I have problems with this:
I need to prove that in the polynomial ring the radical of an ideal generated by monomials is also generated by monomials.
I found a proof on internet that uses the convex hull of the multidegrees of the monomials, but I want a proof that uses less terminology. For example, can be proved in a simple way the following:
Given a monomial $u=x^a = \prod\limits_{k = 1}^n {x_k ^{\alpha_k}}$ we define $\sqrt u = \prod\limits_{k = 1}^n {x_k }$. How can I prove that if $G(I)$ is a minimal set of generators of $I$ ( I proved that this set it's also a monomial), then the set $\left\{\sqrt u: u \in G\left(I\right) \right\}$ generates the radical of $I$?
Best Answer
Claim: If $m_1,\ldots,m_s$ are monomials in $K[x_1,\ldots,x_n]$, then $$\sqrt{\langle m_1,\ldots,m_s\rangle} = \langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle.$$
Proof: Put $k_i:=(\text{greatest exponent of any variable in }m_i)\in\mathbb{N}$, i.e. if $m_i=x_{j_1}^{a_1}\cdots x_{j_l}^{a_l}$ then $k_i=\max\{a_1,\ldots,a_l\}$. Now put $k:=k_1+\cdots+k_s-s+1$. We have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq \langle m_1,\ldots,m_s\rangle$, because every term of $\prod_{j=1}^k(f_{1,j}\sqrt{m_1}+\cdots+f_{s,j}\sqrt{m_s})$ has the form $f\sqrt{m_1}^{\beta_1}\cdots\sqrt{m_s}^{\beta_s}$ where $b_j\in\mathbb{N}_0$ and $\beta_1+\cdots+\beta_s=k$, which means at least one $\beta_j\geq k_j$. Therefore $$\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq\langle m_1,\ldots,m_s\rangle\subseteq\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle/\sqrt{~},$$ $$\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}\subseteq\sqrt{\langle m_1,\ldots,m_s\rangle}\subseteq\sqrt{\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle}.$$ Thus it remains to show that $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}=\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle$, i.e. that squarefree monomial ideals are radical.
If $\sqrt{m_1}=x_{j_1}\cdots x_{j_l}$, we have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_{r=1}^l\langle x_{j_r},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle$, because by point e) from my post here, $$\langle x_{j_1},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle\cap\langle x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle= \sum\sum\langle\mathrm{lcm}(\ast,\ast)\rangle= \langle x_{j_1}x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s} \rangle.$$ Next, $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_r\bigcap_{r'}\langle x_{j_r},x_{j_{r'}},\sqrt{m_3},\ldots,\sqrt{m_s}\rangle$, and so on. Therefore $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle = \bigcap_\lambda\mathfrak{p}_\lambda$ for some ideals $\mathfrak{p}_\lambda$, generated by variables. But $\mathfrak{p}_\lambda$ are prime by b), hence $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle} = \sqrt{\cap_\lambda\mathfrak{p}_\lambda}=\cap_\lambda\sqrt{\mathfrak{p}_\lambda}=\cap_\lambda\mathfrak{p}_\lambda$, since prime ideals are radical. $\blacksquare$