x_{n-1}]/<x_j^2\mid j=1,\ldots, n-1>$, and let $\alpha_k, \beta_k$ be elements of $H^2(B_{n-1})$ for $k=1,2$. From the assumption we have $\alpha_1+\alpha_2=\beta_1+\beta_2$ and $\alpha_1\alpha_2=\beta_1\beta_2=0$.

In general, for two elements $u=\sum_{i=1}^{n-1}u_ix_i$ and $v=\sum_{i=1}^{n-1}v_ix_i$ of $H^2(B_{n-1})$, the identity $uv=0$ holds if and only if $u_jv_i+u_iv_j=0$ for any $j\neq i$. From this, we can see easily that if $uv=0$, one of the following three possibilities follows.

1.  If at least three coefficients in $u$ are non-zero, then $v=0$.

2.  If exactly two coefficients in $u$, say $u_i$ and $u_j$, are non-zero, then so is $v$ with $v_jv_i\neq 0$ and $u_jv_i+u_iv_j=0$.

3.  If only one coefficient in $u$, say $u_j$, is non-zero, then so is $v$ with $v_j\neq 0$.

Suppose $\alpha_1$ has at least three non-zero coefficients. Then (1) implies that $\alpha_2=0$ and $\alpha_1=\beta_1+\beta_2$. If $\beta_1\neq 0$ and has at most two non-zero coefficients then so is $\beta_2$ with non-zero coefficients at the same places as $\beta_1$ by (2) and (3), which is a contradicts to the assumption that $\alpha_1$ has at least three nonzero coefficients because $\alpha_1+\alpha_2=\beta_1+\beta_2$. So $\beta_1$ is either $0$ or has at most three non-zero coefficients. Therefore by (1) either $\beta_1=0$ or $\beta_2=0$, and two bundles $\xi_1$ and $\xi_2$ are isomorphic.

Suppose that 