# Proving inequality about dimension of quotient vector spaces

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• elias001
elias001
TL;DR Summary
solution verification for showing ##d_1+\cdots +d_n-n+1 \leq {\text{dim}}_k k[x_1,\ldots,x_n]/\mathfrak{a}##
The following are from Froberg's "Introduction to Grobner bases" , and Hungerford's undergraduate "Abstract Algebra" text, and also a continuation of this [post](https://math.stackexchange.com/questions/4947061/questions-about-rhs-inequality-of-d-1-cdots-d-n-n1-leq-textdim-k-k?noredirect=1#comment10585303_4947061)

##\textbf{Background}##
##\textbf{Theorem 1:}## ##k[x_1,\ldots,x_{n-1},x_n]\backsimeq (k[x_1,\ldots,x_{n-1}])[x_n]##
##\textbf{Theorem 2:}## Let ##K## be an extension field of ##F## and ##u\in K## an algebraic element over ##F## with minimal polynomial ##p(x)## of degree ##n##. Then ##(1)## ##F(u)\equiv F(x)/(p(x))##
##(2)## ##\{1_F,u,u^2,\ldots,u^{n-1}\}## is a basis of the vector space ##F(u)## over ##F##.
##(3)## ##[F(u):F]=n##.

##\textbf{Assumed Exercise:}## Let ##\mathfrak{a}## be an ideal generated by monomials in ##k[x_1,\ldots,x_n]##. Show that ##k[x_1,\ldots,x_n]/\mathfrak{a}## is a finite dimensional vector space over ##k## if and only if for each ##i## there is a ##d_i>0##, such that ##{x_i^{d_i}}\in \mathfrak{a}##.

##\textbf{Exercise:}## Let ##\mathfrak{a}## be a monomial ideal such that ##\{x_i^{d_i},\cdots,{x_n^{d_n}}\}## is a part of a minimal system of generators for ##\mathfrak{a}##, where ##d_i## are positive. Show that
$$(LHS)\quad d_1+\cdots d_n-n+1 \leq {\text{dim}}_k k[x_1,\ldots,x_n]/\mathfrak{a}\leq d_1d_2\cdots d_n\quad (RHS)$$
##\textbf{Questions}##

I am trying to do the above **Exercise** using the [solution](https://math.stackexchange.com/questions/4946665/showing-that-kx-1-ldots-x-n-mathfraka-is-a-finite-dimensional-vector-spa?noredirect=1#comment10577454_4946665) to **Assumed Exercise** above and basic abstract algebra knowledge.

Below is my attempted solution to the left hand side inequality:

##\textbf{Attempted Solution for (LHS) inequality:}##

From **Assumed Exercise** above, we know that ##k[x_1,\ldots,x_n]/\mathfrak{a}## is a finite dimensional (quotient) vector space, hence we need to count the minimum number of basis elements. We can set the ideal ##\mathfrak{a}\subset k[x_1,\ldots,x_n]## to be ##\mathfrak{a}=\langle x_1,\ldots,x_n \rangle##, and ##x_i^{d_i}\in \langle x_i \rangle## for ##i=1,2,3,\ldots, n##. Also, by **Theorem 1** above, ##k[x_1,\ldots,x_{n-1},x_n]\backsimeq (k[x_1,\ldots,x_{n-1}])[x_n]##, we can reduce to the case from ##n## indeterminates to the case of a single indeterminate for the vector space ##k[x_1,\ldots,x_n]/\mathfrak{a}##.

So let ##F= (k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]), u=x_i, F(u)=(k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}])(x_i),## like the hypothesis of **Theorem 2** above. Using ##(2)## of **Theorem 2**, each ##x_i\in \mathfrak{a}## is a monomial of maximal (is it maximum or total?) degree ##d_i##. Using ##(2)## of **Theorem 2** one more time, there exist ##\{1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]},x_i,x_i^2,\ldots,x_i^{{d_i}-1}\}## as a basis of the vector space ##F(u)=(k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}])(x_i)## over ##\mathfrak{a}##. There are ##d_i## number of elements in the basis ##\{1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]},x_i,x_i^2,\ldots,x_i^{{d_i}-1}\}##.

Since there are ##i=n## indeterminates of the ##x_i## and for each ##x_i##, there are ##d_i## number of basis elements.

Hence there are ##d_1+d_2+\cdots +d_n## total number of basis elements. However each basis contains the identity element ##1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]}##, and they are all equal to each other, since the identity element of a vector space is unique. Hence, the identity element has been double counted ##n-1## times. So the dimension of ##k[x_1,\ldots,x_n]/\mathfrak{a}## is at least ##d_1+d_2+\cdots +d_n-(n-1)=d_1+d_2+\cdots +d_n-n+1##

##\textbf{Can someone comment}## on whether I proved the ##\text{LHS}## of the above **Exercise** correctly please? I am not sure of the following two counting argument step is correct. It has to do with my intepretations of meaning of ##d_i## for arriving at ##d_1+\cdots +d_n## and also and my understanding of the property of the identity elemetn to arrive at subtracting ##n-1##.

>##\textbf{(1)}## Using ##(2)## of **Theorem 2**, each ##x_i\in \mathfrak{a}## is a monomial of maximal (is it maximum or total?) degree ##d_i##. Using ##(2)## of **Theorem 2** one more time, there exist ##\{1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]},x_i,x_i^2,\ldots,x_i^{{d_i}-1}\}## as a basis of the vector space ##F(u)=(k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}])(x_i)## over ##\mathfrak{a}##. There are ##d_i## number of elements in the basis ##\{1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]},x_i,x_i^2,\ldots,x_i^{{d_i}-1}\}##.

Since there are ##i=n## indeterminates of the ##x_i## and for each ##x_i##, there are ##d_i## number of basis elements.

>##\textbf{(2)}## However each basis contains the identity element ##1_{F=k[x_1,x_2,\ldots x_{i-1} x_{i+1}\ldots x_{n}]}##, and they are all equal to each other, since the identity element of a vector space is unique.

Moderator Note: reformatted according to our version of MathJax, see
https://www.physicsforums.com/help/latexhelp/

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