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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.

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You kept $f_4$ but you should have computed its remainder when "divided" by $f_1$ and $f_2$. Notice its leading term is divisible by $f_2$'s leading term, for example. You should review how Buchberger's algorithm works. –  Mariano Suárez-Alvarez Jul 17 '12 at 19:34
    
Mariano, thanks for the comment. If $f_4$ is redundant, then shouldn't $S(f_3,f_4)=0$? –  Derek Allums Jul 17 '12 at 19:55
    
Instead of worrying about that why don't you simply see if $f_4$ is redundant? –  Mariano Suárez-Alvarez Jul 17 '12 at 20:05
    
@MarianoSuárez-Alvarez So I should compute $S(f_1,f_4)$ and $S(f_2,f_4)$? I just did this and got nonzero remainders. I should have gotten 0 for both right? –  Derek Allums Jul 17 '12 at 22:10
    
As I said, you need to reduce $f_4$ modulo the earlier $f_i$'s. As I said in my first comment, its eading term is divisible by $f_2$'s, so this reduction will change it. Did you review the algorithm as I suggested? From where? –  Mariano Suárez-Alvarez Jul 17 '12 at 22:32

1 Answer 1

up vote 1 down vote accepted

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.

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