A Gröbner Basis Computation Gone Bad Here is the problem statement:
Consider the polynomial ideal $I =  \langle b-r_1-r_2, c-r_1r_2 \rangle  \subset \mathbb{Q}[r_1,r_2,b,c].$ Show that $I \cap \mathbb{Q}[b,c] = \langle 0 \rangle$. 
First, if you can solve this without the use of the theory of Gröbner bases, I would love to see a solution, but I am explicitly trying to solve this using some Gröbner basis.
Now, let me give my calculations, which do not yield the desired result. I will use the notation in Dummit and Foote. Put the lexicographic order $r_1 > r_2 > b > c$ on the polynomial ring and let $f_1 = -r_1-r_2 + b$ and $f_2 = -r_1r_2  + c$. Then,
\begin{align*}
S(f_1,f_2) &= \frac{r_1r_2}{-r_1}(-r_1-r_2 + b) - \frac{r_1r_2}{-r_1r_2} (-r_1r_2 + c)\\
&= -r_2(-r_1-r_2 + b) + (-r_1r_2 + c)\\
&=r_2^2-r_2b + c =: f_3.
\end{align*}
Next,
\begin{align*}
S(f_1,f_3) &= \frac{r_1r_2^2}{-r_1}(-r_1-r_2+b) - \frac{r_1r_2^2}{r_2^2}(r_2^2 - r_2b + c)\\
&=-r_2^2(-r_1-r_2 + b) - r_1(r_2^2-r_2b + c)\\
&=r_1r_2b-r_1c+r_2^3-r_2^2b =: f_4
&\\
& \\
S(f_2,f_3) &= \frac{r_1r_2^2}{-r_1r_2}(-r_1r_2 + c) - \frac{r_1r_2^2}{r_2^2}(r_2^2 - r_2b + c) \\
&=-r_2(-r_1r_2 + c)-r_1(r_2^2 - r_2b + c)\\
&=r_1r_2b - r_2c - r_1c =:f_5.
\end{align*}
Feeling like there was still hope this would work, I stopped with:
\begin{align*}
S(f_3,f_4) &= \frac{r_1r_2^2b}{r_2^2}(r_2^2 - r_2b+c) - \frac{r_1r_2^2b}{r_1r_2b}(r_1r_2 - r_1c + r_2^3 - r_2^2b)\\
&=r_1b(r_2^2-r_2b+c)-r_2(r_1r_2b-r_1c + r_2^3-r_2^2b)\\
&=-r_1r_2b -r_1r_2c + r_1bc -r_2^4 + r_2^3b =:f_6.
\end{align*}
So, my question is: what's going on here?  According to Mathematica, the Gröbner basis with this ordering is 
$$
\lbrace c - b r_2 + r_2^2, -b + r_1 + r_2 \rbrace
$$
Notice the first element of this set if what I call $f_3$ and the second element is $f_1$. Thus, I would think my computation should have terminated at $S(f_1,f_3)$ and $S(f_2,f_3)$.
I suspect I am making an arithmetic error, but by now I have some pretty bad tunnel vision so I can't spot it.  Of course, I could also be making a more serious error somewhere, e.g., taking the wrong least common multiple. 
Any help is appreciated. 
 A: As Mariano said in the comments, I "need to reduce $f_4$ modulo the earlier $f_i$'s." I thought I was doing this by calculating the $S(f_i,f_4)$, but all this is doing is cancelling the leading terms.  What I should have done is apply the General Polynomial Division algorithm on page 320 of Dummit and Foote to reduce $f_4:=S(f_1,f_3)$ and $f_5:=S(f_2,f_3)$ modulo $\lbrace f_1,f_2,f_3 \rbrace$ before moving on.  Had I done this, I would have found these to be congruent to $0$ modulo the $f_i$'s and by Proposition 26 in Dummit and Foote, a Groebner basis is $\lbrace f_1,f_2,f_3 \rbrace$. In brief, I misapplied the algorithm by confusing "If $S(f_i,f_j) = 0\ldots$" with "If $S(f_i,f_j) \equiv 0 \mod \lbrace f_1,\ldots,f_n \rbrace \ldots$"
Here is what the calculation should have looked like.  First, notice $f_3:=S(f_1,f_2) \equiv 0 \mod F=\lbrace f_1,f_2,f_3 \rbrace$ trivially since $f_3 \in F$. To conclude that $F$ is a Groebner basis for $I$, we need to show that $$f_4 := S(f_1,f_3), \;\;f_5 := S(f_2,f_3)  \equiv 0 \mod F.$$ Using the notation on page 320 of Dummite and Foote, we notice that $LT(f_4)$ is divisible by $LT(f_2)$ so $q_2 = a_2 = -b$.  Then,  $$f_4 \mapsto f_4 - a_2f_2 = f_4-r_1r_2 + bc = -r_1c+r_2^3-r_2^2b+bc.$$ Next notice that $LT(-r_1c+r_2^3-r_2^2b+bc)$ is divisible by $LT(f_1)$ so $q_1 = a_1 = c$ and 
$$
-r_1c+r_2^3-r_2^2b+bc \mapsto -r_1c+r_2^3-r_2^2b+bc-c(-r_1-r_2+b) = r_2^3-r_2^2b+cr_2.
$$
Finally, notice 
$$
r_2^3-r_2^2b+cr_2 = r_2f_3,
$$
so $q_3a_3=r_2$ and thus, 
$$
f_4 = q_1f_1 + q_2f_2 + q_3f_3 = cf_1 - bf_2 + r_2f_3 \in \langle f_1,f_2,f_3 \rangle.
$$
That is, $f_4 = S(f_1,f_3) \equiv 0 \mod F$.  A similar computation shows that $f_5= cf_1-bf_2$ so $f_5 = S(f_2,f_3) \equiv 0 \mod F$.  Again, by Proposition 26, we conclude that $F$ is a Groebner basis for $I$ under the ordering $r_1 > r_2 > b >c$ (Note: $f_2 \in \langle f_1,f_3 \rangle$ so I really didn't even need to find $S(f_2,f_3)$ to begin with, i.e., $F$ is not a reduced basis but $F' = \lbrace f_1,f_3 \rbrace$ is, although this doesn't really matter for this problem).  
To answer the main question, apply Proposition 29 in Dummit and Foote, which says in particular that $I \cap \mathbb{Q}[b,c] = 0$ iff $F \cap \mathbb{Q}[b,c] = 0$, which is true.
