Computing with ideals: over $K$ or over $\mathbb{Q}\subseteq K$? does it matter? pt.2 Definitions: Let $R$ be any commutative ring with $1$, and let $S\subseteq R$ and $I\unlhd R$. The ideal generated by $S$ is $\langle S\rangle_R:=\bigcap\{J; S\subseteq J\unlhd R\}$, i.e. the intersection of all ideals of $R$ that contain $S$. If $R'\subseteq R$ are commutative rings and $S\subseteq R'$, then clearly $\langle S\rangle_{R'}\subseteq\langle S\rangle_R$. We also have $\langle S\rangle_R\!=\!\langle S'\rangle_R\Leftrightarrow S\!\subseteq\!\langle S'\rangle_R, S'\!\subseteq\!\langle S\rangle_R$, a useful criterium when checking concrete cases. An easy exercise shows that $\langle S\rangle_R\!=\!\sum RS$, i.e. the set of all finite sums of elements of the form $rs$, where $r\!\in\!R, s\!\in\!S$. In particular, $\langle s_1,\ldots,s_k\rangle_R\!=\!Rs_1+\ldots+Rs_k$.
The radical of $I$ is $\sqrt{I}:=\{r\in R; \;\exists n\!\in\!\mathbb{N}: r^n\!\in\!I\}\unlhd R$. Clearly $I\subseteq\sqrt{I}$ and $I\subseteq J\Rightarrow \sqrt{I}\subseteq\sqrt{J}$. The colon ideal is $I:S:=\{r\!\in\!R;\; rS\!\subseteq\!I\}\unlhd R$. Clearly $I:S=I:\langle S\rangle_R$.

Motivation/Example: We have the following exercise. Suppose $K$ is a field of characteristic $0$ (this implies $\mathbb{Q}\subseteq K$). For polynomials $f,g,p,q\!\in\!K[x,y,z]$, given by $f\!=\!x^4\!+\!x^3y\!+\!x^3z^2\!-\!x^2y^2\!+\!x^2yz^2\!-\!xy^3\!-\!xy^2z^2\!-\!y^3z^2$, $\,g\!=\!x^4\!+\!2x^3z^2\!-\!x^2y^2\!+\!x^2z^4\!-\!2xy^2z^2\!-\!y^2z^4$, $\,p\!=\!x^2\!+\!xy\!+\!xz\!+\!yz$, $\,q\!=\!x^2\!-\!xy\!-\!xz\!+\!yz$, compute the following:


*

*find the Grobner basis of $\langle f,g\rangle_{K[x,y,z]}+\langle p,q\rangle_{K[x,y,z]}=\langle f,g,p,q\rangle_{K[x,y,z]}$ w.r.t. degrevlex.

*find some generating set of $\langle f,g\rangle_{K[x,y,z]}\cap\langle p,q\rangle_{K[x,y,z]}$;

*find some generating set of $\sqrt{\langle f, g\rangle_{K[x,y,z]}}$;

*find some generating set of $\langle f,g\rangle_{K[x,y,z]}:\langle p,q\rangle_{K[x,y,z]}$.


Solution (when $K=\mathbb{Q}$): These polynomials are quite large, so computing by hand is highly impractical. Therefore we use a computer. The SINGULAR code 
ring R=0,(x,y,z),dp;     LIB "primdec.lib";
ideal I=x4+x3y+x3z2-x2y2+x2yz2-xy3-xy2z2-y3z2, x4+2x3z2-x2y2+x2z4-2xy2z2-y2z4;
ideal J=x2+xy+xz+yz, x2-xy-xz+yz;
groebner(I+J);
intersect(I,J);
radical(I);
quotient(I,J);

tells us that: 


*

*the Grobner basis of $\langle f,g,p,q\rangle_{\mathbb{Q}[x,y,z]}$
w.r.t. the monomial order degrevlex is $\{xy\!+\!xz,\,
x^2\!-\!xy\!-\!xz\!+\!yz,\, y^2z\!+\!yz^2\}$;

*$\langle f,g\rangle_{\mathbb{Q}[x,y,z]}\cap\langle
p,q\rangle_{\mathbb{Q}[x,y,z]}$ $=$ $\langle
-\!2x^3z^2\!-\!2x^2yz^2\!+\!2xy^2z^2\!+\!2y^3z^2\!-\!2x^4\!-\!2x^3y\!+\!2x^2y^2\!+\!2xy^3$,
$
-2x^2z^4\!+\!2y^2z^4\!-\!4x^3z^2\!+\!4xy^2z^2\!-\!2x^4\!+\!2x^2y^2\rangle_{\mathbb{Q}[x,y,z]}$;

*$\sqrt{\langle f, g\rangle_{\mathbb{Q}[x,y,z]}}=\langle
x^2z^2\!-\!y^2z^2+x^3\!-\!xy^2\rangle_{\mathbb{Q}[x,y,z]}$;

*$\langle f,g\rangle_{\mathbb{Q}[x,y,z]}:\langle
p,q\rangle_{\mathbb{Q}[x,y,z]}$ $=$ $\langle
x^3z^2\!+\!x^2yz^2\!-\!xy^2z^2\!-\!y^3z^2\!+\!x^4\!+\!x^3y\!-\!x^2y^2\!-\!xy^3,$ 
$x^2z^4\!-\!y^2z^4\!-\!2x^2yz^2\!+\!2y^3z^2\!-\!x^4\!-\!2x^3y\!+\!x^2y^2\!+\!2xy^3\rangle_{\mathbb{Q}[x,y,z]}$.


Question: can we now deduce that these facts also hold for ideals in $K[x,y,z]$?

QUESTIONS: Let $k\subseteq K$ be fields, $\mathbf{x}=(x_1,\ldots,x_n)$, and $G,S,S',S''\subseteq k[\mathbf{x}]\leq K[\mathbf{x}]$. From the first post, we know that $\langle S\rangle_{k[\mathbf{x}]}=\langle S\rangle_{K[\mathbf{x}]}\cap k[\mathbf{x}]$ $(\ast)$. How can I prove:


*

*$G$ is a Grobner basis of $\langle G\rangle_{k[\mathbf{x}]}$
$\;\Longleftrightarrow\;$ $G$ is a Grobner basis of $\langle
G\rangle_{K[\mathbf{x}]}$, w.r.t. a monomial order on $K[\mathbf{x}]$. 

*$\langle S'\rangle_{k[\mathbf{x}]}\cap\langle S''\rangle_{k[\mathbf{x}]}=\langle
    S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\langle S'\rangle_{K[\mathbf{x}]}\cap\langle S''\rangle_{K[\mathbf{x}]}=\langle
    S\rangle_{K[\mathbf{x}]}$

*$\sqrt{\langle S'\rangle_{k[\mathbf{x}]}}=\langle S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\sqrt{\langle S'\rangle_{K[\mathbf{x}]}}=\langle S\rangle_{K[\mathbf{x}]}$

*$\langle S'\rangle_{k[\mathbf{x}]}:S''=\langle
S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\langle S'\rangle_{K[\mathbf{x}]}:S''=\langle
S\rangle_{K[\mathbf{x}]}$



What I have done so far:
(1) We must prove that $I\!:=\!\langle ldm(g); g\!\in\!\mathcal{G}\rangle_{k[\mathbf{x}]}\!=\!\langle ldm(f); f\!\in\!\langle\mathcal{G}\rangle_{k[\mathbf{x}]}\rangle_{k[\mathbf{x}]}\!=:\!\overline{I}$ iff $J\!:=\!\langle ldm(g); g\!\in\!\mathcal{G}\rangle_{K[\mathbf{x}]}\!=\!\langle ldm(f); f\!\in\!\langle\mathcal{G}\rangle_{K[\mathbf{x}]}\rangle_{K[\mathbf{x}]}\!=:\!\overline{J}$.
$(\Leftarrow)$: $(\subseteq)$: $\checkmark$. $(\supseteq)\!:$ If $h\!\in\!\overline{I}$, then $h\!\in\!\overline{J}$, so by assumption $h\!\in\!J$ holds, hence by $(\ast)$ we have $h\!\in\!I$. $(\Rightarrow)$: $(\subseteq)$: $\checkmark$. $(\supseteq)$: ???.
(2) $(\Leftarrow)$ By $(\ast)$, we have $\langle S'\rangle_{k[\mathbf{x}]}\cap\langle S''\rangle_{k[\mathbf{x}]}=\langle S'\rangle_{K[\mathbf{x}]}\cap\langle S''\rangle_{K[\mathbf{x}]}\cap k[\mathbf{x}]=\langle S\rangle_{K[\mathbf{x}]}\cap k[\mathbf{x}]=\langle S\rangle_{k[\mathbf{x}]}$.
$(\Rightarrow)$ $(\supseteq)$: From $\langle S'\rangle_{K[\mathbf{x}]}\cap\langle S''\rangle_{K[\mathbf{x}]}\supseteq\langle S'\rangle_{k[\mathbf{x}]}\cap\langle S''\rangle_{k[\mathbf{x}]}\supseteq S$ follows $\langle S'\rangle_{K[\mathbf{x}]}\cap\langle S''\rangle_{K[\mathbf{x}]}\supseteq\langle S\rangle_{K[\mathbf{x}]}$.
$(\subseteq)$: ???
(3) $(\Leftarrow)$: $(\subseteq)$: If $f\in\sqrt{\langle S'\rangle_{k[\mathbf{x}]}}\subseteq\sqrt{\langle S'\rangle_{K[\mathbf{x}]}}=\langle S\rangle_{K[\mathbf{x}]}$, then by $(\ast)$ it follows that $f\in\langle S\rangle_{k[\mathbf{x}]}$ holds. $(\supseteq)$: If $f\in\langle S\rangle_{k[\mathbf{x}]}\subseteq\langle S\rangle_{K[\mathbf{x}]}=\sqrt{\langle S'\rangle_{K[\mathbf{x}]}}$, then $f^m\in\langle S'\rangle_{K[\mathbf{x}]}$ for some $m$, so by $(\ast)$, $f^m\in\langle S'\rangle_{k[\mathbf{x}]}$.
$(\Rightarrow)$: ???
(4) $(\Leftarrow)$: $(\subseteq)$: If $f\in\langle S'\rangle_{k[\mathbf{x}]}:S''$, then $fS''\subseteq\langle S'\rangle_{k[\mathbf{x}]}\subseteq\langle S'\rangle_{K[\mathbf{x}]}$, so $f\in\langle S'\rangle_{K[\mathbf{x}]}:S''=\langle   S\rangle_{K[\mathbf{x}]}$, and by $(\ast)$, $f\in\langle S\rangle_{k[\mathbf{x}]}$.
$(\supseteq)$: If $f\in\langle
S\rangle_{k[\mathbf{x}]}\subseteq\langle
S\rangle_{K[\mathbf{x}]}=\langle S'\rangle_{K[\mathbf{x}]}:S''$, then $fS''\subseteq\langle S'\rangle_{K[\mathbf{x}]}$, and clearly $fS''\subseteq k[\mathbf{x}]$, so by $(\ast)$, $fS''\subseteq\langle S'\rangle_{k[\mathbf{x}]}$, i.e. $f\in\langle S'\rangle_{k[\mathbf{x}]}:S''$. $(\Rightarrow)$: ???

 A: Partial answer too long to fit in comments. 
(2) See the edidted answer to your first post. 
(3) ($\Rightarrow$). Let $I'$ be the ideal of $k[{\bf x}]$ generated by $S'$. The question is whether $\sqrt{I'K[{\bf x}]}=\sqrt{I'} K[{\bf x}]$. It is easy to see that
$$ I'K[{\bf x}]\subseteq \sqrt{I'}K[{\bf x}]\subseteq \sqrt{I'K[{\bf x}]}.$$
Therefore, the desired equality is equivalent to $\sqrt{I'}K[{\bf x}]$ being
radicial, or that the quotient $K[{\bf x}]/\sqrt{I'}K[{\bf x}]$ is reduced. The latter is same as $(k[{\bf x}]/I')\otimes_k K$. It is known that the tensor product is reduced when $K/k$ is a separable extension (e.g. if $k$ has characteristic $0$ or is a finite field). So $\Rightarrow$ holds when $K/k$ is separable. Otherwise it is false (Consider $S'=\{ x^p - \lambda \}$ if $k$ has characteristic $p>0$ and $\lambda\in k \setminus k^p$, $K=k[\lambda^{1/p}]$). 
(4) ($\Rightarrow$). The colon ideal $J=\langle S\rangle_{k[{\bf x}]}$ is the annihilator ideal of the $k[{\bf x}]$-module $M:=(I'+I'')/I'$. We want to know whether $JK[{\bf x}]$ is the annihilator of $(I'K[{\bf x}]+I''K[{\bf x}])/(I'K[{\bf x}])$. The latter is equal to $M\otimes_{k[{\bf x}]} K[{\bf x}]$. We observe that the annihilator ideal $J$ fits into an exact sequence 
$$ 0 \to J\to k[{\bf x}]\to \mathrm{End}_{k[{\bf x}]}(M) $$ 
of $k[{\bf x}]$-modules, where the last map sends $a$ to the multiplication-by-$a$ map. 
As $k[{\bf x}]\to K[{\bf x}]$ is flat and $M$ is finitely generated, the above exact sequence remains exact after tensoring $\otimes_{k[{\bf x}]} K[{\bf x}]$. But we have a canonical isomorphism 
$$ \mathrm{End}_{k[{\bf x}]}(M) \otimes_{k[{\bf x}]} K[{\bf x}]\to \mathrm{End}_{K[{\bf x}]}(M\otimes_{k[{\bf x}]}K[{\bf x}])$$
(see e.g. here). So
the new exact sequence is 
$$ 0\to J\otimes_{k[{\bf x}]} K[{\bf x}]=JK[{\bf x}]\to K[{\bf x}]\to \mathrm{End}(M\otimes_{k[{\bf x}]} K[{\bf x}])$$ 
This shows that $JK[{\bf x}]$ is the annihilator of $M\otimes_{k[{\bf x}]} K[{\bf x}]$ and we are done. This is a little complicated especially if you are not familiar with commutative algebra. I don't know whether there is a more elementary proof (probably using a basis of $K$ as $k$-vector space). 
