The Kahler differential is zero While studying for my commutative algebra exam, I have come across this problem.

Let $K$ be a field. Let $A$ be a $K-$algebra, finitely generated as a $K$-vector space. Prove that $\Omega^1_{A/K} = 0$ if and only if $A\simeq K_1\times\dots\times K_n$ where each extension $K\leq K_i$ is finite and separable.

I have managed to prove one implication but I do not really have ideas for the other.
"$\Rightarrow$" Assume  $A\simeq K_1\times\dots\times K_n$ where each extension $K\leq K_i$ is finite and separable. Then, from the primitive element theorem, there exist $\alpha_i\in K_i$ such that $K_i=K[\alpha_i]$. Let $P_i\in K[X], P_i=\sum_{i=1}^{n_i}a_i^jX^j$ be the minimal polynomials of $\alpha_i$ respectively. Let $d:K_1\times\dots\times K_n\rightarrow M$ be a derivation. 
$d(P_i(\alpha))=d(\sum_{i=1}^{n_i}a_i^j\alpha^j)=\sum_{i=1}^{n_i}a_i^jd(\alpha^j)=\sum_{i=1}^{n_i}(a_i^jj\alpha^{j-1}d(\alpha_i))=d(\alpha_i)\sum_{i=1}^{n_i}(a_i^jj\alpha^{j-1})$. Since the extensions are separable, the coefficient of $d(\alpha_i)$ (which is $P'_i(\alpha)$) is different from $0$, so we get $d(\alpha_i)=0,\ \forall i=\overline{1,n}$, which makes any derivation to be the the zero map, thus $\Omega_{A/K}=0$.
 For the other part, since $A$ is finitely generated as a vector space, it is a $K-$vector space with a finite base: $\{x_1,\dots,x_n\}$ and a bilinear product.
 A: I am showing only the non trivial implication, you showed the other.
1) First suppose that $K$ is algebraically closed.
1)a) Let $\mathfrak{m}$ a maximal ideal of $A$. $B = A / \mathfrak{m}$ is a field ($\mathfrak{m}$ is maximal) and is a finite extension of $K$ as $A$ is a finite dimension over $K$. As $K$ is algebraically close, $B$ is of dimension $1$ over $K$ i.e. $B = K$. This means that for every $a\in A$ there exists a unique $\lambda\in K$ such that $a - \lambda \in \mathfrak{m}$. Note $D(a)$ the class modulo $\mathfrak{m}^2$ of that $a - \lambda \in \mathfrak{m}$. We see that the map $a\mapsto D(a)$ is an $K$ derivation of $A$ into the $A$-module $M = \mathfrak{m} / \mathfrak{m}^2$. The universal property of $\Omega^1_{A/K}$ and the fact that $\Omega^1_{A/K} = 0$ imply that $D = 0$, so that $M = \mathfrak{m} / \mathfrak{m}^2 = 0$, that is, $\mathfrak{m} = \mathfrak{m}^2$ in $A$. Now as $A$ is of finite dimension over $K$, the ideal $\mathfrak{m}$ is finitely generated, say by elements $m_1,\ldots,m_d$. As $\mathfrak{m} = \mathfrak{m}^2$ you can write each $m_i$ as $m_i = \sum_{j=1}^d \lambda_{ij} m_j$. Let $Y \in\mathbf{M}_d(A)$ such that $Y_{ij} = \delta_{ij} - x_{ij}$ for each $i,j$. (That is, the matrix $Y$ is $Y = I_d - X$ where $X$ is the matrix such that $X_{ij} = x_{ij}$ for each $i,j$.) Standard linear algebra fact, you can fin a matrix $Z \in\mathbf{M}_d(A)$ such that $Z Y = \textrm{det}(Y) I_d$. This implies that $\textrm{det}(Y) m_i = 0$ for each $i$, so that $D\mathfrak{m} = 0$. But the matrix $Y$ is equal to $I_d$ modulo $\mathfrak{m}$ by construction, so that if we set $e = 1 - D$, we have $e\in\mathfrak{m}$ and $e m = m$ for each $m\in\mathfrak{m}$. Therefore $e^2 = 1$ and $\mathfrak{m} = A e$. But then $A = (1-e)A \oplus Ae$, the second factor being by what preceeds equal to $\mathfrak{m}$. As $\mathfrak{m}$ is of dimension $1$ over $K$, the previous direct sum decomposition shows $A \simeq K\times A/\mathfrak{m}$ as $K$-algebras. As $\Omega^1_{A/\mathfrak{m}}$ is isomorphic to a quotient of $\Omega^1_{A/K}$ which is zero, $\Omega^1_{A/\mathfrak{m}} = 0$.
1)b) Let's do a pause here, and think about what have we shown up to here. We have show that whenever $A$ is of finite dimension $n$ over $K$ with $\Omega^1_{A/K} = 0$, one has an isomorphism of $K$-algebras $A \simeq K \times C$ where $C$ is a $K$-algebra of dimension $n-1$ over $K$ such that $\Omega^1_{C/K} = 0$. This allows to show by induction that under the hypothesis 


*

*$K$ is algebraically closed

*$A$ is a $K$-algebra of finite dimension $n$ over $K$

*$\Omega^1_{A/K} = 0$


then one has an isomorphism of $K$-algebras $A \simeq \underbrace{K\times \ldots \times K}_{\textrm{$n$ times}}$
2) General case, $K$ is not algebraically closed anymore. Let $L$ be an extension of $K$, and let $E$ be an algebraic closed field containing $L$. Then $A_{(L)} := A \otimes_K L$ is isomorphic to a subring of $A_{(E)} := A \otimes_K E$. (Simple algebra.) Then $A_{(L)} := A \otimes_K L$ is an $L$-algebra of finite dimension over $L$. Now, $\Omega^1_{A_{(E)}/E} \simeq \Omega^1_{A/K} \otimes_A A_{(E)}$ (check that the latter verifies $\Omega^1_{A_{(E)}/E}$'s universal property !) is $0$ as $\Omega^1_{A/K} = 0$. This shows that $\Omega^1_{A_{(E)}/E} = 0$ and the case one shows that $A_{(E)} := A \otimes_K E \simeq \underbrace{E\times \ldots \times E}_{\textrm{$n'$ times}}$, showing that $A_{(E)}$ is a reduced ring (a ring in which there's no non-zero nilpotent element). As $A_{(L)} := A \otimes_K L$ is isomorphic to a subring of $A_{(E)}$, $A_{(L)}$ is also a reduced ring.
Lemma. Let $A$ a (commutative) finite dimensional $L$-algebra. Then $A$ is reduced if and only if we can find finite extensions $L_i / L$ such that $A \simeq L_1 \times \ldots \times L_k$ as $L$ algebras.
Proof of lemma. If we have such an isomorphism, $A$ is trivially reduced. Conversely, if $A$ is reduced, we can suppose that $A$ is not a field (because a field is trivially reduced.) Then, it suffices to show that one can find two $L$-algebras $A_1,A_2$ such that $A \simeq A_1 \times A_2$ as $L$-algebras. Because then the $A_i$ will be reduced and we could use induction hypothesis on them to show that $A$ is reduced. Among non-zero ideals of $A$ that are different from $A$, choose an ideal $I$ such that $I$'s dimension over $K$ is minimal. For any $x\not=0$ in $I$ we have $x^2\not=0$ as $A$ is not reduced. Therefore $I^2 \not = 0$. But $I^2 \subseteq I$ and the minimality of $I$ implies that $I^2 = I$. As we did in 1)a) we can find $e\in A$ such that $e\not= 0,1$ such that $e^2 = 1$, and as in 1)b), conclude, that there is an isomophism $A \simeq A_1 \times A_2$ as $L$-algebras, proving the lemma.
Now this lemma implies that $A_{(L)}\simeq L_1 \times \ldots \times L_k$ as $L$ algebras where $L_i / L$ are finite extension. Now you can easily work out that as in beginning of (2) that we have $\Omega^1_{A_{(L)}/L} = 0$. And then you can easily work out that $\Omega^1_{L_i/L} = 0$ for each $i$. Then I leave you showing that $L_i / L$ are separable with "differential" argument. (Easy exercise.)
Finally, taking $L = K$ gives you what you want to show. (I proved something stronger.)
