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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:

  1. 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.
  2. find some generating set of $\langle f,g\rangle_{K[x,y,z]}\cap\langle p,q\rangle_{K[x,y,z]}$;
  3. find some generating set of $\sqrt{\langle f, g\rangle_{K[x,y,z]}}$;
  4. 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:

  1. the Grobner basis of $\langle f,g,p,q\rangle_{\mathbb{Q}[x,y,z]}$ w.r.t. the monomial order degrevlex is $\{xy\!+\!xz,\, x^2\!-\!xy\!-\!xz\!+\!yz,\, y^2z\!+\!yz^2\}$;
  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]}$;
  3. $\sqrt{\langle f, g\rangle_{\mathbb{Q}[x,y,z]}}=\langle x^2z^2\!-\!y^2z^2+x^3\!-\!xy^2\rangle_{\mathbb{Q}[x,y,z]}$;
  4. $\langle f,g\rangle_{\mathbb{Q}[x,y,z]}:\langle p,q\rangle_{\mathbb{Q}[x,y,z]}$ $=$ $\langle x^3z^2\!+\!x^2yz^2\!-\!xy^2z^2\!-\!y^3z^2\!+\!x^4\!+\!x^3y\!-\!x^2y^2\!-\!xy^3,$ $x^2z^4\!-\!y^2z^4\!-\!2x^2yz^2\!+\!2y^3z^2\!-\!x^4\!-\!2x^3y\!+\!x^2y^2\!+\!2xy^3\rangle_{\mathbb{Q}[x,y,z]}$.

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:

  1. $G$ 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 $K[\mathbf{x}]$.
  2. $\langle S'\rangle_{k[\mathbf{x}]}\cap\langle S''\rangle_{k[\mathbf{x}]}=\langle S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\langle S'\rangle_{K[\mathbf{x}]}\cap\langle S''\rangle_{K[\mathbf{x}]}=\langle S\rangle_{K[\mathbf{x}]}$
  3. $\sqrt{\langle S'\rangle_{k[\mathbf{x}]}}=\langle S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\sqrt{\langle S'\rangle_{K[\mathbf{x}]}}=\langle S\rangle_{K[\mathbf{x}]}$
  4. $\langle S'\rangle_{k[\mathbf{x}]}:S''=\langle S\rangle_{k[\mathbf{x}]}$ $\;\Longleftrightarrow\;$ $\langle S'\rangle_{K[\mathbf{x}]}:S''=\langle S\rangle_{K[\mathbf{x}]}$

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. $f\in\langle S'\rangle_{k[\mathbf{x}]}:S''$. $(\Rightarrow)$: ???


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Hmm, I really didn't expect this to be such an unpopular/uninteresting question. –  Leon Lampret Dec 31 '11 at 17:58
3  
I do not know enough to follow the question or judge the technical aspects of it. However, the post looks dense and filled with symbols. May be adding a few English sentences in between could help? (Also, if it is possible to split the post into more than one question, that will be better too, I feel.) –  Srivatsan Dec 31 '11 at 21:52
1  
1 follows easily from Buchburger's algorithm. –  Aaron Jan 1 '12 at 0:36
    
@Srivatsan: thank you for suggestions. The reason why my post was so dense is because I wanted it to be concise and not too scary (people might think that if the question is long, then so must be the answer). Regardless, I have expanded it quite a bit, showing some background. I would still like to avoid splitting it into several questions, because this will probably not be the last in the series (I have trouble with SINGULAR operations preimage, sat, eliminate, is_injective, is_surjective, minAssGTZ, $\ldots$). –  Leon Lampret Jan 1 '12 at 23:12
    
@Leon Thanks for the edit. My suggestions were mostly generic; I am sure you can judge better than me, to what extent they apply to this question. :-) Good luck with your question anyway! –  Srivatsan Jan 1 '12 at 23:21

1 Answer 1

up vote 2 down vote accepted

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).

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