Determining if two ideals in $\mathbb{Q}[X,Y]$ are equal

Solution 1:

Consider $f\colon \Bbb Q[X,Y]\to \Bbb Q[T]$ given by $X\mapsto T$, $Y\mapsto 2T-2$. Observe that $f$ maps $X-1$, $Y$ to $T-1$, $2(T-1)$ whereas it maps $XY+X-1$, $2X-Y-2$ map to $2T^2-T-1$, $0$. The ideals generated in $\Bbb Q[T]$ clearly differ.

Solution 2:

Note that if $Y \in J$, then there would exist polynomials $P(X,Y),Q(X,Y) \in \mathbb{Q}[X,Y]$ such that $$Y = P(X,Y) \cdot (XY+X-1)+Q(X,Y)(2X-Y-2).$$

Let's find a contradiction with this equality. Consider the system formed by the equations $XY+X-1=0$ and $2X-Y-2=0$. Its solutions are $(x,y) = (1,0)$ and $(x,y) = (-1/2,-3)$. The previous polynomial equation must be satisfied for each $(x,y) \in \mathbb{Q}^2$. In particular, if we take $(x,y) = (-1/2,-3)$, we obtain $-3=0$, which is obviously false in $\mathbb{Q}$.

Solution 3:

You say that $\{XY+X-1,2X-Y-2\}$ is not a Gröbner basis with respect to the lex order (I'll assume that $Y>X$, as mentioned in the comments). In this case, the leading term of $XY+X-1$ is $YX$ and the leading term of your linear form is $-Y$, so replace it by $Y+2-2X$.

Reducing $XY+X-1$ with respect to $Y\leadsto 2X-2$ gives you $X(2X-2)+X-1 = 2X^2-X-1$, which in particular explains the result in the other answer. You now have the ideal

$$J = (2X^2-X-1,Y-2X+2)$$

and since there is no overlapping between the leading terms, you know that this is a Gröbner basis. In particular, since $Y=2X-2$ in the quotient, you see that $J$ defines the finite dimensional algebra

$$ \mathbb Q[X]/(2X^2-X-1) $$ and this is generated by $1$ and $X$ over $\mathbb Q$: it is of dimension two.

On the other hand, the ideal $I$ has as quotient $\mathbb Q$ as $Y=0$ and $X=1$, so that the dimension is $1$. Hence, your two ideals cannot be equal, as they define algebras of different dimensions.