Why is $(XY-1)$ contained in $(X-a, Y-b)$ with $ab=1$?

Question

This is probably a very trivial question, so I apologize in advance.

Let $K$ be an algebraically closed field and $R=K[X,Y]$ the polynomial ring in two variables. I want to show that every ideal $(X-a,Y-b)$ where $a,b\in K$ with $ab=1$ contains the ideal $(XY-1)$.

Fix $a,b\in K$ with $ab=1$. I must show $XY-1\in (X-a,Y-b)$. Multiplying $X-a$ and $Y-b$ gives $XY+1-(aY+bX)$ but I don't come any further.

Answer 1

Just keep going with using $X-a$ and $Y-b$ to translate stuff.

$(X-a)(Y-b)+b(X-a)+a(Y-b)=XY-aY-bX+1+bX-1+aY-1=XY-1$

Answer 2

${\rm mod}\ x\!-\!a,\,y\!-\!b\!:\,\ x\equiv a,\,y\equiv b\,\Rightarrow\, xy-1\equiv ab-1\equiv0 $

Answer 3

Maybe you want to know more, namely: $$f(X,Y)\in(X-a,Y-b) \text{ iff } f(a,b)=0.$$

This solves instantly your question: $f(a,b)=0$ for $f(X,Y)=XY-1$ means $ab=1$.