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?



Are zero dimensional ideals radical?
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.


