I am working on two questions for my Commutative Algebra assignment and am struggling to finish them.
$1.$ Let $S=K[x_1,...,x_n]$, $I\subset S$ an ideal and $<$ a term order. I first showed that $S/I$ is a vector space over $K$ and now I have to show that the monomials that are not in $in_<(I)$ form a $K$-basis for $S/I$.
I started with the following: let $M$ be the set of all monomials and $B=M-J$, where $J=in_<(I)$. I want to show that $B$ is a basis. To show that the monomials span $S/I$, I wrote $f+I\in S/I$ as $f_1+f_2+I$, where $f_1$ is a sum of monomials in $B$ and $f_2$ is the sum of monomials in $J$. Now I conclude that $f+I=f_1+I$ if $f_2\in I$, but is this true? If $x^u\in in_<(I)$, we do not have necessarily $x^u\in I$, right? And what about linear independence?
$2.$ Show that if $x^u\in in_<(I)$, then there is $f\in I$ such that $x^u=in_<(f)$.
I don't really know what to do here. Should I fix a finite basis for I? Or a Grobner basis and work with that? The suggestion is to use $in_<(fg)=in_<(f)in_<(g)$ but I don't see how to use it.
Any suggestion is appreciated! Many thanks
For the first question, let $f\in S/I$. As you did, write $f$ as $f=f_1+f_2$ where $f_1$ is a sum of monomials in $B$ and $f_2$ is the sum of monomials in $J$. Let's focus on $f_1$. For each $x^u \in in(I)$, there is $f\in I$ such that $x^u = in(f)$, let's say
$$f = x^u + \sum_i k_i x^{\alpha_i}$$
where $x^{\alpha_i}<x^u$ for each $i$. Then $x^u+I = -\sum_i k_i x^{\alpha_i} +I$. You can iterate this process to show that $f_1$ is equivalent to some polynomial that is a sum of monomials in $B$ (you need to use the fact that $<$ has a minimum).
Linear independence is easier, since a linear combination of monomials of $B$ that lies in $I$ leads to an element of $I$ whose leading term lies in $B$.
For the second exercise, try to prove the following lemma
Use the fact that $I$ is an ideal and this lemma to get such $f$.