I'm reading characteristic classes form the book Differential forms in Algebraic Topology by Bott and Tu. The Chern classes are defined as follows:


$E\xrightarrow{\rho} M$ is a vector bundle and $E\_p$ be the fiber over $p$. Then $P(E)$, the projectivization of $E$ is a vector bundle with fiber $P(E\_p)\colon=\{\text{1-dim subspaces of $E\_p$}\}$ over $\ell\_p\in P(E)$. It's then discussed that the first Chern class $x$ of the dual of the universal subbundle over $P(E)$ restricted to a fiber is the first Chern class of the fiber. Therefore the $1,x,\ldots,x^{n-1}$ are global classes on $P(E)$, whose restriction to each fiber $P(E)$ freely generates the cohomology of the fiber and hence by Leray-Hirsch $H^\*(P(E))$ is a free module over $H^\*(M)$ with basis $\{1,x,\ldots,x^{n-1}\}$ and then the Chern classes are the coefficients in the expression for $x^n$.


Also by Leray-Hirsch, $H^\*(P(E))\cong H^\*(M)\otimes H^\*(Fiber)$. The fiber $P(E\_p)$ has cohomology $\frac{\mathbb{R}[x]}{x^n}$, as $E\_p$ is a vector space of complex dimension $n$. My question is, why $x^n\neq 0$.


Note: for the trivial bundle $M\times V$, $P(E)=M\times P(V)$ and by Künneth formula $H^\*(P(E))\cong H^\*(M)\otimes H^\*(P(V))$ and then $x^n=0$ as $H^\*(P(V))=\frac{\mathbb{R}[x]}{x^n}$. What is the difference in the general case and the trivial bundle?



Chern classes of complex vector bundle
Bertram already mentioned this in the comments but I thought I'd write an answer for completness's sake.


The Leray-Hirsch theorem says that $H^{\*}P(E)\cong H^{\*}M\otimes H^{\*}(Fiber)\ \ $ $\textit{as $H^{\*}M$ modules}$.


So if $x$ is the first chern class of the tautological line bundle over $P(E)$, there's no reason to expect $x^{n}=0$, even if its pullback to the fiber satisfies that equation. Because if the fiber "twists around," there's no reason to expect that the behavior of a class on one fiber is indicative of its behavior globally. 


On the other hand, when the bundle is trivial, the Kunneth formula (in this setting) tells you that $H^{\*}P(E)\cong H^{\*}M\otimes H^{\*}(Fiber)\ \ $ $\textit{as rings}$.


Here's another fun example to illustrate the difference, where you can even draw pictures! You'll have to use cohomology over $\mathbb{Z}/2$ and you'll be calculating Stiefel-Whitney classes instead of chern classes, but the procedure is exactly the same: Leray-Hirsch, read off coefficients from a basis. Form $E\rightarrow S^1$ by taking the Mobius strip as a vector bundle over $S^1$ plus a copy of the trivial line bundle. Then $P(E)$ will be a familiar non-orientable surface (Klein bottle) whose cohomology ring you can compute from a delta-decomposition. Comparing it to the cohomology ring of the torus (which would be $P(E)$ for the trivial rank 2 bundle) where every positive degree class squares to zero, you can see exactly how the "twist" in the Klein bottle allows one of those classes to no longer square to zero! 


