Suppose \(\succ\) is a monomial ordering on \(\mathbb{N}\). Since \(\succ\) is a total order, either \(0 \succ 1\) or \(1 \succ 0\). In the first case, we have \(a \succ 1+a\) for all \(a \in \mathbb{N}\). However, this has no smallest element in \(\mathbb{N}\), so this case does not occur. The case \(1 \succ 0\) remains. Now \(a+1 \succ a\) for all \(a \in \mathbb{N}\). By transitivity, we see that \(x \succ y\) iff \(x > y\). Thus, this is the unique monomial order on \(\mathbb{N}\) (and thus on \(k[x]\)).

Using the division_algorithm routine provided by the instructor, we can find these in Sage:

sage: R.<x,y> = PolynomialRing(QQ,2, order="lex")
sage: f = x^3*y^2 + x*y^3-y+1
sage: gs = [x*y^2-x,x-y^3]
sage: division_algorithm(f,gs)
([x^2 + x*y + y^2 + y + 1, x^2 + x*y + y + 1], y^4 + y^3 - y + 1)
sage: R.<x,y> = PolynomialRing(QQ,2, order="deglex")
sage: f = x^3*y^2 + x*y^3-y+1
sage: gs = [x*y^2-x,x-y^3]
sage: division_algorithm(f,gs)
([x^2 + y, 0], x^3 + x*y - y + 1)

Thus the remainder is \(y^4 + y^3 - y + 1\) with lex and \(x^3 + xy - y + 1\) with grlex.

Consider \(g=3xf_1-2yf_2 = -3x^2+2y^2+2y\), which is in the ideal by construction. All the terms in \(g\) have smaller degree than the leading terms of \(f_1, f_2\) so the remainder on division by \(f_1,f_2\) is itself.

Observe that \[\langle \operatorname{LT}(g_1), \operatorname{LT}(g_2), \operatorname{LT}(g_3) \rangle = \langle xy^2, xy, x \rangle = \langle x \rangle.\] However, if \(g= g_2-yg_3 = y^2z^4-z^2\) then \(\operatorname{LT}(g)=y^2z^4 \notin \langle x \rangle\) but \(g \in I\).

Suppose that \(G\) is a Gröbner basis of \(I\). Let \(f \in I\). Then \(\operatorname{LT}(f) \in \langle \operatorname{LT}(I) \rangle\). By definition of Gröbner basis, \[\langle \operatorname{LT}(g_1), \ldots, \operatorname{LT}(g_t)\rangle = \langle \operatorname{LT}(I) \rangle.\] Since \(\langle \operatorname{LT}(I) \rangle\) is a monomial ideal, we see that \(\operatorname{LT}(f)\) must be divisible by some \(\operatorname{LT}(g_i)\) by Proposition 2.4.2.

Conversely, suppose the leading term of any element of \(I\) is divisible by one of the \(\operatorname{LT}(g_i)\). If \(G\) is not a Gröbner basis, then there must exist an \(f \in I\) such that \(\operatorname{LT}(f) \notin \langle \operatorname{LT}(g_1), \ldots, \operatorname{LT}(g_t)\rangle\). This is a contradiction, so \(G\) is a Gröbner basis.

There exists a Gröbner basis \(G = \{ g_1, \ldots, g_t \}\) for \(I\) since every nonzero ideal has a Gröbner basis. Now by Proposition 2.6.1, there exists a \(g \in I\) and a unique \(r \in k[x_1,\ldots,x_n]\) such that \(f = g+r\) and no term of \(r\) is divisible by any of \(\operatorname{LT}(g_1), \ldots, \operatorname{LT}(g_t)\).

Since \(f\) is given and \(r\) is unique, we see that \(g=f-r\) is unique and thus the expression \(f=g+r\) is unique.

Suppose \(t\) is a term in \(r\) that is divisible by an element \(g \in \operatorname{LT}(I)\). By Exercise 2.5.5, \(g\) is divisible by one of the \(\operatorname{LT}(g_i)\). Thus \(t\) is divisible by \(\operatorname{LT}(g_i)\); a contradiction. Thus no term of \(r\) is divisible by any of \(\operatorname{LT}(g_1), \ldots, \operatorname{LT}(g_t)\).

Note that there is a specific formula in the text for the \(S\)-polynomial.

1. \(S(f,g)=(\frac{yz}{4})f-(x)g=-3 x^{2} z^{4} - \frac{7}{4} y^{3} z\)
2. \(S(f,g)=(z^2)f-(\frac{x^3y}{3})g=\frac{1}{3} x^{3} y^{2} - z^{4}\)
3. \(S(f,g)=f-\frac{1}{2}g=2 x y z - 2\)

Set \(g_1=y-x^2\), \(g_2=z-x^3\) and \(I=\langle g_1, g_2\rangle\). Observe that \(\operatorname{LT}(g_1)=x^2\) and \(\operatorname{LT}(g_2)=x^3\), so \(\langle \operatorname{LT}(g_1), \operatorname{LT}(g_2) \rangle = \langle x^2 \rangle\). Note that \[xy-z=(x)(y-x^2)-(1)(z-x^3)=xg_1+g_2 \in I,\] so \(\operatorname{LT}(xy-z)=xy \in \operatorname{LT}(I)\). But \(xy \notin \langle \operatorname{LT}(g_1), \operatorname{LT}(g_2) \rangle\). Thus, it is not a Gröbner basis.