A-basis of $\Omega_A^1$ In the lecture notes of a course in Algebraic Geometry that I am following they are calculating a basis of $\Omega_A^1$ for $A=k[x_1,...,x_n]/(f_1,...,f_r)$. We get 
\begin{align*}\Omega^1_A = (\bigoplus_{i=1}^n A\cdot dx_i) /(A\cdot df_1 +···+A\cdot df_r)\end{align*} 
where  $df_i = \sum (\delta  f_i/\delta x_j)dx_j$. 
My question is why this is the case. And how do we proceed to get the basis? 
We defined $\Omega^1_A$ as an A-module that together with a  k-derivation $d: A\to \Omega^1_A$, has the universal property: For any A-module M and any derivation D: A → M there exists a unique A-linear map $\phi$ such that $\phi \circ d= D$
 A: Edit: I wrote everything for $\mathbb{Z}$ derivations, for some reason. You can make the appropriate changes.
I think you are asking why the presentation you gave for a module $\Omega_A^1$ (along with the derivation from $d: A \to \Omega_A$ that formally differentiates a polynomial) satisfies the given universal property. So I will attempt to answer that question.
First, you need to check that $d$ is a derivation. This is straightforward.
Next, given any other derivation $d' : A \to M$, where $M$ is an $A$ module, we need to produce this unique $A$-linear $\phi : \Omega_A \to M$. To solve these kind of problems (showing that something has a universal property), you need to figure out what features about the universal map $\phi$ are forced, then use them to define $\phi$, then check that the result is well-defined and satisfies the properties you desire.
We know that $\phi \circ d = d'$, hence $\phi$ must send $d(x_i)$ to $d'(x_i)$. We can construct $\phi$ like this on the free module $\oplus A dx_i$, by extending $A$-linearly, and to check that it factors through your construction, we need to check that each of the $df_i$ is killed by $\phi$. 
But as $f_i = 0$ in $A$, we have $d'(f_i) = 0$ (from $\mathbb{Z}$-linearity of the derivation, for example). Now, $df_i = \Sigma \frac{\partial f_i}{\partial x_j} dx_j$, so $\phi$ sends $df_i$ to $\Sigma \frac{\partial f_i}{\partial x_j} d'x_j$ (since $\phi$ is $A$-linear, and we know how it behaves on the generating set $dx_j$). So we only need to check that $d'(f_i) = \Sigma \frac{\partial f_i}{\partial x_j} d'x_j$. I claim that this is true for any derivation, namely we have the following lemma:
If $D: A \to M$ is a derivation, then the following hold:


*

*For $a \in A$, $D(a^n) = n a^{n-1} Dx$. (The "power rule".)

*If $a_i \in A$, then $D(a_1 \ldots a_n) = \Sigma a_1 \ldots \hat{a_i} \ldots a_n  D( a_i )$.

*$D(a_1^{m_1} \ldots a_n^{m_n}) = \Sigma m_i (a_1^{m_1} \ldots a_n^{m_n} / a_i) D(a_i)$. (All $m_i \geq 1$.)

*If $f$ is a polynomial expression in $a_i$ (I mean $f = \Sigma a_1^{i_1} \ldots a_n^{i_n}$, some element of the ring $\mathbb{Z}[a_1, \ldots, a_n]$), then $D(f) = \Sigma \frac{\partial f}{\partial a_j} Da_j$, where here $\partial/ \partial{a_j}$ means a formal derivative in the polynomial ring $\mathbb{Z}[a_1, \ldots, a_n]$. (The chain rule.)


(For proofs, use linearity, the Leibnitz rule, induction, etc. It should agree with your calculus intuition.)
Therefore, from  $d'(f_i) = 0$ it follows that $\phi (d(f_i)$. Hence $\phi$ factors (uniquely, from the universal property of quotients) through the module $\oplus A dx_i / (df_j)$. It follows from general nonsense about universal property any other module with this universal property is isomorphic to $\Omega_A$ via a unique-isomorphism (in the category of derivations $D: A \to N$).
You can generalize this all to $R$-linear derivations, where $R$ is some base ring over which $A$ is given the structure of an $A$-algebra. This is done by replacing $\mathbb{Z}$ with $R$ throughout, and insisting that all derivations have $dr = 0$, for all $r \in R$ (which is equivalent to them being $R$ linear, as you can check with the product rule). This is worthwhile, because it captures the very useful notion of a "relative differential."
