Monomial ideal as a vector space

Question

I'm to prove the following statement:

Let $ K $ be a field.

And ideal $ I $ in $ K[x_1, \dots, x_n] $ is monomial (generated by monomials in $ x_1, \dots, x_n $) $ \iff $ it is spanned on monomials as a vector space over $ K $.

and

The minimal set of generators of a monomial ideal is uniquely defined and it is finite.

As to the first: I always believed that every polynomial can be expressed as a sum of monomials in $ x_1, \dots, x_n $, so basically every ideal in $ K[x_1. \dots, x_n] $ is spanned by some monomials. How should I understand it?

Answer 1

There are two basic remarks about monomial ideals:

1. A monomial belongs to $I$ iff it is divisible by one of the generators of $I$;

2. A polynomial belongs to $I$ iff all its monomials belong to $I$.

With these two properties at hand one can try to prove what you want.

"$\Leftarrow$" This should be clear.

"$\Rightarrow$" $I$ must be spanned over $K$ by all monomials which are divisible by a monomial in a system of generators of $I$.

(For the second maybe you should make it clear what means "minimal".) First note that a monomial ideal is finitely generated by Hilbert Basis Theorem. Furthermore, from 2. above one can choose a finite system of generators consisting of monomials, say $m_i$. Let's assume we can't get rid any of them. In particular, they can't divide each other. If consider another minimal system of monomial generators, say $n_j$, then from 1. above some $n_j$ divides $m_i$, and some $m_k$ divides $n_j$. This is possible only if $k=i$ and then $m_i=n_j$, and so on.