I'm wondering how can i make sense of a partial solution to a two-system of polynomial equations. Let $\{f,g\}\subseteq K[x,y]$ be polynomials in a polynomial ring over the field $K$. Given the equations: $$f(x,y)=a$$ $$g(x,y)=b$$ Every solution is contained in the affine variety $\mathcal{V}(I_0)≠\emptyset$ of the ideal $$I_0=\langle f(x,y)-a,g(x,y)-b\rangle$$ Suppose i have a function $t\in K[x,y]$ such that: $$I_1=\langle f(t,y)-a,g(t,y)-b\rangle=$$ $$\langle a-a,g(t,y)-b\rangle=\langle g(t,y)-b\rangle$$ Is there a relationship between $I_0$ and $I_1$?
Given these conditions, i thought about the possibility of the ideals satisfying: $I_1\subseteq I_0$, as in the following example: let $$I_0=\langle x+y-a,xy-b\rangle$$ And let $t=a-y$, then $$I_1=\langle ay-y^2-b\rangle$$ We can show that $I_1\subseteq I_0$ by showing that $r\in I_0$ satisfies: $$r=-y(x+y-a)+xy-b=$$ $$ay-y^2-b$$ So $$I_1=\langle r\rangle\subseteq I_0$$ But is this generally the case? How can i prove it? Maybe the ideals satisfy some other relation (i.e not inclussion)?
Actually, i found something interesting... We can extend the polynomial ideal $I_0\subseteq K[x,y]$ to the field of algebraic functions with degree one (in two variables) by noting: $$I_0\subseteq A[x,y]=\left\{x\in K[x,y]^2|x=\frac{n(x,y)}{d(x,y)};n\in K[x,y],d\in K[x,y]/\{0\}\right\}$$ By performing this extension, we can define a function $\eta\in A[x,y]$ as the solution to the equation: $$\eta(f(x,y)-a)+g(x,y)-b=g(t,y)-b$$ Which then shows: $$\eta=\frac{g(t,y)-g(x,y)}{f(x,y)-a}$$ (remember, $t\in K[x,y]$) this is utterly unexpected for me, because this shows that the ideal inclussion $I_1\subseteq I_0$ depends explicitly in the base set of the ideals, and we have the implication statement: $$I_0\subseteq A[x,y]\Rightarrow I_1\subseteq I_0$$ But the same statement for ideals in the ring $K[x,y]$ might not be true (i.e when $\eta\notin K[x,y]$)