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Consider a finite dimensional $\mathbb R$-algebra $\mathbb F$. Assume that $\mathbb F$ is associative, commutative, and has the unit $1$.

Define $\mathbb F_z$ as the set of zero divisors of $\mathbb F$ and $\mathbb F^\ast$ as the set of units (invertible elements) of $\mathbb F$.

We have that $\mathbb F$ is the disjoint union of $\mathbb F_z$ and $\mathbb F^\ast$ (see here).

Furthermore, $\mathbb F_z$ is closed (and therefore $\mathbb F^\ast$ is open).

Question: Does the set $\mathbb F_z$ have empty interior?

My attempt so far: I believe the result is true and follows from the Baire theorem: If $\langle \cdot, \cdot\rangle$ is an inner product on $\mathbb F$, then we can consider the sphere $\mathbb S$ on $\mathbb F$. We have $$\mathbb F_z = \cup_{a \in \mathbb S} \{x \in \mathbb F: ax = 0\}.$$

I think that we can reduce the above union to a countable union by replacing $\mathbb S$ with a countable dense subset of the sphere. If that was the case, $\mathbb F_z$ would be written as a countable union of closed sets of empty interior and by the Baire theorem it would have empty interior.

Hugo
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    Hint: An element $a$ of $\mathbb F$ can be viewed as a linear endomorphism of $\mathbb F$ (namely, left multiplication by $a$), and thus has a determinant. It is a unit if this determinant is nonzero, and is a zero-divisor if this determinant is zero. – darij grinberg May 13 '21 at 12:20
  • @darijgrinberg That is essentially the proof that $\mathbb F = \mathbb F_z \sqcup \mathbb F^\ast$. How does it imply $\mathbb F_z$ have empty interior? – Hugo May 13 '21 at 12:24
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    You have to show that any open neighborhood contains an element with nonzero determinant. Well, if $a \in \mathbb{F}$ is any element, what can you say about $\det\left(a+\epsilon\cdot 1\right)$ for small $\epsilon > 0$ ? – darij grinberg May 13 '21 at 12:34
  • The zero divisors (and zero itself) form a linear subspace, which obviously has empty interior – SVG May 13 '21 at 12:43
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    @Serge Zero divisors do not form a linear space. Take the product algebra $\mathbb R \times \mathbb R$. The zero divisors are $\mathbb R \times 0 \cup 0 \times \mathbb R$. – Hugo May 13 '21 at 12:46
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    @darijgrinberg That works! The real polynomial $f(x):= \det(a+x.1)$ is zero only for finite $x$'s. If $a$ is zero divisor, then $f(0)=0$. Therefore, for a small neighborhood $N$ of $0 \in \mathbb R$, we have $f(x) \neq 0$ for $x \in N\setminus{0}$. Hence, a+x.1 is unit for all $x \in N\setminus{0}$.

    Thank you very much!

     – Hugo May 13 '21 at 12:56
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    @Hugo C Botós. True, I misstated my comment. I meant a union of linear spaces. But it really is not entirely obvious that it's not a dense one. – SVG May 13 '21 at 12:57

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