Let $k$ be a field and let $A$ be a commutative algebra over $k$. I want to calculate the exterior algebra $\Lambda_A^\bullet A$. We have $\Lambda_A^0 A = \Lambda_A^1 A= A$, and $\Lambda_A^k A = 0$ for $k > 0$. This is because $A$ is a free $A$-module so all wedges $x \wedge y$ for $x,y \in A$ disappear.
Thus $\Lambda_A^\bullet A \cong A \oplus A$ as $A$-modules, and the $A$-algebra multiplication is given by $$(a,b)\cdot (c,d) = (ac, ad + bc).$$ The unit is $(1,0)$.
However in general this does not seem to satisfy the universal property. The universal property says that for any linear map $f \colon A \to X$ where $X$ is an $A$-algebra such that $f(a)^2 = 0$ for all $ a \in A$, there should be a unique $A$-algebra homomorphism $\phi \colon A \oplus A \longrightarrow X$ such that $\phi (0, b) = f(b)$. Since this thing has to be unital there is only one possible thing it could be, the map $\phi(a,b) = a + f(b)$.
The problem comes in checking this is an algebra map. We have $$ \phi ((a,b)\cdot(c,d)) = \phi (ac, ad + bc) = ac + f(ad + bc) $$ and $$ \phi (a,b) \phi(c,d) = (a + f(b))(c + f(d)) = ac + a f(d) + f(b)c + f(b)f(d) $$ But the two expressions are equal if and only if $f(b)f(d) = 0$. However I can't see why this is true in general unless $2$ is invertible and $X$ is commutative.
What is my error?
We have $f(1)^2=0$, hence by $A$-linearity $f(b) f(d) = bd f(1)^2 = 0$.
By the way, the exterior algebra of $A$ over $A$ is $A[x]/(x^2)$, the algebra of dual numbers, with $x$ in degree $1$.