• But since $x^{\beta}$ occurs on the right hand side, there must a term on the left hand side which is $x^{\beta}$ with non-zero coefficient. So we must have some $\alpha_*, j_*$ such that $x^{\alpha_*[j_*] + \alpha_*} \sim x^\beta$, or $\alpha_* + \alpha_*[j_*] = \beta$, or $\alpha_* \leq \beta$, which means that $x^\beta$ lies in the ideal as it can be generated by scaling $x^\alpha_* \in I$.

§ Polynomial in monomial ideal is linear combination of ideal elements

• If $f \in I$, then this means that $f = \sum_i x^{\alpha} c_i$ for polynomials $c_i \in K[x, y]$.
• Expanding $c_i$ into monomials $c_{ij}$, we see that each of the terms on the RHS is some monomial $x^{\alpha_i}c_{ij}$ which is a multiple of $x^{\alpha_i}$, and thus lives in the ideal $I$.
• So, $f$ is a linear combination of $x^{\alpha_i} c_{ij}$ which live in the ideal.
• MORAL : A monomial ideal is determined by its monomials. Any polynomial in the monomial ideal is generated by monomials in the ideal.

§ Dickson's Lemma: monomial ideals are finitely generated

• Induction on the number of variables. $n = 1$ is done since $k[x]$ is a PID, needs only a single generator.
• Let's have $n+1$ variables, which we write as $K[x_1, \dots, x_n, y]$ with $y$ being the new variable we add (for induction).
• Suppose $I \subseteq K[x_1, \dots, x_n, y]$ is a monomial ideal. We must find a generating set for $I$.
• Let $J$ be the ideal generated by the ideal $I$ where we set $y$ to $1$. One way to think about this is to write $J \simeq I/(y-1)$.
• Alternatively, being very explcit, we define $J \equiv \{ x^\alpha : \exists k, x^\alpha y^k \in I\}$. That is, $J$ consists of all $x^\alpha$ such that for some $k$, $x^\alpha y^k \in I$.
• Philosophically, $J$ is the projection of $I$ onto the $\{ x_i \}$.
• Our inductive hypothesis says that $I$ is finite generated by $I \equiv \langle x^{\alpha(1)}, \dots, x^{\alpha(n)} \rangle$.
• For each $i \in [1, n]$, we know that we have $x^{\alpha(i)}y^{m(i)} \in I$ for some $m(i)$. Let $M \equiv \max_i m(i)$ be the largest of all $m_i$.
• Now consider the slices of $J$ at $y^k$. That is, we wish to generate $y^k \cdot I$ for all $k \in [0, m]$. Define $J_k \equiv \langle y^k x^{\alpha(i)} \rangle$.
• By our induction hypothesis, each of the $J_k$ is finitely generated.
• Thus, the full $J$ is generated by the collection of all generators for each $J_k$ for $0 \leq k \leq M$. To compute $M$, we finitely generated $I$.
• See that every monomial in $I$ is divisible by the generator of some $J_k$ for some $k$. Suppose some $x^\beta y^b \in I$. If $b \geq M$, then we find some $x^\alpha(i)|x^\beta$ in $J$. Then, we consider $x^\alpha(i) y^{m(i)}$ which will definitely divide $x^\beta y^b$ since $m(i) \leq M \leq b$, and then $y^{m(i)}|y^b$.
• If we have $b < M$, then we consider the ideal $J_b$. Then the monomial $x^\beta y^b$ will be generated by monomials in $J_k$.
• Thus, since (1) every monomial in $I$ lies in some $J_k$, and vice versa, (2) monomial ideals are determined by their monomials, and (3) The $J_k$ are finitely generated, we have shown that $I$ is finitely generated by the union of generators of the $J_k$, $\cup \texttt{gen}(J_k)$

§ Ideal of leading terms

• For any ideal $I$, define the ideal of leading terms $LT(I)$ to be the ideal conisting of elements as the leading term of elements of $I$. So, $LT(I) \equiv \{ LT(f) : f \in I \}$. Check that this is an ideal. ( $0 = LT(0)$, $1 = LT(1)$, $LT(f+g) = LT(f)+LT(g)$, and $LT(fg) = LT(f) LT(g)$).
• Suppose we have an ideal $I \equiv \langle f_1, f_2, \dots, f_n \rangle$. Now we have two ideals that we wish to compare: $LT(I)$, the ideal of leading terms, and $\langle LT(f_1), \dots, LT(f_n) \rangle$, the ideal generated by the leading terms of the generators of $I$.
• We must always have $\langle LT(f_1), \dots LT(f_s) \rangle \subseteq LT(I)$by the definition of $LT(I)$ which contains all leading terms.
• However, $LT(I)$ can be larger.
• A generating set for $I$ given by $I \equiv \langle f_1, f_2, \dots, f_s \rangle$is a Grober basis iff it is true that $LT(I)$ equals $\langle LT(f_1), LT(f_2), \dots, LT(f_s) \rangle$.

§ Proof of hilbert basis theorem

• We wish to show that every ideal $I$ of $k[x_1, \dots, x_n]$ is finitely generated.
• If $I = \{ 0 \}$ then take $I = (0)$ and we are done.
• Pick polynomials $g_i$ such that $(LT(I)) = (LT(g_1), LT(g_2), \dots, LT(g_t))$. This is always possible since $(LT(I))$ is a monomial ideal, which is finitely generated by Dickinson's Lemma.
• We claim that $I = (g_1, g_2, \dots, g_t)$.
• Since each $g_i \in I$, it is clear that $(g_1, \dots, g_t) \subseteq I$.
• Conversely, let $f \in I$ be a polynomial.
• Divide $f$ by $g_1, \dots, g_t$ to get $f = \sum_i a_i g_i + r$ where no term of $r \in K[x_1, \dots, x_n]$ is divisible by any of $LT(g_1), \dots, LT(g_t)$. We claim that $r = 0$.
• See that $r = f - \sum_i a_i g_i$. We have $r \in I$, since $f \in I$ and the $g_i$ live in $I$.
• Thus, we must have $LT(r) \in LT(I)$ (by the definition of $LT(I)$).
• If $LT(r)$ is nonzero, then since (1) $LT(I) = \langle LT(g_1), \dots, LT(g_n) \rangle$, and (2) $LT(I)$ is a monomial ideal, $LT(r)$ must be divisible by one of the generators!
• This contradicts the assumpion that $r$ is a reminader --- a remainder is by definition not divisible by any $LT(g_i)$.
• Thus, we have shown that if $f \in I$, then $f \in (g_1, \dots, g_n)$.

§ References

• Cox, Little, O'Shea: computational AG.