Question about repeated quotient groups

Question

I was wondering if there was a simpler representation for the quotient group $(K[x,y]/\langle xy\rangle)/\langle x, y-1\rangle$. Where $K[x,y]$ is the polynomial ring with variables $x,y$ over the field $K$. I was thinking that maybe it's the same as the quotient group $K[x,y]/\langle xy,x,y-1\rangle$, since mapping an element in $K[x,y]$ to the first quotient group will remove any multiples of $xy$, and the second quotient group mapping will map the new element to one without any multiples of the ideal $\langle x, y-1\rangle$.

Answer 1

Yes the quotient can be simplified as you wrote because of the third isomorphism theorem.

Moreover, $(xy,x,y-1)=(x,y-1)$ so we can write it as $K[x,y]/(x,y-1)$. By the first isomorphism theorem, this is essentially the image of the map $K[x,y]\to K$ which evaluates $x$ at the value $0$ and evaluates $y$ at the value $1$. IOW, the quotient is $\cong K$ itself.