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 $I$?
Let $J=\oplus J_i$ be a $\mathbb{Z}$ or $\mathbb{N}$ graded module over a graded ring $R=\oplus R_i$ (this question is such a case). Denote by $J^\ast$ the ideal generated by homogenous elements of $J$.
Whenever we write something in its unique grade decomposition, we will write $x=\sum x_i$ where $x_i$ is homogeneous of grade $i$.
Key idea: By definition, an element $x\in J$ is in $J^\ast$ iff $x_i\in J$ for all $i$.
Proposition: $(\sqrt{J})^\ast=\sqrt{J^\ast}$.
Lemma: If $z$ is a homogenous element of $J$, then $z\in \sqrt{J}\iff z\in\sqrt{J^\ast}$.
Proof of lemma: $z\in \sqrt{J}\iff \exists n, z^n\in J\iff \exists n, z^n\in J^\ast\iff z\in \sqrt{J^\ast} $ QED.
Proof of proposition: By the lemma, any homogenous element of $\sqrt{J}$ is in $\sqrt{J^\ast}$, and so $(\sqrt{J})^\ast\subseteq\sqrt{J^\ast}$.
We now proceed to show equality via a contradiction. Suppose there exists an $x=\sum x_i\in \sqrt{J^\ast}\setminus(\sqrt{J})^\ast $. If one of the $x_i$ is in $\sqrt{J}$, then $x-x_i$ is still in $\sqrt{J^\ast}\setminus(\sqrt{J})^\ast $. WLOG then, replace $x$ with what is left after subtracting all of the $x_i$ that are already in $\sqrt{J}$.
Suppose momentarily that $x$ has a strictly positive grade term, and let $x_{\max}$ be the term of highest positive grade. By definition, there exists an $n$ such that $x^n\in J^\ast$, so every homogenous term of $x^n$ is in $J$. But the largest grade term in $x^n$ is $x_{\max}^n$, so $x_{\max}^n\in J$, and consequently $x_{\max}\in\sqrt{J}$. This is a contradiction since $x_{max}$ is not in $\sqrt{J}$. Therefore $x$ has no strictly positive grade term.
If the grading is By a similar argument, $x$ has no strictly negative grade term. Thus, $x$ is homogeneous of grade zero, and hence is in $(\sqrt{J})^\ast$: a contradiction!