Calculating Grobner Bases In this question, $ℚ[x,y,z]$ is endowed with the lexicographic order with $x > y > z$.
Set $u:= x^2 + 2yz^2$ and $v:= y^2 - 3xz$. Denote by $J$ the ideal of $ℚ[x,y,z]$ generated by $u$ and $v$.


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*Compute a reduced Gröbner basis of $J$ by using Buchberger's algorithm.

*Does $f:= x^2 - xz + 4$ belong to $J$?

*Prove that $V(J)$ is infinite.

*Does $g:= (x + y + z)^{2014}$ belong to $J$?


$\mathbb{Q}[x,y,z]$ is endowed with lex: $x > y > z$.
$J = \langle x^2 + 2yz^2, xz - \frac{1}{3}y^2 \rangle = \langle u, v \rangle$.
1) The Buchberger's algorithm would go as follows:
$$S(u,v) = x^2 z + 2yz^3 - x^2 z + \frac{1}{3}xy^2 = \frac{1}{3}xy^2 + 2yz^3 = \frac{1}{3} p$$
Since $p$ cannot be reduced any further with respect to $u$ and $v$ then $G_J = \{ x^2 + 2yz^2, xz - \frac{1}{3}y^2 \} \cup \{ xy^2 + 6yz^3 \}$ and $J = \langle u,v,p \rangle$.
$$S(v,p) = xy^2 z - \frac{1}{3}y^4 - xy^2 z - 6yz^4 = -\frac{1}{3}y^4 - 6yz^4 = -\frac{1}{3}q$$
Again since $q$ cannot be reduced any further with respect to $u,v$ and $p$ then $G_J = \{ x^2 + 2yz^2, xz - \frac{1}{3}y^2, xy^2 + 6yz^3 \} \cup \{ y^4 + 18yz^4 \}$ and $J = \langle u,v,p,q \rangle$.
It is easy to check that any other $S$-polynomial reduces to $0$ modulo $J$. So $G_J$ is a Groebner basis of $J$. It is also straightforward to verify that $G_J$ is a reduced Groebner basis of $J$.
2) No, it doesn't. It reduces to $-\frac{1}{3}y^2-2yz^2+4$ with respect to $G_J$.
3) Elimination ideal $J_z = J \cap k[z] = \{ 0 \}$ which means that $\mathbb{V}(J_z) = \mathbb{Q}$ and $\mathbb{V}(J)$ is a curve in $\mathbb{Q}^3$.
4) $g \in J \subseteq \sqrt{J} \iff g_{red} \in \sqrt{J}$, where $g_{red} = x + y + z$.
To check radical membership you can use Proposition 8 (Chapter 4, paragraph 2) of "Ideals, Varieties and Algorithms" by Cox, Little, O'Shea. It says that:
$$g_{red} \in \sqrt{J} \iff 1 \in \langle u, v, 1 - w g_{red} \rangle = \hat{J} \subseteq \mathbb{Q}[x,y,z,w].$$
Computing reduced Groebner basis of $\hat{J}$ with respect ot grlex ($x>y>z>w$) you'll get 5 nonconstant polynomials, which means $1 \notin \hat{J}$ and $g \notin J$. (I do not recommend you to compute Groebner basis of $\hat{J}$ with respect to lex ordering in this case, because it has enormous polynomials).
In fact, in this particular case you have $J = \sqrt{J}$, so it is easier to reduce $g_{red} = x + y + z$ with respect to $G_J$. However, it is not easy to show that $J = \sqrt{J}$.
