Are zero dimensional ideals radical?

Question

I have a question about Theorem 3.7.25. of Computational commutative algebra I by M. Kreuzer and L. Robbiano.

Let $K$ be a perfect field, $I \subseteq K[x_1, \ldots, x_n]$, be a zero dimensional radical ideal in normal $x_n$ position, let $g_n \in K[x_n]$ be the monic generator of the elimination ideal $I \cap K[x_n]$, and let $d = \deg(g_n)$.

Then, the reduced Groebner basis of the ideal $I$ with respect to lex is of the form $\{x_1 − g_1,\ldots , x_{n−1} − g_{n−1}, g_n\}$, where $g_1,\ldots, g_{n−1} \in K[x_n]$.

My question is if we have a zero-dimensional ideal $I$ with the reduced Groebner basis of the form $\{x_1 − g_1,\ldots, x_{n−1} − g_{n−1}, g_n\}$, can we claim that it's a radical ideal in general? If not, what are the conditions we need to be sure $I$ is radical?

Answer 1

From the comments, I got my answer and I will write it here for future reference. In general, if $ I = (x_1 - g_1, \ldots, x_{n-1}-g_{n-1}, g_n)$ with $g_1, \ldots, g_n \in K[x_n]$, then $K[x_1, \ldots, x_n]/I \cong K[x_n]/(g_n)$. So $I$ is radical iff $K[x_n]/(g_n)$ is reduced iff $g_n$ is separable.

Also, from the same book, Proposition 3.7.15 (Seidenberg’s Lemma): Let $K$ be a field, let $P = K[x_1, \ldots, x_n]$, and let $I \subseteq P$ be a zero-dimensional ideal. Suppose that, for every $i \in \{1, \ldots, n\}$, there exists a non-zero polynomial $g_i \in I \cap K[x_i]$ such that $gcd(g_i , g_i^{\prime} ) = 1$. Then $I$ is a radical ideal.