The image of $f$ in $A$ in its resdue field

Question

Let $A = k[x,y]/(xy,y^2)$, where $k$ is a field, and take $x+1+I \in A$. I know that $Q = (x+2+I, y+I)$ is a prime (maximal) ideal in $A$. Could someone please show me exactly how the image of $x+1+I$ in the residue field at $Q$ is $0$? (This is related to a question I asked earlier.) Thank you!

Answer 1

While the residue field is defined as the quotient $A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}$, it turns out to be isomorphic to the field of fractions of $A/\mathfrak p$ (try proving this yourself! Use the natural map $A \to A_{\mathfrak p}$). This makes it easy to see that something is zero in the residue field if and only if it's contained in the prime ideal $\mathfrak p$ (it's also not hard to prove this directly from the $A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}$ definition of the residue field).

So $x + 1 \in A$ doesn't go to zero in the residue field at $Q$ because it's not contained in $Q$. It should go to $-1$ in the residue field.