MathB.in - Share Mathematics with LaTeX and Markdown

Question: Let $K$ be a field. An ideal $I$ of $K[X_1,...,X_n]$ is called a monomial ideal if $I$ can be generated by a set of monomials (that is, a set of elements of the form $cX_1^{m_1}...X_n^{m_n}$ for $c \in K$ and nonnegative integers $m_i$). Show that if $I$ is a monomial ideal and $f$ is an element of $I$, then every monomial appearing in $f$ is also an element of $I$.

Solution:
Since $K$ is noetherian, by the Hilbert basis theorem so is $K[X_1,...,X_n]$, and in particular any ideal of $K[X_1,...,X_n]$ is finitely generated. Let $f \in I$ and suppose $cX_{i_1}^{a_1}...X_{i_k}^{a_k}$ , $a_i \geq 1$ and $c \neq 0$ is a monomial of $f$, suppose for the sake of a contradiction, that this monomial is not in $I$. Then if $M$ is a set of monomials generating $I$, $M$ does not contain any monomial in $H = \{cX_{i_1}^{b_1}...X_{i_k}^{b_k} : 0 \leq b_i \leq a_i, c \neq 0 \}$ or else $I$ would contain $cX_{i_1}^{a_1}...X_{i_k}^{a_k}$. Since $I$ is an ideal of a noetherian ring, it is finitely generated, say by the polynomials $P_1,...,P_s$. Moreover no monomial in $H$ may appear in a $P_j$, since $M$ is a generating set for $I$ and does not contain any monomial in $H$, and $P_j \in (M)$. But then any polynomial of the form $U_1P_1 + ... U_sP_s$ does not contain a monomial in $H$, this is a contradiction since $f$ must be of this form but contains the monomial $cX_{i_1}^{a_1}...X_{i_k}^{a_k}$.

Sat, 12 Dec 2020 19:41 GMT