# Showing maximality of elements of a primary decomposition of 0

For reference, this problem is Eisenbud "Commutative Algebra with a View Toward Algebraic Geometry" problem 3.10(b)

Let $R=k[a,b]/I$ where $I=(a)\cap(a,b)^2$.

We have already shown that for any nonzero $\lambda\in k$, there exists a (minimal) primary decomposition $$0=(a)\cap(a+\lambda b^n)$$ where $(a)$ is already prime and $(a+\lambda b^n)$ is $(a,b)$-primary. We also showed that if $(a)\cap J=0$ and $(a+\lambda b^n)\subseteq J\ne R$ then $J$ is also $(a,b)$-primary.

We are attempting to prove that, in fact, these $(a+\lambda b^n)$ are maximal among all ideals $J$ such that $(a)\cap J=0$ but need some help continuing from this point.

We have tried naively to take an element in $J$ and show it must be a multiple of $a+\lambda b^n$. Taking note of how $I$ is defined, we got to the point that an arbitrary element $x\in J$ must be of the form $$x=za+\sum_{i=1}^mc_ib^i$$ and that an element in $y\in(a+\lambda b^n)$ is of the form $$y=d_0a+\sum_{i=0}^kd_i\lambda b^{i+n}$$ (note there is no constant term in $x$). Our tact has been to try to prove that $z=c_n$ that $c_i=0$ for all $i<n$, but we have had no luck so far.

Can anyone please shed some light on the subject? Thanks in advance.

Consider $R/(a+\lambda b^n)\cong k[b]/(b^{n+1})$ since in this quotient $b^n=-\frac{1}{\lambda}a$ so $b^{n+1}=0.$ But every ideal of this ring are of the form $(b^m)$ for some $m\le n$, so leveraging the correspondence (lattice isomorphism) theorem, $J=(a+\lambda b^n,b^m)$ whence $a\in J$.
This contradicts the assumption that $J\cap (a)=0$.