I am not familiar with polynomial ring.
Suppose we have an ideal $\langle f_1, f_2, \ldots, f_n \rangle$ of the polynomial ring $k[x_1, x_2, \ldots, x_n]$. Suppose there is some syzygy relation between the generators $f_i$.
Of course one wants to make the basis really independent, namely to get rid of the syzygy relation.
Is this always possible with the Groebner basis?
Since $\langle f_1, \ldots, f_n \rangle$ is usually not a free $k[x_1, \ldots, x_n]$-module, then there is usually no such basis. Consider for example $\langle x, y \rangle \trianglelefteq k[x,y]$. Then $yx - xy = 0$, so there is a $k[x_1, \ldots, x_n]$-linear relation, i.e., a syzygy, between the generators $x, y$. Similarly, for any generators $f_1, f_2$ of $\langle x, y \rangle$, we have $f_2 f_1 - f_1 f_2 = 0$. Since $\langle x,y \rangle$ can't be generated by fewer than $2$ elements, this shows that there is no $k[x_1, \ldots, x_n]$-basis for $\langle x, y \rangle$.
The way you "get rid" of syzygies is by using free resolutions and Hilbert's Syzygy Theorem.