When is a local algebra reduced? Let $k$ be a field and let $A$ be a local $k$-algebra which has finite dimension over $k$. Let $\mathfrak{m}$ be the maximal ideal of $A$ and let $k' = A / \mathfrak{m}$ be the residue field.
For every $a \in A$ we define $\mathrm{Tr}(a) \in k$ the trace of the $k$-linear endomorphism $A \to A$ given by the multiplication by $a$. The map $\mathrm{Tr} \colon A \to k$ is $k$-linear. Consider the symmetric $k$-bilinear form $\langle \cdot,\cdot \rangle \colon A \times A \to k$ given by $\langle a, b \rangle = \mathrm{Tr}(ab)$.
Now consider the following statements: 


*

*$A$ is reduced, or equivalently $\mathfrak{m} = 0$, i.e. $A = k'$;

*$\langle \cdot, \cdot \rangle$ is non-degenerate;

*$k' / k$ is a finite separable extension of fields.

*$\mathrm{Tr} \colon A \to k$ is surjective.


(Some assertions contained in this paragraph are wrong.) I know a proof of the facts "(2) implies (1)" and "(2) iff (3) iff (4)". (1) does not implies (2) because it suffices to pick $A$ as an inseparable finite field extension of $k$. Is there some fact which involves the field extension $k' / k$ and which is equivalent to (1)?
My question comes from studying the property of the discriminant and the ramification in finite extensions of Dedekind domains. Lemma 7.4.14 at page 289 of Qing Liu's Algebraic geometry and arithmetic curves concerns the problem above; but, in my opinion, the condition "every nilpotent element $\epsilon$ of $A$ verifies $\epsilon^m = 0$ for some integer $m$ prime to $\mathrm{char}(k)$" is empty. Am I wrong? It seems to me that the following proposition holds:
Proposition Let $n \geq 2$ be an integer and let $A$ be a noetherian ring with nilradical $\mathcal{N}$. There exists an integer $m$ prime to $n$ such that $\mathcal{N}^m = 0$.
EDIT. I hope to do true assertions, unlike my original post. (2) implies (1) because a nilpotent element of $A$ belongs to the radical of the form $\langle \cdot, \cdot \rangle$. If (1) holds, then (2), (3) and (4) are equivalent. (4) holds if an only if $\langle \cdot, \cdot \rangle$ is non zero. So it is obvious that (2) implies (4). Now consider the further statement:
5 . $k'/k$ is separable and every nilpotent element of $A$ has nilpotence order prime to $\mathrm{char} (k)$. 
(4) and (5) are equivalent by the lemma cited above.So (4) implies (3).
Now my questions are:


*

*Is there exists a proof of the lemma cited above that does not require scheme theory?

*Are there any other implications among the statements above?

*How the notions smooth and étale are involved with this matter?

 A: Yes, I also think that the condition you quote is empty and that your Proposition is true:
If $\mathcal N^r=0$ (and such an $r$ exists by noetherianity), any prime $p\geq r$ not dividing  $n$ will do .  
More generally, a finite dimensional algebra over a field is separable ( or equivalently: étale) if and only if it  isomorphic to the product of finitely many  separable (automatically finite dimensional) extension fields.
If the algebra is local, there can of course be only one factor in the product! 
Edit Since Andrea asks, yes there are proofs not involving scheme theory. You can look one up in Bourbaki's Algebra, Chapter 5, §8, Proposition 1. 
Beware that for Bourbaki "extension" is used only for algebras that are fields.
You will see that the surjectivity of the trace on $A$, your point 4, implies that $A$ is separable (=étale) if $A$ is a field, but not otherwise. This explains how Hagen's counterexample is possible. 
A: I am confused: what do you want to show? 
After all tame ramification must be allowed in Liu's lemma. See the following example: Take a field $k$ with $\mathrm{char}(k)\neq 2$ and consider the extension $k[[t]]\subset k[[\sqrt{t}]]$ of discrete valuation rings. Let $A:=k[[\sqrt{t}]]/tk[[\sqrt{t}]]$, then a basis of $A$ as a $k$-vector space is given by the residues $b_1:=1+tk[[\sqrt{t}]]$ and $b_2:=\sqrt{t}+tk[[\sqrt{t}]]$.
The coordinate matrix of the left multiplication by the element $a_1b_1+a_2b_2$, $a_1,a_2\in k$ looks like this:
$
A=\left(\begin{array}{cc}a_1 & 0\\
a_2 & a_1\end{array}\right)
$
Hence $\mathrm{Tr}(a_1b_1+a_2b_2)=2a_1$ thus showing surjectivity. So reducedness of $A$ is not necessary for the equivalence of (3) and (4).
A: First, any nilpotent element of $A$ has trace $0$ over $k$. So as you noticed, (2) implies that $A$ is a field. Then the classical theory (see Georges's answer) implies that $A$ is a finite separable field extension of $k$. Of course, the converse is also true. 
Now we consider Condition (4). As $A$ is Artinian, the separable closure $k''$ of $k$ in $k'$ can be lifted in $A$ and $A$ becomes a $k''$-algebra. Moreover 
$$\mathrm{Tr}_{k''/k}(\mathrm{Tr}_{A/k''}(A))=\mathrm{Tr}_{A/k}(A),$$ 
and $\mathrm{Tr}_{k''/k}(k'')=k$. So (4) is equivalent to $\mathrm{Tr}_{A/k''}(A)=k''$.
Note that $\mathrm{Tr}_{A/k''}(1)=\dim_{k''} A$. So (4) holds if $\mathrm{char}(k)=0$ or is positive and prime to $\dim_{k''} A$. These conditions imply $k''=k'$ because $\dim_{k''} A$ is divisible by $[k':k'']$ (use the chain of the quotients $\mathfrak m^r/\mathfrak m^{r+1}$ which are $k'$-vector spaces). So the remaining case is when $\mathrm{char}(k)=p>0$ and $p$ divides $\dim_{k''} A$. Then let me prove that $\mathrm{Tr}_{A/k''}(A)=0$. We already have $\mathrm{Tr}_{A/k''}(k'')=0$ (because $p$ divides $\dim_{k''} A$) and $\mathrm{Tr}_{A/k''}(\mathfrak m)=0$. Let $a\in A$. As $k'/k''$ is purely inseparable, there exists $r\ge 1$ such that $a^{p^r}\in k'' + \mathfrak m$.
Then 
$$(\mathrm{Tr}_{A/k''}(a))^{p^r}=\mathrm{Tr}_{A/k''}(a^{p^r})=0.$$ 
So the trace of $a$ is zero. 
Conclusion: (4) is equivalent to 

$k'/k$ is separable and $\dim_k A/[k':k]$ is prime to $\mathrm{char}(k)$. 

In particular, the cited Lemma is correct only when $A$ is generated by one element (which is enough for the purpose of Proposition 7.4.13). 
Hope you agree with the above proof.
