Proving the Conjecture: Fields and Nilpotent Elements

[Bashyboy]

Homework Statement

Through my exploration/experimentation, I came up with this little conjecture: Let $F$ be a field and $x$ some element in the field. If there exists a natural number $m$ such that $x^m = 0$, then $x=0$. In other words, a field contains no nontrivial nilpotent elements.

Homework Equations

The Attempt at a Solution

Suppose that such a natural number $m$ exists for which $x^m = 0$, or $x \cdot x^{m-1}$. Since fields do not contain zero divisors, either $x=0$ or $x^{m-1}$. If the latter holds, we are finished, so assume $x^{m-1} = 0$ or $x \cdot x^{m-1} =0$. This implies $x^{m-2} = 0$, etc.

It seems that I could repeat this process until I obtain $x=0$, suggesting that the conjecture is true (unless I made some elementary mistake, which is not unlikely!). However, is there anyway of making this process/argument more rigorous?

 

[PeroK]

Homework Statement

Through my exploration/experimentation, I came up with this little conjecture: Let $F$ be a field and $x$ some element in the field. If there exists a natural number $m$ such that $x^m = 0$, then $x=0$. In other words, a field contains no nontrivial nilpotent elements.

Homework Equations

The Attempt at a Solution

Suppose that such a natural number $m$ exists for which $x^m = 0$, or $x \cdot x^{m-1}$. Since fields do not contain zero divisors, either $x=0$ or $x^{m-1}$. If the latter holds, we are finished, so assume $x^{m-1} = 0$ or $x \cdot x^{m-1} =0$. This implies $x^{m-2} = 0$, etc.

It seems that I could repeat this process until I obtain $x=0$, suggesting that the conjecture is true (unless I made some elementary mistake, which is not unlikely!). However, is there anyway of making this process/argument more rigorous?

It's a bit simpler to consider the inverse of any non-zero element.

 

[Bashyboy]

So, are you saying I do the following: Suppose there exists a nonzero element $x \in F$ such that $x^m = 0$ for some $m \in \mathbb{N}$. Since $F$ has multiplicative inverses, $x^{-m}$ exists and therefore $x^{-m} x^m = x^{-m} 0$ or $1 = 0$, which is a contradiction. Hence, $x = 0$.

Does that sound right?

 

[fresh_42]

If $x^m=0$ then you can't invert it. But you can invert each $x$ $m-$times under the assumption it isn't zero.

 

[Bashyboy]

But you can invert each $x$ $m$-times under the assumption it isn't zero.

Do you mean something like $x^m = 0$ is $\underbrace{xx...x}_{|m|-times} = 0$ and multiplying the equation by $x^{-1}$ $|m|$-times gives us $(\underbrace{x^{-1}x^{-1}...x^{-1}}_{|m|-times} (\underbrace{xx...x}_{|m|-times}) = 0$ or $1 = 0$, the desired contradiction.

 

[fresh_42]

Do you mean something like $x^m = 0$ is $\underbrace{xx...x}_{|m|-times} = 0$ and multiplying the equation by $x^{-1}$ $|m|$-times gives us $(\underbrace{x^{-1}x^{-1}...x^{-1}}_{|m|-times} (\underbrace{xx...x}_{|m|-times}) = 0$ or $1 = 0$, the desired contradiction.

Yes, because the assumption on $x$ (+ associativity) allows you to do that while $x^m=0$ doesn't.
However, your original proof by induction on $m$ has been valid, too. Except you had some wording issues in the middle:

Since fields do not contain zero divisors, either $x=0$ or $x^{m-1}$. If the latter holds, we are finished, so ...

It should have been:

Since fields do not contain zero divisors, either $x=0$ or $x^{m-1}=0$. If the former holds, we are finished, so ...

and now you could have used your induction hypothesis, that $x^{m-1}=0$ already implies $x=0$. (The induction base $x^1=0$ is trivial.)

Edit: Induction is the mathematical formalism for "and so on".

 

[PeroK]

Alternatively, if $x^m = 0$ then we can take $m$ to be the lowest natural number for which this holds. You can then derive a contradiction from applying $x^{-1}$ once.

 

[Bashyboy]

Alternatively, if $x^m = 0$ then we can take $m$ to be the lowest natural number for which this holds. You can then derive a contradiction from applying $x^{-1}$ once.

Oh, yes! The well-ordering property, right? Very clever! So, I just need to prove that if $x \neq 0$, then $x^{-1} \neq 0$. Here is my proof. Suppose that $x \neq 0$ yet $x^{-1} = 0$. Then $xx^{-1} = 1$ would imply $x \cdot 0 = 1$ or $0 = 1$, a contradiction. Does this seem right?

 

[PeroK]

Oh, yes! The well-ordering property, right? Very clever! So, I just need to prove that if $x \neq 0$, then $x^{-1} \neq 0$. Here is my proof. Suppose that $x \neq 0$ yet $x^{-1} = 0$. Then $xx^{-1} = 1$ would imply $x \cdot 0 = 1$ or $0 = 1$, a contradiction. Does this seem right?

If $x^{-1} = 0$ then $xx^{-1} = 0$ which is absurd. Also, it would mean that $0$ is invertible, as $0^{-1} = x$.

PS there's no reason to consider $x^{-1} = 0$ in any case.