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My question

Let $K$ be a field and let $I \subseteq K[x_1,\dots, x_n]$ be an ideal. Now does the following equality always hold?

$$ (I \cdot K[x_1,\dots, x_n,y]) \ \cap K[x_1,\dots,x_n] \quad = \quad I $$

Main thoughts

I am aware that for many rings $R$ (in stead of $K[x_1,\dots, x_n]$) the inclusion ''$\subseteq$'' doesn't need to hold. I think in this particular case''$\subseteq$'' does hold, even though adding a polynomial variable to a ring does not give an integral extension. I find it hard to show that "$\subseteq$" is true. Could someone please show me how this works?

Those polynomial rings are Noeterian, so we can write $$ I \ = \ (h_1,\dots, h_n)_{K[x_1,\dots,x_n]}. $$ An element $ f \in I \cdot K[x_1, \dots, x_n]$ can therefore be written as $$ f_1h_1 + \dots + f_nh_n \quad \text{ for some } \quad f_i \in K[x_1,\dots, x_n, y]. $$ We need to show that is $f$ has no $y$'s in it, then the $f_i$ don't have $y$'s either.

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    Why not just write it as a polynomial in $y$ and show that the coefficient of $y^k, k>0$ must be zero? – 2017-01-24
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    If $I \subset R$ then $I. R[x] \cap R =I. R \cap R = I$ – 2017-01-24
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    If $R \subset S$ is a ring extension, then $I \subset I^{ec}$ always holds, and $I = I^{ec}$ for all ideals $I$ of $R$ iff $S$ is a faithfully flat $R$-algebra. Here $S = R[y]$ is a free $R$-algebra, where $R = k[x_1,\dots,x_n]$, so it is faithfully flat, and your claim always holds. See http://math.stackexchange.com/questions/194261 – 2017-01-24

2 Answers 2

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If $R$ is any unital commutative ring and $I$ is an ideal in $R$, then an element in $I\cdot R[x]$ has the form $\sum a_i f_i(x)$ where the $a_i$ are in $I$. Writing $f_i(x)=\sum b_{i,j} x^j$ we get that $\sum a_i f_i(x)=\sum_j (\sum_i a_i b_{i,j})x^j$. If this element is in $R$, i.e. a constant, then the coefficient of $x^j, j>0$ are zero, so it has the form $\sum a_i b_{i,0}$ where the $a_i\in I$ and $b_{i,0}\in R$, so it is in $I\cdot R$. Since $I$ is an ideal, this element must be in $I$, so you get that $I\cdot R[x]\cap R\subseteq I$ (and the other inclusion is trivial).

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Choose a monomial order on $K[x_1, \dotsc, x_n,y]$, where the monomials are first ordered by $y$ and then by a monomial order on $K[x_1, \dotsc, x_n]$. Note that this is an elimination ordering for $x_1, \dotsc, x_n$.

Any Groebner basis for $I \subset K[x_1, \dotsc, x_n]$ will be a Groebner basis for $I \cdot K[x_1, \dotsc, x_n,y] \subset K[x_1, \dotsc, x_n,y]$ by the Buchberger criterion.

Thus by the elimination theory, this will also be a Groebner basis for $(I \cdot K[x_1, \dotsc, x_n,y]) \cap K[x_1, \dotsc, x_n]$, hence we have $$(I \cdot K[x_1, \dotsc, x_n,y]) \cap K[x_1, \dotsc, x_n]=I$$.