I have to prove that in the ring $k[x,y,z]$ the radical of $(xy,yz,xz)$ is $(xy,yz,xz)$ itself. Can anyone give any pointers. I have sort of proved it by showing that $(xy,yz,xz)$ is the radical of $(xy,(x-y)z)$. But I would like alternate proofs. Thanks.
It would be nice if somebody can explain me the square free formulation of monomial radicals. Thank you.
I assume that $k$ is a field.
Let $I$ be a monomial ideal in $R=k[x_1,x_2,\ldots, x_n]$, generated by $\{\mathbf{x}^{\alpha_j}\}_{j=1}^m$. Here, $\alpha_j \in \mathbb{Z}_{\geq 0}^n$, and $\mathbf{x}^{\alpha_j} = x_1^{\alpha_{j1}} x_2^{\alpha_{j2}}\cdots x_n^{\alpha_{jn}}$.
Lemma: The set of monomials $$\{\mathbf{x}^\alpha : \mathbf{x}^{\alpha_j} \not\mid \mathbf{x}^\alpha \text{ for all }j \in \{1,\ldots,m\}\}$$ is a $k$-basis for $R/I$.
Proof: The ideal $I$ is spanned over $k$ by monomials, since it is spanned over $R$ by monomials, and $R$ is generated by monomials. The monomials in $I$ are exactly the monomials divisible by $\mathbf{x}^{\alpha_j}$ for some $j$. If $\mathbf{x}^{\alpha}$ is thus divisible, it is in $I$; conversely, if $f_j \in R$, then all of the monomials $\mathbf{x}^\alpha$ in $\sum_j f_j \mathbf{x}^{\alpha_j}$ are divisible by some $\mathbf{x}^{\alpha_j}$. Since $R$ has a $k$-basis of all monomials, $R/I$ has the desired basis.
Theorem: $I$ is radical if and only if $I$ is generated by square-free monomials.
Proof: Suppose $I$ is generated by $\{\mathbf{x}^{\alpha_j}\}_{j=1}^m$, which are all square-free. Let $f \in R/I$ be nonzero. We may write $f = \sum c_\alpha \mathbf{x}^\alpha$ in $R/I$, where the sum is over $\alpha$ such that $\mathbf{x}^\alpha$ is not divisble by any $\mathbf{x}^{\alpha_j}$. Pick some monomial $\mathbf{x}^\alpha$ such that $c_\alpha \neq 0$ and $\mathbf{x}^\alpha$ has maximal degree of all those in $f$. Then the coefficient of $\mathbf{x}^{\ell\alpha}$ in $f^\ell$ is $(c^{\alpha})^\ell$, by maximality of degree. I claim that $\mathbf{x}^{\ell\alpha} \notin I$. For if $\mathbf{x}^{\ell\alpha} \in I$, then $\mathbf{x}^{\alpha_j} \mid \mathbf{x}^{\ell\alpha}$, but since $\mathbf{x}^{\alpha_j}$ is square-free, this implies $\mathbf{x}^{\alpha_j} \mid \mathbf{x}^{\alpha}$, a contradiction. Since $k$ is a field, we conclude $f^\ell \neq 0$ in $R/I$. This shows $I$ is radical.
Conversely, if $I$ is radical, then for $f \in I$, the maximal squarefree $g$ such that $g \mid f$ is also contained in $I$, for $g^\ell$ is divisble by $f$ for large enough $\ell$. Hence, we may replace any generating set with a squarefree generating set.
Since you're just asking for alternate proofs...
As $k$-vector spaces, $$k[x,y,z]/(xy,yz,xz) \cong k \oplus xk[x] \oplus yk[y] \oplus zk[z]$$ which is a ring isomorphism with the induced multiplication on the right-hand side. Then using this induced multiplication, we can see the resulting ring has no nonzero nilpotents by looking at the highest powers of $x, y, z$ in each element.
