Let $X = Gr(2,4)$ the complex Grassmannian of $2$-planes in $V = \Bbb C^4$ and $S$ the tautological bundle, $Q$ the quotient bundle. The cohomology ring is generated by $c_1(S), c_2(S)$ with relations $c(S)c(Q) = 1$, coming from the short exact sequence of vector bundles $0 \to S \to V \to Q \to 0$.
One should get $c_1(S)^4 = 1$ purely from these relations but I'm not able to do so. Can someone explain how to do it ?
First of all, $c_1(S)^4 = 2$, not 1. The computation itself is quite easy. Let me denote $a_i := c_i(S)$, $b_i = c_i(Q)$. Then the relations are $$ a_1 + b_1 = a_2 + a_1b_1 + b_2 = a_1b_2 + a_2b_1 = a_2b_2 = 0. $$ The first gives $b_1 = -a_1$, the second gives $b_2 = a_1^2 - a_2$, and the last two give $$ a_1^3 = 2a_1a_2, \qquad a_1^2a_2 = a_2^2. $$ A combination of the last two equalities gives $ a_1a_2^2 = a_1(a_1^2a_2) = (a_1^3)a_2 = 2a_1a_2^2, $ hence $a_1a_2^2 = 0$. This means that the cohomology ring is spanned over $\mathbb{Z}$ by $1$, $a_1$, $a_2$, $a_1a_2$, $a_2^2$, and that $$ a_1^4 = 2a_2^2 $$ which translates into the required equality $c_1(S)^4 = 2$.