family of linear polynomials such that $p_{k}\left( 0,\ldots,0\right) =0$. Let $\alpha_{i},\beta_{i}\in\mathbb{R}$, $a_{i},b_{i}\in\mathbb{Z}^{+}$ be scalars for $i=1,\ldots,n-1$. We define the polynomials $f_{i}\in\mathbb{R}\left[ x_{1},\ldots,x_{n-1}\right] $ for $i=1,\ldots,n-1$, as follows \begin{align*} f_{1} & =\alpha_{1}p_{1}^{a_{1}}-\beta_{1}p_{2}^{b_{1}}\\ f_{2} & =\alpha_{2}p_{2}^{a_{2}}-\beta_{2}p_{3}^{b_{2}}\\ & \cdots\\ f_{n-1} & =\alpha_{n-1}p_{n-1}^{a_{n-1}}-\beta_{n-1}p_{n}^{b_{n-1}}% \end{align*} Is the ideal $\left\langle f_{1},f_{2},\ldots,f_{n-1}\right\rangle $ zero dimensional?

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    $\begingroup$ How is the dimension of an ideal $I$ defined? The Krull dimension of the quotient? $\endgroup$
    – YCor
    May 2, 2015 at 11:42

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Looks like this is false.

Let $p_k=(x_1,x_1+x_2,x_2+x_3,x_2+x_2+x_3)$ and all other constants $1$. It is of dimension $1$ according to sage.

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