Showing identity of ideals in $\mathbb{Q}[x,y]$ Problem: Let $R=\mathbb{Q}[x,y]$ and consider the ideal
  $$I=\{f\in R:f(0,0)=f(1,1)=0\}.$$ Prove that $I=\langle x-y,y-x^2\rangle.$


Since $x-y$ and $y-x^2$ belong to $I$ we have $\langle x-y,y-x^2\rangle\subset I.$
To show the other relation I don't know even how to start.
To start take any element $f$ of $I$ which is of the form $a_0+a_1x+a_2x^2+....+a_nx^n$ where $a_i\in \mathbb Q[y]$ and $f(0,0)=f(1,1)=0$

Again note that $a_i=b_0+b_1 y+b_2y^2+.....+b_ny^n$ where $b_i\in \mathbb Q$.

Apply the two conditions you will that $b_0=0$ from one of them .

See  if you can make use of the other one and make $f$ of the form $f=(x-y)g(x,y)+(y-x^2)h(x);g(x,y),h(x,y)\in \mathbb Q[x,y]$
To start take any element $f$ of $I$ which is of the form $a_0+a_1x+a_2x^2+....+a_nx^n$ where $a_i\in \mathbb Q[y]$ and $f(0,0)=f(1,1)=0$

Again note that $a_i=b_0+b_1 y+b_2y^2+.....+b_ny^n$ where $b_i\in \mathbb Q$.

Apply the two conditions you will that $b_0=0$ from one of them .

See  if you can make use of the other one and make $f$ of the form $f=(x-y)g(x,y)+(y-x^2)h(x);g(x,y),h(x,y)\in \mathbb Q[x,y]$