A Grobner basis $\{g_1,g_2,\dots,g_r\}$ for an ideal $I$ is the set of polynomials such that $I=\langle g_1,g_2,\dots,g_r\rangle$. Also, if you take any polynomial in the ideal $I$, the leading term of that polynomial will be divisible by the leading term of some $g_i$ for $1\leq i\leq r$.
The book "A First Course in Abstract Algebra" by Fraleigh states:
You may wonder how any basis $\{ g_1,g_2,\dots,g_r\}$ can fail to be a Grobner basis for $I=\langle g_1,g_2,\dots,g_r\rangle$. However, cancellation of power products can occur in addition.
Fraleigh makes it sound like it is rare to see a basis of an ideal not being a Grobner basis for the same ideal. I don't understand this at all. Take the ideal $\langle x^2y,x^2z\rangle$ in $F[x,y,z]$. Clearly, the basis elements cannot be reduced any further. Now take the polynomial $f=x^2y+x^2z$. Clearly, the power product of the leading term of neither $x^2y$, nor $x^2z$ divides the leading term of $x^2(y+z)$, which is $x^2(y+z)$ itself. Hence, $\{x^2y,x^2z\}$ is not a Grobner basis for $I$.
Am I going wrong somewhere? Why is it usually expected that the power product of one of the basis elements divides the leading term of every polynomial in the ideal?
Thanks in advance!