# Prove $\langle x \rangle + \cap P_i = \cap (\langle x \rangle + P_i)$ for monomials ideals.

I'm trying to follow the proof of this fact on this book on Gröbner basis. At one point of the proof they use this equality that I do not fully understad. I wonder if it is evident:

Given prime monomial ideals $P_i$ (which are generated by sets of variables) in $K[X_1,\ldots,X_n]$ and another variable $x$ (that may or not be a generator in some ideals $P_i$) then I would need to show that $\langle x \rangle + \cap P_i = \cap (\langle x \rangle + P_i)$ where intersections are finite.

The inclusion $\subseteq$ is clear. For the other one I'm having more problems. The intuition is that each $\cap P_i$ is computed by taking $lcm(X_i,X_j) = X_iX_j$ and therefore consists of monomials. Can you help me to prove the other inclusion?

• This is quite easy. First of all assume that you have only two prime ideals. If a monomial $m$ belongs to $(x,P_1)\cap(x,P_2)$ then it is divisible by $x$ (and you are done) or by some generator of $P_1$ and $P_2$. – user26857 Feb 2 '18 at 17:05

In general, if $J$ is a monomial ideal and $(I_i)_i$ is a family of monomial ideals of $K[x_1,\ldots,x_n]$ then $$\bigcap_{i} (J+I_i) = J + \bigcap_i I_i.$$ This is easy to prove. The inclusion $\supseteq$ is clear. Conversely, let $u\in \bigcap_{i} (J+I_i)$ be a monomial. If $u\in J$ then there is nothing to prove. If $u\notin J$, since $u\in J+I_i$, then $u\in I_i$. Hence $u\in \bigcap_i I_i$.