Let $I $ denote the ideal in $Q [x,y,z]$ generated by $[x^2+y^2,xz−y,z^3−zy^3,xy+zy^2]$
Compute a generating set for $ I∩Q[y].$
Compute a generating set for $ I∩(y). $
Compute a generating set for $ (I:y) $
For $ I∩(y). $
I have set $(p1,p2,p3,p4)$ and the $ [x,0,z] $
Then I have obtained $ P1x^2 + P2xz + P3z^3 = 0 $
Is that the right step, and what do I do next?
I don't actually know where to start for the first generating set.
Thanks :)
On
In such cases i let the computer algebra system of my preference, today sage, to make the computations. Let us see the results:
sage: R.<x,y,z> = PolynomialRing(QQ)
sage: J = R.ideal( [ x^2+y^2, x*z-y, z^3-z*y^3, x*y+z*y^2] )
sage: J.elimination_ideal( [x,z] )
Ideal (y^5 + y^2) of Multivariate Polynomial Ring in x, y, z over Rational Field
The message is as follows. First of all, the element $Y = y^2+y^5$ is in the posted ideal $J$:
sage: y^2+y^5 in J
True
The representation of this element w.r.t. the generators of $J$ is
sage: Y = y^2+y^5
sage: Y.lift(J)
[y^3, y*z^2 - y, -x*y, -x*y^2 + z]
Indeed, in $$ y^3(x^2+y^2)+y(z^2-1)(xz-y)-xy\cdot z(z^2-y^3) + (z-xy^2)\cdot y(x+zy) $$ there survive only the monomials in powers of $y$.
sage: y^3*(x^2+y^2)+y*(z^2-1)*(x*z-y)-x*y*z*(z^2-y^3) +(z-x*y^2)*y*(x+z*y)
y^5 + y^2
How to get this using Gröbner bases?
Unfortunately, the order of the letters x,y,z in the alphabet is not the order we need for our purposes, so let us consider the "same problem", but we eliminate $x,y$, and keep $z$. The corresponding code needs the lex order on the polynomial ring:
sage: R.<x,y,z> = PolynomialRing(QQ, order='lex')
sage: J = R.ideal( [ x^2+y^2, x*z-y, z^3-z*y^3, x*y+z*y^2] )
sage: J.elimination_ideal([x,y])
Ideal (z^5 + z^3) of Multivariate Polynomial Ring in x, y, z over Rational Field
sage: J.groebner_basis()
[x^2 + y^2, x*y + y^2*z, x*z - y, y^3 + z^4, y^2*z^2 + y^2, y*z^4 + y*z^2, z^5 + z^3]
sage: (z^3+z^5).lift(J)
[0, -y^2*z, z^2 + 1, y*z^2]
Note that $z^3+z^5$ is the generator of the elimination ideal, after eliminating $x,y$, and the "last element" in the Gröbner basis.
The same with the elimination of $y,z$, we use the order invlex, the reverse lexicographic order:
sage: R.<x,y,z> = PolynomialRing(QQ, order='invlex')
sage: J = R.ideal( [ x^2+y^2, x*z-y, z^3-z*y^3, x*y+z*y^2] )
sage: J.elimination_ideal([y,z])
Ideal (x^8 + x^2) of Multivariate Polynomial Ring in x, y, z over Rational Field
sage: J.groebner_basis()
[z^3 - x^3, y*z^2 - x^4, x*z - y, y^2 + x^2, x*y + x^5, x^8 + x^2]
sage: (x^2+x^8).lift(J)
[-x^4*y^2 - x^2*y + x^6 + 1, -y*z^2 + x^3*y*z + y, x^3*y^2 + x*y, -x^3*z^2 - z + x^3*y^3 + x*y^2 + x^3]
Similarly, $x^2+x^8$ is the generator of the elimination ideal, after eliminating $y,z$, and the "last element" in the Gröbner basis.
Now i need some tiny contortion to set the order of the letters $x,y,z$ which places $y$ at the end, after $z$.
sage: R.<x,z,y> = PolynomialRing(QQ, order='lex')
sage: J = R.ideal( [ x^2+y^2, x*z-y, z^3-z*y^3, x*y+z*y^2] )
sage: J.elimination_ideal([x,z])
Ideal (y^5 + y^2) of Multivariate Polynomial Ring in x, z, y over Rational Field
sage: J.groebner_basis()
[x^2 + y^2, x*z - y, x*y + z*y^2, z^3 - z*y^3, z^2*y - y^4, y^5 + y^2]
This computes the ideal $J\cap \Bbb Q[y]$. (I was using $J$ instead of $I$, since $J$ is a free letter in sage.)
I do not know the notation $(I:y)$, but i think the above addresses by explicit computation the difficulties of the question, and a way to proceed.
Keeping in mind that it is an exam question, I propose the following solution:
Let's name the binomials:$$f_1=x^2+y^2 \\ f_2=xz-y \\ f_3=z^3-zy^3 \\ f_4=xy+zy^2 \\ I=\langle f_1, f_2, f_3, f_4\rangle$$
I. A generating set for $I \cap \mathbb{Q}[y]$
It means that we have to eliminate $x$ and $z$ from the equations $f_i=0, i=1,..,4$. This can be done by considering the $lex$ order $x>z>y$ and simply finding the Gröbner basis: I used Buchberger's algorithm by finding the S-polynomial ($S(f_i,f_j)$) of each pair in $F=(f_1,f_2,f_3,f_4)$ and its remainder on division by F (denoted by $\overline{S(f_i,f_j)}^F, i\neq j$). If the remainder is not zero, it is added to $F$ until all $\overline{S(f_i,f_j)}^F=0$.
$$S(f_1,f_2) = xy+zy^2 =f_4 \in I \implies \overline{S(f_1,f_2)}^F=0 \\ S(f_1,f_3) = x^2zy^3+z^3y^2 =zy^3f_1 +y^2f_3 \in I \implies \overline{S(f_1,f_3)}^F=0\\ S(f_1,f_4) = xzy^2-y^3 = y^2f_2 \in I \implies \overline{S(f_1,f_4)}^F=0\\ S(f_2,f_3) = xzy^3-z^2y = y^3f_2 + (y^4-z^2y) \notin I \implies \overline{S(f_2,f_3)}^F=-z^2y+y^4 \\ $$ Thus, we must add $f_5=-z^2y+y^4$ to our ideal $I$ and call it $I'$. Now, $F' = (f_1,f_2,f_3,f_4,f_5)$ $$S(f_1,f_5)=x^2y^4+y^3z^2y^4f_1-y^2f_5 \in I' \implies \overline{S(f_1,f_5)}^{F'}=0\\ S(f_2,f_4)=-z^2-y^2=yf_5+(-y^5-y^2) \notin I' \implies \overline{S(f_1,f_5)}^{F'}=-y^5-y^2 $$ Thus, we must add $f_6=-y^5-y^2$ to our ideal $I'$. Hence $F''=(f_1,...,f_6)$ It turns out that the remaining S-polynomials yield $\overline{S(f_1,f_6)}^{F''}=\overline{S(f_2,f_5)}^{F''}=\overline{S(f_2,f_6)}^{F''}=\overline{S(f_3,f_4)}^{F''}=\overline{S(f_3,f_5)}^{F''}=\overline{S(f_3,f_6)}^{F''}=\overline{S(f_4,f_5)}^{F''}=\overline{S(f_4,f_6)}^{F''}=\overline{S(f_5,f_6)}^{F''}=0$.
Therefore the Gröbner basis (It is easy to verify that it is also the reduced Gröbner basis by observing the leading terms) of I is $$G=\langle x^2+y^2, xz-y, z^3-zy^3, xy+z^2, z^2y-y^4, y^5+y^2\rangle$$ resulting in $I \cap \mathbb{Q}[y] =\langle y^5+y^2 \rangle$.
II. A generating set for $I \cap \langle y \rangle$
Like you started, we are looking for a polynomial $f \in I$ and $f \in \langle y \rangle$ at the same time. Let $$f=p_1f_1+p_2f_2+p_3f_3+p_4f_4 = q_1y \tag{1}$$ where $p_1,p_2,p_3,p_4$ and $q_1$ are functions of $x,y$ and $z$. Our aim is to find $p_i,i=1,..,4$. $\{(y,0,0,0),(0,y,0,0),(0,0,y,0),(0,0,0,1)\}$ are trivial solutions. When $y=0$, equation $(1)$ becomes $$r_1x^2+r_2xz+r_3z^3=0 \tag{2}$$ where $r_i$ are functions of $x,z$. The generating set of all solutions of equation $(2)$ is $\{(z,-x,0),(0,z^2,-x),(z^3,0,-x^2)\}$.
Substituting these 7 solutions in equation $(1)$ gives us the following polynomials $$g_1=x^2y+y^3\\ g_2=xyz-y^2\\ g_3=z^3y-zy^4\\ g_4=xy+zy^2\\ g_5=xy+zy^2=g_4\\ g_6=xzy^3-yz^2\\ g_7=x^2zy^3+z^3y^2=zy^2g_1+yg_3\\$$
Hence, the generating set for the given ideal is $\langle x^2y+y^3, xyz-y^2,z^3y-zy^4, xy+zy^2, xzy^3-yz^2 \rangle$. Note that this is not a Gröbner basis and finding it shouldn't be hard by following the same procedure for the first question. It turns out to be$$I \cap \langle y \rangle = \langle y^5+y^2, z^2y-y^4, xy+zy^2 \rangle$$
III. A generating set for $I:y$
There is a theorem that states if $\{h_1,...,h_p\}$ is a basis of the ideal $I \cap \langle g \rangle$, then $\{\dfrac{h_1}{g},...,\dfrac{h_p}{g}\}$ is a basis of $\langle I:g \rangle$. Therefore, $$I:y = \langle \frac{y^5+y^2}{y}, \frac{z^2y-y^4}{y}, \frac{xy+zy^2}{y} \rangle\\ = \langle y^4+y, z^2-y^3, x+zy \rangle$$
I have only briefly read the Cox-Little-O'shea book and I have come up with this solution based on my understanding. Nonetheless, there might be a simpler solution or a way to shorten these computations (probably by using improvements of the Buchberger's algorithm). I will be happy to know if someone can suggest improvements or shortcuts to these approaches.