Kähler differential over a field I have been working with Kähler differentials, and I have $\Omega^1_{B/k}$, where B is a commutative $k$-algebra, and $k$ is a field. I was wondering that for $d(b)=0$, does this imply that $b\in k$? I know that there are examples, however, like in $\Omega^1_{\mathbb{R}/\mathbb{Z}}$ we have that $d(\alpha)=0$ iff $\alpha$ is algebraic. But since I am over a field, I thought maybe this will force $b\in k$ in my case? Any hint or guidance would be most helpful!
 A: $da=0$ does not imply that $a\in k.$ Here is my example.
Let $B=k^{\{0,1\}},$ i.e. functions on $\{0,1\}$ with values in $k.$ Product in algebra $B$ is given by pointwise multiplication, i.e. $p:B\otimes B\to B$ is given by formula
$$p(f\otimes g)(x)=f(x)g(x), x=0,1.$$
Consider elements $a,b\in B$ such that $$a(0)=0, a(1)=1,b(0)=1,b(1)=-1$$
We see that $ab=-a,a^2=a$ and $b+2a=1.$
From construction of Kahler differential we know that $\Omega_{B/k}=I/I^2$ where $I=\ker{p}.$ Consider
$$b(1\otimes a-a\otimes 1)(1\otimes a-a\otimes 1)=b(1\otimes a^2-2a\otimes a+a^2\otimes 1)=b(1\otimes a-2a\otimes a+a\otimes 1)=b\otimes a+ 2a\otimes a-a\otimes 1=(b+2a)\otimes a - a\otimes 1=1\otimes a-a\otimes 1.$$
From construction of Kahler differential we also know that $da=[1\otimes a-a\otimes 1].$ We proved that $(1\otimes a-a\otimes 1)\in I^2,$ so $da=0.$ On the other hand we easily see that $a\notin k.$
$k\hookrightarrow B$ is obviously indentified as constant functions.

My idea is based on the fact from calculus which says, that functions with property $df=0$ are locally constant. Here I split domain into just two points 0 and 1.
A: If $B$ is a separable field extension of $k$ then in fact $\Omega^1_{B/k} = 0$: indeed, if $x \in B$ has minimal polynomial $f \in k[t]$, separable by assumption, then $f(x) = 0$ implies $f'(x)\,dx = 0$ and since $B$ is a field, $dx = 0$.
