# Is it possible to prove the zero product property in a vector space the other way around.

EDIT: problem solved don't bother reading.

In a vector space $V = (V, F)$, the zero product property states that for all $\lambda \in F$ and $\underline{v} \in V$, if $\lambda \cdot \underline{v} = \underline{0}$ then $\lambda = 0$ or $\underline{v} = \underline{0}$.

The standard proof goes as follows: Assume $\lambda \neq 0$. Then since $F$ is a field $\lambda^{-1}$ exists. Then we have $\lambda \cdot \underline{v} = \underline{0} \implies \lambda^{-1} \cdot (\lambda \cdot \underline{v}) = \lambda^{-1} \cdot \underline{0} \Longrightarrow (\lambda^{-1}\lambda)\cdot \underline{v} = \underline{v} = \underline{0}$, giving the result.

The proposition is of the form $p \rightarrow (q \, \vee \, r)$ and the standard proof does it using the logical equivalence $p \rightarrow (q \, \vee \, r) \equiv (p \, \wedge \, \neg q) \rightarrow r$. My question is whether it is possible to do this proof the other round, i.e. using the equivalence $p \rightarrow (q \, \vee \, r) \equiv (p \, \wedge \, \neg r) \rightarrow q$, that is, assuming that $\lambda \cdot \underline{v} = \underline{0}$ and $\underline{v} \neq \underline{0}$ and concluding that $\lambda = 0$.

I want to do this because I am investigating weaker settings e.g. modules over rings of various types. The reason the standard proof works is because non-zero elements in a field have a special attribute: they have inverses. This leads me to believe that if it were possible to prove this the other way around, you would have to use some special attribute of non-zero vectors in a vector space. The only one I could think of is that the zero vector is the unique solution to $\underline{v} + \underline{v} = \underline{v}$, and hence non-zero vectors do not add with themselves to make themselves, however this property didn't seem to help.

A proof would be greatly appreciated, as would a proof that you can't prove the zero product property in this way.

EDIT: I've figured out that what I'm looking for isn't possible; I was hoping for a proof where the use of multiplicative inverses of scalars wasn't used, however such a proof cannot exist because if we don't allow ourselves inverses, i.e. if we work with a module over a commutative ring with identity $M = (M, R)$, the result is no longer true. Indeed as N8tron says if we take $M = \mathbb{Z}_4$ and also $R = \mathbb{Z}_{4}$ we have $2 \cdot \underline{2} = \underline{0}$ but $2 \neq 0$ and $\underline{2} \neq \underline{0}$. So in conclusion, proofs using the alternate logical equivalence probably exist, but they will all necessarily use the existence of multiplicative inverses of non-zero scalars, which is what I was hoping to avoid.

• It would really be helpful to know what kind of modules you're dealing with if $R= \Bbb Z / 4 \Bbb Z$ then take $R$ as an $R$-module then $2$ is a zero divisor both as a vector and a scalar. Jun 5, 2018 at 21:52
• Do your rings have zero divisors? Jun 5, 2018 at 22:22
• Note that even if the ring has no zero divisors, it's still possible to have torsion elements in modules, e.g. $\mathbb{Z} / p \mathbb{Z}$ as a $\mathbb{Z}$-module has $p \cdot (1 + p \mathbb{Z}) = 0$ although $p \ne 0$ in $\mathbb{Z}$ and $1 + p \mathbb{Z} \ne 0$ in the module. Jun 5, 2018 at 23:22