\magnification=\magstep1
\noindent
4.2/15.  An {\it alternate proof}:  We need to show that $I = \langle xy, xz, yz\rangle$
is a radical ideal, or equivalently that if $f^m \in I$ for some $m \ge 1$, then 
$f \in I$.  
\bigskip
\noindent
An equivalent statement is the contrapositive form of this implication:  
\medskip
\item{$\bullet$} If $f \notin I$, then $f^m \notin I$ for {\it all} $m \ge 1$.  
\medskip
\noindent
(Note that the existential quantifier ``for some'' flips to ``for all'' when
we negate the quantified statement.)
\bigskip
\noindent
Now, to start, the set $G = \{xy,xz,yz\}$
is easily seen to be a Gr\"obner basis for $I$ by Buchberger's Criterion:  The
$S$-polynomials of all pairs equal $0$, hence they give zero remainders on 
division by $G$.  Since the condition of Buchberger's Criterion
is satisfied, $G$ is a Gr\"obner basis for $I$.
(Note:  This is true for {\it any monomial ideal}, in fact.  The
finite generating set promised by Dickson's Lemma is also a Gr\"obner basis for
the monomial ideal!)  
\bigskip
\noindent
Hence $f \notin I$ means that $f$ contains a term that 
is not divisible by any of $xy, xz, yz$.  The terms that are not divisible by
any of these three monomials are the ``pure powers''
$$x^k, y^k, z^k$$
for $k \ge 0$.  So if $f \notin I$, say $f$ contains a term $c x^k$ for 
some $k$, and {\it in addition, assume that this is the highest pure power
of $x$ appearing in $f$ with $c \ne 0$}.  If $m \ge 1$, then 
when we expand out the power we get
$$f^m = (c x^k + \cdots )^m = c^m x^{km} + \cdots , $$
and the $x^{km}$ must be the highest pure power of $x$ that is contained
in $f^m$.  Hence, since $c^m \ne 0$, we have $f^m \ne I$.  Since this is
true for all $m \ge 1$, we see that $f \notin I$ implies $f^m \notin I$
for all $m \ge 1$, and that is what we wanted to show in this case.
The cases where $f$ contains pure powers of $y$ and/or $z$ are similar.  
\bye
