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I'm trying to prove that if $ K$ is a field and $ I $ is a monomial ideal in $ K[x_1, \dots, x_n] $, then

$$\sqrt{I} = I \iff I ~\text{is generated by square-free monomials}$$

So I tried to do the ,,$ \Rightarrow"$ by contradiction:

suppose $ x_1^{q_1} \dots x_k^2 \dots x_n^{q_n} $ is one of the generators of $I$. Somehow this should lead to the fact that there exists some $ f \in K[x_1, \dots, x_n], f \notin I $ and $ n\in\mathbb{N} $ such that $ f^n \in I $, but I can't see how? Maybe $ f=x_1^{q_1} \dots x_k \dots x_n^{q_n} $ and $ n = 2 $, but how do I prove $ f\notin I $?

user26857
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Jytug
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1 Answers1

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Take a minimal system of (monomial) generators for $I$. If $x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ with $q_i\ge 1$ is one of them, let's suppose that some $q_t\ge 2$. Then $x_{i_1}\cdots x_{i_r}\in\sqrt I$, and $x_{i_1}\cdots x_{i_r}\notin I$.
If $x_{i_1}\cdots x_{i_r}\in I$, then there is a monomial $m$ in the minimal system of generators of $I$ such that $m\mid x_{i_1}\cdots x_{i_r}$. Since $x_{i_1}\cdots x_{i_r}\mid x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ we must have $m=x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$, but this is not possible because $x_{i_t}$ appears in $m$ at most once.

user26857
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  • And by minimal you mean that there is no other generator that divides this one? – Jytug Nov 17 '14 at 21:32
  • It might be useful to note that an element $f$ is in a monomial ideal iff each of its monomial terms is divisible by at least one monomial generator. – Pedro Nov 18 '14 at 01:14
  • Why do $m\mid x_{i_1}\cdots x_{i_r}$ and $x_{i_1}\cdots x_{i_r}\mid x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ imply $m=x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ instead of just $m\mid x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$? –  Dec 14 '15 at 12:15
  • @Exterior If $m\mid x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ and $m\ne x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ then one can remove $x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ from the system of generators, contradicting the minimality. – user26857 Dec 14 '15 at 12:27
  • @user26857 is it moreover true that the radical of a f.g ideal in a UFD is precisely the ideal generated by the prime decompositions of the generators, only without multiplicities? –  Dec 14 '15 at 12:31
  • @Exterior This holds for monomial ideals, but not in such a generality: see $I=(y,x^2-y)$. – user26857 Dec 14 '15 at 17:49