The radical of a monomial ideal is also monomial

Question

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 $I$?

Answer 1

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. $\blacksquare$

Answer 2

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!

Answer 3

You can do it as follows:

Given $F \in K[X_1,\ldots,X_n] \setminus \{0\}$ one defines its support, $\mathrm{supp}(F)$ as the set of monomials of its distributive form. It is useful (not only for this exercise) to prove that:

If $I$ is a non-zero ideal of $K[X_1,\ldots,X_k]$, then $I$ is monomial $\iff \forall F \in I$, $\mathrm{supp}(F) \subseteq I$.

In this case, you use this characterization and perform induction in the Newton diagram of polynomial (non-null powers of polynomial), $N(F)$.

The case $N(F) = 1$ is direct. The case $N(F) > 1$ you use the induction hypothesis by removing the leading term of your polynomial.