$\DeclareMathOperator\Fl{Fl}$It is known that $H^\*(\Fl(m)) \cong R^{\mathbb Z}(m)$, where $\Fl(m)$ denotes the variety of complete flags in $\mathbb C^m$, and $R^{\mathbb Z}(m)$ is the coinvariant algebra, which may be defined by $$\frac{\mathbb Z[x\_1,\dotsc,x\_m]}{e\_1(x\_1,\dotsc,x\_m),e\_2(x\_1,\dotsc,x\_m),\dots,e\_m(x\_1,\dotsc,x\_m)}$$
More generally the coinvariant algebra is just the quotient $\mathbb C[x\_1,\dotsc,x\_m]/I\_m$ (the above just replaces $\mathbb Z$ with $\mathbb C$), where $I\_m$ is the ideal generated by the non-constant, homogeneous $S\_m$-invariants, where $S\_m$ acts on $\mathbb C[x\_1,\dotsc,x\_m]$ in the natural way. The above definition presented is where we choose the elementary symmetric polynomials $e\_i$ as a basis of $I\_n$. The isomorphism in that case is realized by sending the generator $x\_i$ of $R^{\mathbb Z}(m)$ to $-c\_1(U\_i/U\_{i-1})$, where $U\_i$'s are the standard bundles over $\Fl(m)$. These successive quotients filter the trivial bundle, $\mathbb C^m \times \Fl(m)$, by line bundles, thus the $i$th chern class of the trivial bundle is given by $e\_i(x\_1,\dotsc,x\_m)$ and must vanish hence the isomorphism. I would say this constitutes a geometric understanding of why the relations $e\_i$ appear.


If we chose a different basis for $I\_m$, the power sum symmetric polynomials, there is of course another isomorphism $H^\*(\Fl(m)) \cong \mathbb Z[x\_1,\dotsc,x\_m]/(p\_1(m),p\_2(m),\dotsc,p\_m(m))$. What is the geometric significance of the $p\_i(m)$'s, if any? Can they be realized in the same way as above but corresponding to different bundles?


The power sum symmetric polynomials are defined by
$$
p\_k(n) = x^k\_1 + x^k\_2 + \dotsb + x^k\_n.
$$



Alternative bases of symmetric polynomials in cohomology ring of flag varieties and coinvariant algebras
I assume we compute cohomology with $\mathbb{Q}$ coefficients. Let $x\_i$ map to $c\_1(U\_i/U\_{i-1})$, like you suggest. Then the polynomial $p\_k(n)$ gets sent to $k!$ times the $k$-th graded piece of the Chern character of the trivial bundle.


