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$\DeclareMathOperator\Fl{Fl}$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 $$ p_k(n) = x^k_1 + x^k_2 + \dotsb + x^k_n. $$

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    $\begingroup$ A small point, but you need $\mathbb{Q}$ coefficients to have these generating sets agree in general, I think you only get $2e_2$ in this ideal with $\mathbb{Z}$ coefficients. $\endgroup$
    – Chris H
    Commented Jun 30, 2023 at 21:20

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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.

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