Proof: Define $ψ : Hom_R(R^n,R) → R^n$ by $φ \mapsto (φ(e_1),...,φ(e_n))$. We have $$ψ(φ_1+rφ_2) = ((φ_1+rφ_2)(e_1),...,(φ_1+rφ_2)(e_n)) =$$$$(φ_1(e_1),...,φ_1(e_n)) + r(φ_2(e_1),...,φ_2(e_n)) = ψ(φ_1)+rψ(φ_2)$$so $ψ$ is a homomorphism of $R$-modules. If $ψ(φ)=(0,...,0)$, then for arbitrary $r^*=(r_1,...,r_n)=r_1e_1+...+r_ne_n∈R^n$ we have $φ(r^*)=r_1φ(e_1)+...+r_nφ(e_n)=0$ so that $φ=0$ and now $ψ$ is injective. Again for arbitrary $r^*∈R^n$, define $φ∈Hom_R(R^n,R)$ by $φ(e_i)=r_i$ and extending linearly, so that $ψ(φ)=r^*$ and now $ψ$ is surjective and an isomorphism of $R$-modules.$~\square$
Tuesday, June 4, 2013
R-Module Homomorphisms from Free Modules (10.3.13)
Dummit and Foote Abstract Algebra, section 10.3, exercise 13:
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Let $R$ be a commutative ring. Prove $Hom_R(R^n,R)≅R^n$ as $R$-modules.
Proof: Define $ψ : Hom_R(R^n,R) → R^n$ by $φ \mapsto (φ(e_1),...,φ(e_n))$. We have $$ψ(φ_1+rφ_2) = ((φ_1+rφ_2)(e_1),...,(φ_1+rφ_2)(e_n)) =$$$$(φ_1(e_1),...,φ_1(e_n)) + r(φ_2(e_1),...,φ_2(e_n)) = ψ(φ_1)+rψ(φ_2)$$so $ψ$ is a homomorphism of $R$-modules. If $ψ(φ)=(0,...,0)$, then for arbitrary $r^*=(r_1,...,r_n)=r_1e_1+...+r_ne_n∈R^n$ we have $φ(r^*)=r_1φ(e_1)+...+r_nφ(e_n)=0$ so that $φ=0$ and now $ψ$ is injective. Again for arbitrary $r^*∈R^n$, define $φ∈Hom_R(R^n,R)$ by $φ(e_i)=r_i$ and extending linearly, so that $ψ(φ)=r^*$ and now $ψ$ is surjective and an isomorphism of $R$-modules.$~\square$
Proof: Define $ψ : Hom_R(R^n,R) → R^n$ by $φ \mapsto (φ(e_1),...,φ(e_n))$. We have $$ψ(φ_1+rφ_2) = ((φ_1+rφ_2)(e_1),...,(φ_1+rφ_2)(e_n)) =$$$$(φ_1(e_1),...,φ_1(e_n)) + r(φ_2(e_1),...,φ_2(e_n)) = ψ(φ_1)+rψ(φ_2)$$so $ψ$ is a homomorphism of $R$-modules. If $ψ(φ)=(0,...,0)$, then for arbitrary $r^*=(r_1,...,r_n)=r_1e_1+...+r_ne_n∈R^n$ we have $φ(r^*)=r_1φ(e_1)+...+r_nφ(e_n)=0$ so that $φ=0$ and now $ψ$ is injective. Again for arbitrary $r^*∈R^n$, define $φ∈Hom_R(R^n,R)$ by $φ(e_i)=r_i$ and extending linearly, so that $ψ(φ)=r^*$ and now $ψ$ is surjective and an isomorphism of $R$-modules.$~\square$
Quotients and Finite Generation of Modules (10.3.7)
Dummit and Foote Abstract Algebra, section 10.3, exercise 7:
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If $N$ is a submodule of the $R$-module $M$, prove that if $M/N$ and $N$ are finitely generated, then so is $M$.
Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$
Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$
Saturday, June 1, 2013
R-Module Homomorphisms from R (10.2.9-10)
Dummit and Foote Abstract Algebra, section 10.2, exercises 9-10:
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Let $R$ be a commutative ring with 1.
9. Prove $Hom_R(R,M)≅M$ as $R$-modules.
10. Prove $End(R) ≅ R$ as rings.
Proof: (9) Define a mapping $ψ : Hom_R(R,M) → M$ by $φ \mapsto φ(1)$, and prove it is an $R$-module isomorphism. We have $ψ(φ_1+rφ_2)=φ_1(1)+rφ_2(1)=ψ(φ_1)+rψ(φ_2)$ for $R$-module homomorphicity, and if $ψ(φ)=φ(1)=0$ then $φ(r)=rφ(1)=0$ so that $φ=0$ and $ψ$ is thus injective. We show $ψ$ is surjective by constructing the mapping $φ$ defined by $φ(1)=m$ and $φ(r)=r \cdot m$ and showing $φ∈Hom_R(R,M)$: We have this because $φ(r_1+r^* \cdot r_2)=(r_1+r^* \cdot r_2)m=r_1 \cdot m +r^* \cdot (r_2 \cdot m)=φ(r_1)+r^* \cdot φ(r_2)$.
(10) By the above they are already isomorphic as abelian groups, so it suffices to show $ψ$ in this case fulfills $$ψ(φ_1 \circ φ_2)=(φ_1 \circ φ_2)(1)=φ_1(φ_2(1))=φ_1(φ_2(1) \cdot 1))=$$$$φ_1(1)φ_2(1)=ψ(φ_1)ψ(φ_2)~\square$$
9. Prove $Hom_R(R,M)≅M$ as $R$-modules.
10. Prove $End(R) ≅ R$ as rings.
Proof: (9) Define a mapping $ψ : Hom_R(R,M) → M$ by $φ \mapsto φ(1)$, and prove it is an $R$-module isomorphism. We have $ψ(φ_1+rφ_2)=φ_1(1)+rφ_2(1)=ψ(φ_1)+rψ(φ_2)$ for $R$-module homomorphicity, and if $ψ(φ)=φ(1)=0$ then $φ(r)=rφ(1)=0$ so that $φ=0$ and $ψ$ is thus injective. We show $ψ$ is surjective by constructing the mapping $φ$ defined by $φ(1)=m$ and $φ(r)=r \cdot m$ and showing $φ∈Hom_R(R,M)$: We have this because $φ(r_1+r^* \cdot r_2)=(r_1+r^* \cdot r_2)m=r_1 \cdot m +r^* \cdot (r_2 \cdot m)=φ(r_1)+r^* \cdot φ(r_2)$.
(10) By the above they are already isomorphic as abelian groups, so it suffices to show $ψ$ in this case fulfills $$ψ(φ_1 \circ φ_2)=(φ_1 \circ φ_2)(1)=φ_1(φ_2(1))=φ_1(φ_2(1) \cdot 1))=$$$$φ_1(1)φ_2(1)=ψ(φ_1)ψ(φ_2)~\square$$
Tuesday, May 28, 2013
Alternate Proof of Hilbert's Basis Theorem and Gröbner Bases (9.6.43-44)
Dummit and Foote Abstract Algebra, section 9.6, exercises 43-44:
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43. Let $\mathcal{S}=\{N~|~N~\text{not a finitely generated monomial ideal}\}$ and assume by way of contradiction that $\mathcal{S}≠\varnothing$.
(a) Prove $\mathcal{S}$ has a maximal element.
(b) Show there exist monomials $x,y∉M$ such that $xy∈M$.
(c) Show there exists $M_0 \subseteq M$ with $M_0$ a finitely generated monomial ideal such that $M_0+(x)=M+(x)$ and $M=M_0+(x)(M : (x))$. Deduce $M$ is finitely generated, forcing $\mathcal{S}=\varnothing$.
44. If $\varnothing≠I\subseteq F[x_1,..,x_n]$ is an ideal, prove $LT(I)$ is finitely generated. Conclude $I$ has a Gröbner basis and deduce Hilbert's Basis Theorem.
Proof: (a) Let $\mathcal{C}$ be a typical chain totally ordered by inclusion. Since $M=\bigcup_{N∈\mathcal{C}} N$ is a monomial ideal by 9.6.39, by Zorn's Lemma it suffices to show it is infinitely generated; assume the contrary, so that $M=(n_1,...,n_r)$. Let $n_i$ be maximal with regard to the ordering $n_i∈N_i \geq n_j∈N_j :⇔ N_j \subseteq N_i$, and observe $N_i \subseteq M = (n_1,...,n_r) \subseteq N_i$ so that $N_i=M$ is finitely generated, a contradiction.
(b) By 9.6.42(a), the only prime monomial ideals are finitely generated. Therefore there exist polynomials $x',y'∈F[x_1,...,x_n]$ such that $x',y'∉M$ and $x'y'∈M$. By 9.6.10, there exist monomial terms in $x',y'$ not divisible by any $m_i$ for $M=(m_1,...)$. Showing $x'y'=m∈M$ and expanding $x'$ to subtract the products of $y'$ and monomial terms of $x'$ divisible by monomials of $M$ (which are elements of $M$), we obtain $xy'∈M$ for $x$ composed of terms not divisible by monomials of $M$. Similarly, $xy∈M$. Now $LT(xy)=LT(x)LT(y)∈LT(M) \subseteq M$ (the latter $\subseteq$ being due again to 9.6.10) so that $LT(x)$ and $LT(y)$ are monomials not within $M$ whose product is within $M$.
(c) Fix a monomial ordering $≤$. First note that $m \mid m' ⇒ md=m'$ so that, since $d \geq 1$, we have $md = m' \geq m$, i.e. $m \mid m' ⇒ m ≤ m'$.
Let $x$ be the minimal monomial by $≤$ not within $M$ such that there exists another monomial $y$ not within $M$ fulfilling $xy∈M$. Revisit $M=(m_1,...)$ and let $M^*=(m_1,...,x)$ as before. Since $M \subset M^*$ by the maximality of $M$ we must have $M^*=(T)$ for some finite subset of monomials $T=\{q_1,...,q_t\}$. Reformat this generating subset so that it is minimal, by excluding a generator $n$ if $M^*=(T \setminus \{n\})$. This process is seen to terminate, and in part we are left with $T=\{n_1,...,n_s\}$ where $n_i \not \mid n_j$ for all $i≠j$. Assume $x∉T$: Then by 9.6.10 since $x∈M^*$ we have $n_i \mid x$ so $n_id=x$ for some $i$, and since $n_i∈M⇒x∈M$, yet also $n_i∉M⇒n_i < x \land n_i(dy)∈M$ with $dy∉M$ since $d < x$ as well, we inevitably violate the minimality of $x$, and we must instead have $x∈T$. Rewrite $T=\{n_1,...,n_{s-1},x\}$.
Now let $M_0=(T \setminus \{x\})$. Assume $n_j∈T \setminus \{x\}$ yet $n_j∉M$: Since the elements of $T$ are from $M^*$ generated by $\{m_1,...,x\}$ we must have either $m_i \mid n_j$ for some $i$ or $x \mid n_j$; the latter is excluded by the minimalization of $T$ and the former implies $n_j∈M$. Therefore $M_0 \subseteq M$, and evidently $M_0+(x)=M+(x)=M^*$.
Show $M=M_0+(x)(M : (x))$: ($\supseteq$) By the definition of $(M : (x))$, the direction is clear. ($\subseteq$) We have $m_i∈M \subseteq M_0+(x)$, so that either $n_k \mid m_i ⇒ m_i∈M_0$ or $x \mid m_i ⇒ m_i ∈ (x)(M : (x))$.
Finally, we see that $M_0$ is finitely generated, and also $y∉M$ with $y∈(M : (x))$ so that $(M : (x))$ is finitely generated by the maximality of $M$, and also $(x)(M : (x))$ and now $M_0+(x)(M : (x))=M$ is finitely generated, a contradiction.$~\square$
44. Since $LT(I)$ is a monomial ideal, by the previous it is finitely generated. In the same way as in 9.6.1, we may generate each of the elements of the finite generating set of $LT(I)$ by a finite number of finite combinations of $LT(g_i)$ for $g_i$ elements of $I$, so that by proposition 24 these $g_i$ form a Gröbner basis and as well a finite generating set for the arbitrary ideal $I$, which is Hilbert's Basis Theorem.$~\square$
(a) Prove $\mathcal{S}$ has a maximal element.
(b) Show there exist monomials $x,y∉M$ such that $xy∈M$.
(c) Show there exists $M_0 \subseteq M$ with $M_0$ a finitely generated monomial ideal such that $M_0+(x)=M+(x)$ and $M=M_0+(x)(M : (x))$. Deduce $M$ is finitely generated, forcing $\mathcal{S}=\varnothing$.
44. If $\varnothing≠I\subseteq F[x_1,..,x_n]$ is an ideal, prove $LT(I)$ is finitely generated. Conclude $I$ has a Gröbner basis and deduce Hilbert's Basis Theorem.
Proof: (a) Let $\mathcal{C}$ be a typical chain totally ordered by inclusion. Since $M=\bigcup_{N∈\mathcal{C}} N$ is a monomial ideal by 9.6.39, by Zorn's Lemma it suffices to show it is infinitely generated; assume the contrary, so that $M=(n_1,...,n_r)$. Let $n_i$ be maximal with regard to the ordering $n_i∈N_i \geq n_j∈N_j :⇔ N_j \subseteq N_i$, and observe $N_i \subseteq M = (n_1,...,n_r) \subseteq N_i$ so that $N_i=M$ is finitely generated, a contradiction.
(b) By 9.6.42(a), the only prime monomial ideals are finitely generated. Therefore there exist polynomials $x',y'∈F[x_1,...,x_n]$ such that $x',y'∉M$ and $x'y'∈M$. By 9.6.10, there exist monomial terms in $x',y'$ not divisible by any $m_i$ for $M=(m_1,...)$. Showing $x'y'=m∈M$ and expanding $x'$ to subtract the products of $y'$ and monomial terms of $x'$ divisible by monomials of $M$ (which are elements of $M$), we obtain $xy'∈M$ for $x$ composed of terms not divisible by monomials of $M$. Similarly, $xy∈M$. Now $LT(xy)=LT(x)LT(y)∈LT(M) \subseteq M$ (the latter $\subseteq$ being due again to 9.6.10) so that $LT(x)$ and $LT(y)$ are monomials not within $M$ whose product is within $M$.
(c) Fix a monomial ordering $≤$. First note that $m \mid m' ⇒ md=m'$ so that, since $d \geq 1$, we have $md = m' \geq m$, i.e. $m \mid m' ⇒ m ≤ m'$.
Let $x$ be the minimal monomial by $≤$ not within $M$ such that there exists another monomial $y$ not within $M$ fulfilling $xy∈M$. Revisit $M=(m_1,...)$ and let $M^*=(m_1,...,x)$ as before. Since $M \subset M^*$ by the maximality of $M$ we must have $M^*=(T)$ for some finite subset of monomials $T=\{q_1,...,q_t\}$. Reformat this generating subset so that it is minimal, by excluding a generator $n$ if $M^*=(T \setminus \{n\})$. This process is seen to terminate, and in part we are left with $T=\{n_1,...,n_s\}$ where $n_i \not \mid n_j$ for all $i≠j$. Assume $x∉T$: Then by 9.6.10 since $x∈M^*$ we have $n_i \mid x$ so $n_id=x$ for some $i$, and since $n_i∈M⇒x∈M$, yet also $n_i∉M⇒n_i < x \land n_i(dy)∈M$ with $dy∉M$ since $d < x$ as well, we inevitably violate the minimality of $x$, and we must instead have $x∈T$. Rewrite $T=\{n_1,...,n_{s-1},x\}$.
Now let $M_0=(T \setminus \{x\})$. Assume $n_j∈T \setminus \{x\}$ yet $n_j∉M$: Since the elements of $T$ are from $M^*$ generated by $\{m_1,...,x\}$ we must have either $m_i \mid n_j$ for some $i$ or $x \mid n_j$; the latter is excluded by the minimalization of $T$ and the former implies $n_j∈M$. Therefore $M_0 \subseteq M$, and evidently $M_0+(x)=M+(x)=M^*$.
Show $M=M_0+(x)(M : (x))$: ($\supseteq$) By the definition of $(M : (x))$, the direction is clear. ($\subseteq$) We have $m_i∈M \subseteq M_0+(x)$, so that either $n_k \mid m_i ⇒ m_i∈M_0$ or $x \mid m_i ⇒ m_i ∈ (x)(M : (x))$.
Finally, we see that $M_0$ is finitely generated, and also $y∉M$ with $y∈(M : (x))$ so that $(M : (x))$ is finitely generated by the maximality of $M$, and also $(x)(M : (x))$ and now $M_0+(x)(M : (x))=M$ is finitely generated, a contradiction.$~\square$
44. Since $LT(I)$ is a monomial ideal, by the previous it is finitely generated. In the same way as in 9.6.1, we may generate each of the elements of the finite generating set of $LT(I)$ by a finite number of finite combinations of $LT(g_i)$ for $g_i$ elements of $I$, so that by proposition 24 these $g_i$ form a Gröbner basis and as well a finite generating set for the arbitrary ideal $I$, which is Hilbert's Basis Theorem.$~\square$
Monday, May 27, 2013
Prime and Maximal Monomial Ideals (9.6.42)
Dummit and Foote Abstract Algebra, section 9.6, exercise 42:
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(a) Show $M$ is a monomial prime ideal if and only if $M=(S)$ for some $S \subseteq \{x_1,...,x_n\}$.
(b) Show $(x_1,...,x_n)$ is the only maximal monomial ideal.
Proof: (a) ($\Leftarrow$) Letting $I$ be such that $S=\{x_i~|~i∈I\}$, we have $F[x_1,...,x_n]/M ≅ F[x_j~|~j∉I]$ which is again a polynomial ring in several variables over a field and is thus in particular an integral domain. ($⇒$) Since $M$ is a monomial ideal, let $M=(m_1,...)$ with the monomials countably ordered by $m_1 ≤ m_2$ if $\text{deg}(m_2) < \text{deg}(m_1)$ and decided by lexicographic ordering if $\text{deg}(m_2)=\text{deg}(m_1)$. Trim this generating list by a process of eliminating $m_i$ if it is not generatable by linear combinations of previous monomials. Choose $m∉\{x_1,...,x_n\}$ in the generating set. letting $x_i$ be a variable dividing $m$, we must have $x_i∉M$ else by exercise 10 $x_i$ is divisible by a term in the generating list, forcing $x_i$ in the generating list so that by the ordering $x_i$ appears before $m$ and now $m$ could not have been included. Since $x_i$ was arbitrary, we have the product of the variables dividing $m$ violating the primality of $M$. Therefore $m$ does not exist, and this generating subset is a subset of $\{x_1,...,x_n\}$.
(b) Since maximal ideals are prime ideals, if $M=(S)$ with $S≠\{x_1,...,x_n\}$, by part (a) we have $M \subset (x_1,...,x_n) \subset F[x_1,...,x_n]$, a contradiction.$~\square$
(b) Show $(x_1,...,x_n)$ is the only maximal monomial ideal.
Proof: (a) ($\Leftarrow$) Letting $I$ be such that $S=\{x_i~|~i∈I\}$, we have $F[x_1,...,x_n]/M ≅ F[x_j~|~j∉I]$ which is again a polynomial ring in several variables over a field and is thus in particular an integral domain. ($⇒$) Since $M$ is a monomial ideal, let $M=(m_1,...)$ with the monomials countably ordered by $m_1 ≤ m_2$ if $\text{deg}(m_2) < \text{deg}(m_1)$ and decided by lexicographic ordering if $\text{deg}(m_2)=\text{deg}(m_1)$. Trim this generating list by a process of eliminating $m_i$ if it is not generatable by linear combinations of previous monomials. Choose $m∉\{x_1,...,x_n\}$ in the generating set. letting $x_i$ be a variable dividing $m$, we must have $x_i∉M$ else by exercise 10 $x_i$ is divisible by a term in the generating list, forcing $x_i$ in the generating list so that by the ordering $x_i$ appears before $m$ and now $m$ could not have been included. Since $x_i$ was arbitrary, we have the product of the variables dividing $m$ violating the primality of $M$. Therefore $m$ does not exist, and this generating subset is a subset of $\{x_1,...,x_n\}$.
(b) Since maximal ideals are prime ideals, if $M=(S)$ with $S≠\{x_1,...,x_n\}$, by part (a) we have $M \subset (x_1,...,x_n) \subset F[x_1,...,x_n]$, a contradiction.$~\square$
Saturday, May 25, 2013
Minimal Gröbner Bases' Commonalities (9.6.15)
Dummit and Foote Abstract Algebra, section 9.6, exercise 15:
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Fix a monomial ordering on $R=F[x_1,...,x_n]$.
(a) Prove $\{g_1,...,g_n\} \subseteq I \subseteq R$ for $I$ an ideal and monic $g_i$ is a minimal Gröbner basis for $I$ if and only if $\{LT(g_1),...,LT(g_n)\}$ is a minimal generating set for $LT(I)$.
(b) Prove the leading terms and order of a minimal Gröbner basis of $I$ is uniquely determined.
Proof: (a) $⇒$ By the definition of a minimal Gröbner basis, we have $(LT(g_1),...,LT(g_n))=LT(I)$ and $LT(g_i) \not \mid LT(g_j)$ for any $i≠j$. We can see that $(\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}) \subset I$ since for there to be equality would be, by exercise 10, to presume an element in $\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}$ divides $LT(g_i)$. $\Leftarrow$ By proposition 24 we have $\{g_1,...,g_n\}$ is a Gröbner basis of $I$ and since their leading terms is a minimal generating set for $I$ we must have $LT(g_i) \not \mid LT(g_j)$ for $i≠j$.
(b) Since $LT(I)$ is a monomial ideal, by the previous exercise its minimal generating set of monomials is uniquely determined. By part (a) the leading terms of a minimal Gröbner basis's elements must be uniquely determined, as well as the number of leading terms (which is the order of the basis).$~\square$
(a) Prove $\{g_1,...,g_n\} \subseteq I \subseteq R$ for $I$ an ideal and monic $g_i$ is a minimal Gröbner basis for $I$ if and only if $\{LT(g_1),...,LT(g_n)\}$ is a minimal generating set for $LT(I)$.
(b) Prove the leading terms and order of a minimal Gröbner basis of $I$ is uniquely determined.
Proof: (a) $⇒$ By the definition of a minimal Gröbner basis, we have $(LT(g_1),...,LT(g_n))=LT(I)$ and $LT(g_i) \not \mid LT(g_j)$ for any $i≠j$. We can see that $(\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}) \subset I$ since for there to be equality would be, by exercise 10, to presume an element in $\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}$ divides $LT(g_i)$. $\Leftarrow$ By proposition 24 we have $\{g_1,...,g_n\}$ is a Gröbner basis of $I$ and since their leading terms is a minimal generating set for $I$ we must have $LT(g_i) \not \mid LT(g_j)$ for $i≠j$.
(b) Since $LT(I)$ is a monomial ideal, by the previous exercise its minimal generating set of monomials is uniquely determined. By part (a) the leading terms of a minimal Gröbner basis's elements must be uniquely determined, as well as the number of leading terms (which is the order of the basis).$~\square$
Uniqueness of Minimal Generating Monomial Subsets of Ideals (9.6.14)
Dummit and Foote Abstract Algebra, section 9.6, exercise 14:
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Suppose $I$ is a monomial ideal in $F[x_1,...,x_n]$ minimally generated by $M=\{m_1,...,m_k\}$ for $m_i$ a monomial. Prove this generating subset is unique.
Proof: Note we must necessarily have $m_i \not \mid m_j$ for all $i≠j$ or else it is not a minimally generating set. Now, assume another minimally generating set $N=\{n_1,...,n_l\}$. Since $(M)=(N)=I$ we have by exercise 10 that $LT(n_1)=n_1$ is divisible by, say, $m_1$. Similarly, $m_1$ is divisible by a monomial from $N$. Continue in this fashion so that we must necessarily reach a point in the chain of division such that $m_i$ repeats, so that $m_i \mid ... \mid n_j \mid ... \mid m_i$ and now $m_i=n_j$. Starting again with an element $m$ in $M$ besides $m_i$, we can follow the chain once more which will reveal another equality between elements, which won't be the one between $m_i$ and $n_j$ as then $m \mid m_i$, a contradiction. In this fashion, every element of the minimally-ordered generating set will be paired up with its equal in the opposite set, i.e. will be a subset of the opposite set. By the nature of minimally generated sets, this inclusion cannot be proper, and now $M=N$.$~\square$
Proof: Note we must necessarily have $m_i \not \mid m_j$ for all $i≠j$ or else it is not a minimally generating set. Now, assume another minimally generating set $N=\{n_1,...,n_l\}$. Since $(M)=(N)=I$ we have by exercise 10 that $LT(n_1)=n_1$ is divisible by, say, $m_1$. Similarly, $m_1$ is divisible by a monomial from $N$. Continue in this fashion so that we must necessarily reach a point in the chain of division such that $m_i$ repeats, so that $m_i \mid ... \mid n_j \mid ... \mid m_i$ and now $m_i=n_j$. Starting again with an element $m$ in $M$ besides $m_i$, we can follow the chain once more which will reveal another equality between elements, which won't be the one between $m_i$ and $n_j$ as then $m \mid m_i$, a contradiction. In this fashion, every element of the minimally-ordered generating set will be paired up with its equal in the opposite set, i.e. will be a subset of the opposite set. By the nature of minimally generated sets, this inclusion cannot be proper, and now $M=N$.$~\square$
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