Let $G\_2(\mathbb{C}^{n+1})$ be the complex grassmannian. 


Then the cohomology ring $H^\*(G\_2(\mathbb{C}^{n+1});\mathbb{C})=\mathbb{C}[c\_1,c\_2]/(f\_n,f\_{n+1})$, where $f\_k=\sum\_{i=0}^{[k/2]}(-1)^{k-i}{{k-i}\choose i}c\_1^{k-2i}c\_2^i$. 


I want to compute the cup-length of the first Chern class $c\_1$, i.e., the smallest positive integer $l$ such that $c\_1^l=0$. (For the real case, I found a paper *Cup products in Grassmannians, R.E. Stong, Topology and its Applications, Volume 13, Issue 1, 1982, Pages 103–113*). 


By considering the dimension, I have $l\leq 2n-1$. Tested by computer, I obtain $l=2n-1$ for $n\leq 30$. 


Are there any references or method to prove $l=2n-1$? This question is equivalent to the following question: [a problem about ideals of polynomial rings](https://mathoverflow.net/questions/204730/a-problem-about-ideals-of-polynomial-rings). Thanks!



cup-length of the first Chern class of complex grassmannian
The mere fact that the Grassmannian $G=G\_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such that the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one such class does.


