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\centerline{MATH 392 -- Seminar in Computational Commutative Algebra}
\centerline{Solutions for Midterm Exam}
\centerline{March 22, 2019}
\bigskip
\noindent
I.
\item{A)} (20) Prove that every ideal $I$ in the polynomial ring $k[x]$ (one variable)
is principal (that is, $I = \langle g(x)\rangle$ for some single polynomial).
\medskip
\item{} {\it Solution:}  Let $I$ be an ideal in $k[x]$.  If $I = \{0\}$, then we can take $g$ to be
the zero polynomial.  If $I$ contains nonzero polynomials, let $g(x)$ be any nonzero element of $I$
of minimal degree.  We must show that every $f(x) \in I$ is a multiple of $g(x)$.  So using the 
division algorithm in $k[x]$, write $f(x) = q(x)g(x) + r(x)$, where either $r(x)$ is identically 
zero, or else $\deg(r(x)) < \deg(g(x))$.  We have $r(x) = f(x) - q(x)g(x)$.  $f(x) \in I$ by 
assumption.  $g(x)$ is also in $I$ by construction.  $I$ is an ideal, hence closed
under products by arbitrary polynomials, so $q(x)g(x) \in I$.  Similarly $I$ is closed under
sums, so $r(x) = f(x) - q(x)g(x) \in I$.  But $g(x)$ was a nonzero element of $I$ of minimal
degree and $r(x)$ is either identically 
zero, or else $\deg(r(x)) < \deg(g(x))$.  Since the second alternative is not 
possible, $r(x) = 0$, which shows $f(x) = q(x)g(x)$.  This shows $I \subseteq \langle g(x)\rangle$.
The opposite inclusion is also true because $g(x) \in I$, so $\langle g(x) \rangle \subseteq I$
since $I$ is closed under multiplication by arbitrary polynomials.
\medskip
\item{B)} (10) Find $g(x)$ as in part A for the ideal $I = \langle x^2 + 7x + 10, x^3 + 2x^2 + 4\rangle$
in ${\bf Q}[x]$.
\medskip
\item{} {\it Solution 1:}  The polynomial $g(x)$ must be the gcd of the two generators for $I$.
We see by factoring that $x^2 + 7x + 10 = (x + 2)(x + 5)$ and $x^3 + x^2 + 4 = (x + 2)(x^2 - x + 2)$
and $x^2 - x + 2$ does not factor farther in ${\bf Q}[x]$ because this polynomial has no rational
(or even real) roots.  Hence the gcd is $x + 2$.  
\medskip
\item{} {\it Solution 2:}  We can also find the gcd using the Euclidean Algorithm in 
${\bf Q}[x]$.  Dividing:
$$\eqalign{x^3 + 2x^2 + 4 &= (x - 6)(x^2 + 7x + 10) + 32x + 64\cr
           x^2 + 7x + 10 &= \left({x \over 32} + {5\over 32}\right)(32 x + 64) + 0\cr}$$
The last nonzero remainder is the gcd, up to a constant multiple.  Note $32 x + 64 = 32(x + 2)$
so the monic gcd is $x + 2$ as in the first solution.   
\bigskip
\noindent
II.  
\item{A)} (15) Define:  $G$ is a Gr\"obner basis for an ideal $I \subset k[x_1,\ldots,x_n]$ with respect to a monomial
order $>$.
\medskip
\item{}  {\it Solution:}  A Gr\"obner basis for $I$ with respect to $>$ is a finite collection of polynomials
$G = \{g_1,\ldots,g_t\} \subset I$ such that the monomial ideal $\langle LT_>(I)\rangle$ is generated by 
$LT_>(g_1),\ldots,LT_>(g_t)$.  (Equivalently you could also say it is a finite collection of 
polynomials $G$ as above such that for every nonzero $f \in I$, $LT_>(f)$ is divisible by 
$LT_>(g_i)$ for some $i$.)
\medskip     
\item{B)} (10) Assuming the statement of Dickson's Lemma, prove that Gr\"obner bases exist for every ideal $I$
in $k[x_1,\ldots,x_n]$ and with respect to every monomial order.
\medskip
\item{} {\it Solution:}  We may assume that $I \ne \{0\}$, since in that case we can take $G = \emptyset$. 
Dickson's Lemma is the statement that every monomial ideal in $k[x_1,\ldots,x_n]$
is generated by a {\it finite set} of monomials.  We apply that result to the monomial ideal 
$\langle LT_>(I) \rangle = \langle LT_>(f) \mid f \in I\rangle$.  This says $\langle LT_>(I)\rangle$
is generated by some finite collection of monomials $x^{\alpha(1)}, \ldots, x^{\alpha(t)}$.  Moreover, 
by definition each of those is equal to the leading term of some element in $I$:
$$x^{\alpha(i)} = LT_>(g_i),$$
for some $g_i \in I$.  By the definition (part A), this says $\{g_1,\ldots,g_t\}$ is 
a Gr\"obner basis for $I$.  
\medskip
\item{C)} (5) What else do you need to know in order for the result from part B to give a proof
of the Hilbert Basis Theorem?  (You don't need to give the proof, just say what else must be proved.) 
\medskip 
\item{} {\it Solution:}  The other fact that must be proved to get a proof of the Hilbert
Basis Theorem (the statement that every ideal in $k[x_1,\ldots,x_n]$ has a finite
basis) is that a Gr\"obner basis for $I$ is also an ideal basis for $I$, or equivalently
that if $G$ is a Gr\"obner basis for $I$ and $f \in I$, then the remainder on division
of $f$ by $G$ is zero:  $\overline{f}^G = 0$.  This follows from the definition of 
a Gr\"obner basis and the properties of the Division Algorithm.
\bigskip 
\noindent
III.  
\item{A)} (20)  State and prove the Elimination Theorem.  
\medskip
\item{} {\it Solution:}  Let $I$ be an ideal in $k[x_1,\ldots,x_n]$
and let $I_\ell = I \cap k[x_{\ell + 1}, \ldots, x_n]$ be the elimination ideals for $\ell = 1,\ldots, n-1$.
The Elimination Theorem states that if $G$ is a Gr\"obner basis for $I$ with respect to the lexicographic
order with $x_1 > x_2 > \ldots > x_n$, then 
$G_\ell = G \cap k[x_{\ell + 1}, \ldots, x_n]$ is a Gr\"obner basis of $I_\ell$,
for all $\ell = 1,\ldots, n-1$.   Proof:  We must show that if $f$ is any element of $I_\ell$, then 
$LT_{lex}(f)$ is divisible by one of the leading terms of the elements of $G_\ell$.  But if $f \in I_\ell$,
then $f$ depends only on the variables $x_{\ell+1},\ldots,x_n$, and the same is true for $LT_{lex}(f)$.  
Since $G$ is a Gr\"obner basis for $I$, $LT_{lex}(f)$ is divisible by $LT_{lex}(g)$ for some $g \in G$.
But this means that $LT_{lex}(g)$ can only depend on the variables $x_{\ell + 1}, \ldots, x_n$.  
By the properties of the lex order with 
$$x_1 > x_2 > \cdots > x_\ell > x_{\ell + 1} > \cdots x_n,$$
any monomial containing any of the variables $x_1,\ldots,x_\ell$ is {\it greater than} all monomials
containing only the variables $x_{\ell+1},\ldots,x_n$.  This means that no term in $g$ can contain
any of the variables $x_1,\ldots,x_\ell$.  Hence by definition $g \in G_{\ell}$.  We have shown 
that for every nonzero $f \in I_\ell$, $LT_{lex}(f)$ is divisible by $LT_{lex}(g)$ for some 
$g \in G_\ell$.  This shows that $G_{\ell}$ is a Gr\"obner basis for $I_\ell$ by the definition.
\medskip
\item{B)} (10)  A certain ideal $J \subset {\bf Q}[x,y,z]$ has a Gr\"obner basis
$$B = \{x^3 - 3x^2 + 2x, x^2 y - xy, y^2 - y, z - xy\}$$
with respect to the lexicographic order with $z > y > x$.   What are bases for 
the elimination ideals 
$$J_1 = J \cap {\bf Q}[y,x] \quad {\rm and} \quad J_2 = J\cap {\bf Q}[x]?$$
\item{} {\it Solution:}  $J_1 = \langle x^3 - 3x^2 + 2x, x^2 y - xy, y^2 - y\rangle$
and $J_2 = \langle x^3 - 3x^2 + 2x\rangle$.  
\medskip
\item{C)} (10)  Use the information in part B to determine all of the points in $V(J)$.
\medskip
\item{} {\it Solution:}  We begin by setting the generator of $J_2$ equal to zero and by
factoring we find $x(x - 1)(x - 2) = 0$, so $x = 0$, $x = 1$, or $x = 2$.  
If we substitute $x = 0$ in the rest of the Gr\"obner basis, we find:
$$B|_{x=0} = \{0,0,y^2-y,z\}$$
From this we see $y(y - 1) = 0$ so $y = 0$ or $y = 1$.  And then with either of those $y$-values,
$z = 0$.  
So we have two points with $x = 0$, namely $(0,0,0)$ and $(0,1,0)$.  
If we substitute $x = 1$
in the rest of the Gr\"obner basis, we find:
$$B|_{x=1} = \{0,0,y^2 - y, z - y\}$$
Hence we find two more points $(1,0,0)$ and $(1,1,1)$.
Finally, if we substitute $x = 2$ into the rest of the Gr\"obner basis, we find
$$B|_{x=2} = \{0,2y,y^2 - y,z-2y\}$$
The only solution is $(2,0,0)$. 
This means that $V(J)$ consists of five points in all:
$$V(J) = \{(0,0,0), (0,1,0), (1,0,0), (1,1,1), (2,0,0)\}.$$
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