I have recently been introduced to Kähler differentials. Our lecturer gave a sketch of the proof of how to construct such object. He said that if $A$ is a $k$-algebra, then we can construct the following free module
$$\Omega_A^1:=\bigoplus_{a\in A}A\cdot da / \text{rel}$$
where $d$ is a symbol and rel is the following set of relations:
- $d(a+b)-(da+db)$
- $d(\lambda a)=\lambda da$
- $d(ab)-(adb+dba)$
where $a,b\in A$ and $\lambda \in k$. He also later on said that defined $d:A\longrightarrow \Omega_A^1$ as $a\longmapsto [da]$. He said that checking the universal property is easy. I am struggling with this, however. I have looked on the Stacks Project and what they do makes sense but since their notation is so different from mine, I am not able to conclude my argument. Also I am not 100% sure that I understand their notation. Can someone help me with this?
Edit: What I want, precisely, is a proof of the universal property based on my definition.
Best Answer
For notational clarity, I will write $\partial\colon A\to \Omega^1_A$ for the map $a\mapsto [da]$, and I will reserve $da$ for the (representative of the) class in $\Omega^1_A$. We want to show that, for any $A$-module $M$, there is a natural bijection $$\mathrm{Hom}_A(\Omega^1_A, M)\xrightarrow{\cong}\mathrm{Der}_k(A,M), f\mapsto f\circ\partial.$$ I want to point out that $\Omega^1_A$ is not, as you wrote, a free module, but a quotient of a free module. From the universal property of the quotient and from the universal property of free modules, we see that an $A$-module map $f\colon \Omega^1_A\to M$ is the same data as a choice, for each $a\in A$, of an element $f(da)\in M$ such that the relations specified in the quotient hold, i.e. such that the following hold:
This means that the map $f\circ\partial\colon A\to M$ is a $k$-derivation, since a $k$-derivation is a $k$-linear map (the first two bullets) that satisfies the Leibniz rule (third bullet), so our map $f\mapsto f\circ\partial$ is well-defined.
Given a $k$-derivation $\delta\colon A\to M$, we define $f\colon\Omega^1_A\to M$ by $f(da):=\delta(a)$ (and extend this $A$-linearly to a map from all of $\Omega^1_A$). If this is well-defined, it is an actual $A$-linear map, and then it is immediate that $f\circ\partial=\delta$. Moreover, if $\delta=g\circ\partial$ for some $A$-linear map $g\colon \Omega^1_A\to M$, then $f(da)=g(da)$ for all $a\in A$, so that $f=g$. Therefore, the map $$\mathrm{Hom}_A(\Omega^1_A, M)\to\mathrm{Der}_k(A,M), f\mapsto f\circ\partial$$ would be a bijection. This means that the only thing we still have to show is that the map $f\colon\Omega^1_A\to M$ given by $A$-linear extension of the assignment $f(da):=\delta(a)$ is well-defined. For this, we need to show that this definition of $f$ satisfies the relations in the definition of $\Omega^1_A$, i.e. we need to check the three bullets above.
Therefore, $f$ is well-defined, and with this we win.
Edit: I will give some philosophical justification for calling elements of $\Omega^1_A$ $1$-forms. To be more precise, you can consider them to be $1$-forms on the affine scheme $\mathrm{Spec}(A)$. Given that you are not familiar with scheme theory, I will quickly say what $\mathrm{Spec}(A)$ is supposed to be. In algebraic geometry, we want to think about the ring $A$ as the ring of (globally defined, algebraic) functions on a geometric object $\mathrm{Spec}(A)$ for which it is ''most natural'' to think of its functions as elements of $A$. For instance, $\mathrm{Spec}(\mathbb{R}[x])$ is the real line (actually, there are some extra generic points, but that's not important now), the natural space to consider real polynomial functions in one variable on, and $\mathrm{Spec}(\mathbb{R}[x,y])$ is the real plane (plus some generic points), the natural space to consider two-variable real polynomials on. And a space like $\mathrm{Spec}(\mathbb{R}[x,y]/(y-x^2))$ will be the parabola $y=x^2$ in this plane, as all functions in the ideal $(y-x^2)$ vanish there. So, for every commutative ring $A$ we have a geometric object $\mathrm{Spec}(A)$ called the spectrum of $A$ and we can think about elements of $A$ as being globally defined ''functions'' on $\mathrm{Spec}(A)$.
A general scheme is glued together from these spectra of commutative rings in an analogous way as a smooth real $n$-dimensional manifold is glued from copies of $\mathbb{R}^n$. As a slogan, schemes are like complex manifolds, but while zooming in around a single point on a complex manifold you see complex analysis, if you zoom in on a point of a scheme you will see commutative algebra. We try to give analogues of concepts in differential geometry in order to study algebra via geometric methods. (For our discussion here, you can also think about schemes as being similar to real manifolds, but for actual scheme theory it is more precise to have complex manifolds and holomorphic functions in your mind.)
For the question what $1$-forms are on $\mathrm{Spec}(A)$, let us therefore first look at the analogous situation of smooth manifolds. Here, the analogue of spectra of rings are the smooth manifolds $\mathbb{R}^n$ for some $n\geq 0$. Given such an $n$, let $A=C^\infty(\mathbb{R}^n)$ be the ring of smooth real-valued functions on $\mathbb{R}^n$, which is an $\mathbb{R}$-algebra. It holds that the $A$-module $\Omega^1(\mathbb{R}^n)$ of smooth (globally defined) $1$-forms on $\mathbb{R}^n$ satisfies the universal property of $\Omega^1_A$. Namely, using the global coordinates $(x^1,\ldots,x^n)$ on $\mathbb{R}^n$, the $1$-form $\sum_{i=1}^n f_i dx^i$ in $\Omega^1(\mathbb{R}^n)$ is associated to the element $\sum_{i=1}^n f_i dx^i$ in $\Omega^1_A$. This is a well-defined $A$-linear map because the $dx^i$ form an $A$-basis over $\Omega^1(\mathbb{R}^n)$, and after some checking you see that it is bijective. Therefore the Kähler module describes the correct $1$-forms in the case of the basic building blocks of manifolds.
When going to arbitrary rings and spectra of these rings, there is generally no ''analytic'' way to define $1$-forms, but the beauty of the universal property above is that it is a purely algebraic statement. This makes it a natural approach to define $1$-forms in algebraic geometry.
Of course, you have already some intuitive idea how $1$-forms on, say, the plane $\mathrm{Spec}(\mathbb{R}[x,y])$ should behave. Namely, we want all $1$-forms to be of the form $f\cdot dx+g\cdot dy$ for some $f,g\in\mathbb{R}[x,y]$. Moreover, every $f\in\mathbb{R}[x,y]$ should give via its differential $df$ a $1$-form, given by $df=\frac{\partial f}{\partial x} \cdot dx+\frac{\partial f}{\partial y}\cdot dy$. This differential is $\mathbb{R}$-linear and satisfies the Leibniz rule (and this already means that $dr=0$ for any $r\in\mathbb{R}$). Moreover, we know that these rules enable us to compute all the differentials $df$ for polynomials $f$ in terms of $dx$ and $dy$. This means that these rules are the only ones we need for the differential in order to get the ''expected result''. So the only things we need to axiomatize about the differential is that it is linear over our base field (if you are working over a field), and that is satisfies a Leibniz rule. The module $\Omega^1(\mathrm{Spec}(A))$ of $1$-forms on $\mathrm{Spec}(A)$ is therefore in a way the simplest module that has a differential map $d\colon A\to \Omega^1(\mathrm{Spec}(A))$ in the sense above. But the Kähler module is this simplest way to build such a differential, as this is exactly what the universal property tells you. Therefore, we may take $\Omega^1(\mathrm{Spec}(A))=\Omega^1_A$.
(You can generalize the universal property of $\Omega^1_A$ to allow for $A$ to be a sheaf of rings, thereby giving a coordinate-free way to define $1$-forms on arbitrary schemes (and manifolds).)