Let $A$ be a finite-dimensional associative commutative algebra with unity over a field $k$. Is it possible that every hyperplane of $A$ (where we consider $A$ as a vector space over $k$) contains a non-trivial ideal? When $A$ is semisimple (i.e. has no nilpotents), it's a standard theorem that $A$ is isomorphic to a direct product of fields so it's easy to describe all ideals and explicitly demonstrate a hyperplane which does not contain any of them. How to prove the situation to be impossible in the general case (when $A$ has nilpotents)?
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$k$-algebra is a vector space over $k$, and by a hyperplane I mean the $k$-linear subspace of codimension 1. – Dmitry Feb 10 '13 at 17:19
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Ok, I've added it. – Dmitry Feb 10 '13 at 18:03
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I might be wrong, but consider $A = k \cdot 1 \oplus N$, where $N$ is $2$-dimensional, and $N^2 = 0$. Thus $N$ is the nilradical, and every subspace of $N$ is an ideal of $A$. Then any hyperplane (i.e. any subspace of dimension $2$) of $A$ intersects $N$ non trivially in an ideal of $A$.
PS So $k[x,y]/(x^2,xy,y^2)$ would do, where $k[x,y]$ is a ring of polynomials in $x, y$.
Andreas Caranti
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Thank you! Your example looks correct to me. I should've tried to devise examples harder before running to a conclusion they do not exist. – Dmitry Feb 10 '13 at 18:34
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You're welcome! It's easy in hindsight (which is notoriously 20/20), but I also spent some time trying to lift the result from the semisimple case... – Andreas Caranti Feb 10 '13 at 18:40