# abstract algebra – Doubt in computation of groebner basis and proving a set is a groebner basis.

I have a set $$G$$ , and I am trying to prove that the set is a groebner basis .

From chapter $$2$$ of Ideals,Varieties and Algorithm I have the following theorem,

A basis $$G = {g_1,cdots, g_n }$$ for an ideal I is a Gröbner basis if and only if $$S(g_i, g_j) to_{G} 0$$ for all $$i ne j$$.

My ideal $$I$$ is generated by $$y_1+ y_2*x_3 + y_3*x_3^2$$,$$y_1+ y_2*x_2 + y_3*x_2^2$$,$$y_1 + y_2*x_1+y_2*x_1^2$$ and the monomial order is $$(y_1>y_2>y_3>x_1>x_2>x_3)$$ lexicographic ordering

The set $$G$$ is $${y_3*x_1^2*x_2 – y_3*x_1^2*x_3 – y_3*x_1*x_2^2 + y_3*x_1*x_3^2 + y_3*x_2^2*x_3 – y_3*x_2*x_3^2 , y_2*x_2 – y_2*x_3 + y_3*x_2^2 – y_3*x_3^2 , y_2*x_1 – y_2*x_3 + y_3*x_1^2 – y_3*x_3^2 , y_1+ y_2*x_3 + y_3*x_3^2}$$

I am trying to claim is that this set $$G$$ is the groebner basis of $$I$$

I calculated the $$6$$, S-polynomials of the polynomials in $$G$$ and I have shown that the remainder obtained after dividing these (by the polynomials present in G) is $$0$$ .Is this process enough to conclude that $$G$$ is a groebner basis of $$I$$?Or should I be calculating the leading term of $$S(g_i,g_j)$$ where $$i ne j$$ and $$g_i,g_j in G$$ and then checking whether the leading term of this $$S$$-polynomial is present in the ideal generated by $$langle LT(g_1),Lt(g_2) ,cdots Lt(g_n) rangle$$?