The following question is from Appendix C of Matsumura's Commutative ring Theory, page $283$.
Given an Ideal $I \subset A$ of a ring $A$, the author claims that it might happens that the exterior product $\wedge^2 A=0$, but still $\wedge^2 I \neq 0$. The author given the following example:
Let $k$ be a field and $A=k[x,y]$ be the polynomial ring and $I=xA+yA$ be the ideal. Then he claimed that $\wedge^2 I \neq 0$ by defining a map $\varphi: I \times I \to k=A/I$ by $$\varphi(f,g)=[\partial (f,g)/\partial (x,y)]_{(x,y)=(0,0)}$$ with $\varphi(x,y)=1$ so that $\varphi \neq 0$ and so $\wedge^2I \neq 0$.
But I didn't understand at some places as follows:
At first $\varphi (f,g)$ is the jacobian matrix and so $\varphi(x,y)=$identiry matrix. But it is given $1$.
Also how do we show $\wedge^2 I \neq 0$ but $\wedge^2 A=0$ ?
Any explanation please