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MULTIPLE CHOICE CORRECT ANSWER

Consider the ideal $I := (ux, uy, vx, uv)$ in the polynomial ring $\mathbb{Q}[u, v, x, y]$, where

$u, v, x, y$ are indeterminates. Choose the correct statement(s) from below:

$i):-$ Every prime ideal containing $I$ contains the ideal $(x, y)$;

$ii):-$ Every prime ideal containing $I$ contains the ideal $(x, y)$ or the ideal $(u, v)$;

$iii):-$ Every maximal ideal containing $I$ contains the ideal $(u, v)$;

$iv):-$ Every maximal ideal containing $I$ contains the ideal $(u, v, x, y)$.

ATTEMPT

Let P be the Prime-Ideal containing $I := (ux, uy, vx, uv)$ $I \subseteq P$ $\implies ux,vx,vy \in P$

If $x\notin I$ then since $ux,vx \in P$ we have $u,v \in P \implies (u,v) \in P$

on the other hand

If $u\notin I$ then since $uy,ux \in P$ we have $x,y \in P \implies (x,y) \in P$

Is my partial solution correct?

I had done with Options i) and ii).

What about iii),iv)?

learning_math
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  • (i) is false, since $I\subseteq\langle u,v\rangle.$ Similarly, (ii) is false as $I\subseteq \langle u,x\rangle$ – learning_math Dec 24 '17 at 07:00
  • @learning_math ii) not false I think. –  Dec 24 '17 at 07:47
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    Is the ideal in question really $(ux, uy, vx, uv)$, or is it supposed to be $(ux, uy, vx, \color{red}{vy})$? You say that $vy \in P$, but that doesn't seem to follow from your definition of $I$. – Viktor Vaughn Dec 24 '17 at 16:49
  • @Quasicoherent Sir the question is correct.I matched 3 times.I had mistaken in my attempt.I corrected that. –  Dec 24 '17 at 17:13
  • $\langle u-1,v,x,y\rangle$ is a maximal ideal such that $I\subseteq \langle u-1,v,x,y\rangle$ and $\langle u,v\rangle\not\subset \langle u-1,v,x,y\rangle$ and $\langle u,v,x,y\rangle\not\subset\langle u-1,v,x,y\rangle$. This means that both (iii) and (iv) are false. – learning_math Dec 25 '17 at 23:39
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    A primary decomposition of $I$ (computed in Macaulay2) is $I = (u,v) \cap (u,x) \cap (v,x,y)$. You can use this to analyse each of i) through iv). – Trevor Gunn Dec 26 '17 at 01:31
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    On the other hand, $(u,v) \cap (x,y) = (ux,uy,vx,vy)$ and this ideal satisfies (ii). – Trevor Gunn Dec 26 '17 at 01:39
  • @TrevorGunn, how does observation $(u,v) \cap (x,y) = (ux,uy,vx,vy)$ say that '(iii) Every maximal ideal containing $I$ contains the ideal $(u,v)$', and '(iv) Every maximal ideal containing I contains the ideal $(u,v,x,y)$' are false? (I am taking $I=(ux, uy, vx, vy)$ and not the one that OP posted). – Silent Apr 30 '19 at 03:11
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    @Silent $V(I)$ is a union of two planes and maximal ideals containing $I$ correspond to points of $V(I)$. Just take a point on the $V(x,y)$ plane not contained in the $V(u,v)$ plane. E.g. $(x,y,u,v-1)$. – Trevor Gunn Apr 30 '19 at 04:24
  • @TrevorGunn, Wow! Can I know which book to read to find this kind of relation between ideals to planes? It was totally new and unexpected to me! I want to know more .. – Silent Apr 30 '19 at 06:53
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    @Silent The correspondence between ideals and vanishing loci is the subject of algebraic geometry. J. S. Milne's notes get to what I'm talking about around page 40. Mile's Reid's book around page 60 (he has an extra chapter on elliptic curves that Milne doesn't). You'll want to start reading from the beginning of course. – Trevor Gunn Apr 30 '19 at 12:43
  • Thank you very much, sir! @TrevorGunn – Silent Apr 30 '19 at 12:47

1 Answers1

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Your question is incorrect because all the four options are false.

(i) is false because $I\subseteq \langle u,v\rangle$.

(ii) is false because $I\subseteq \langle u,x\rangle$.

For (iii) and (iv)....

Notice that $\langle u-1,v,x,y\rangle$ is a maximal ideal. For a proof see Magdiragdag's answer.

Notice also that $I\subseteq\langle u-1,v,x,y\rangle$.

(iii) is false because $\langle u,v\rangle\not\subset\langle u-1,v,x,y\rangle$.

(iv) is false as $\langle u,v,x,y\rangle\not\subset\langle u-1,v,x,y\rangle$.

To prove $\langle u,v\rangle$ is prime..

Consider the evaluation map $f:\mathbb{Q}[u,v,x,y]\rightarrow \mathbb{Q}[z_1,z_2]$ defined as $$f(u)=f(v)=0 \text{ and } f(x)=z_1 \text{ and } f(y)=z_2$$ Then $$\frac{\mathbb{Q}[u,v,x,y]}{\langle u,v\rangle}\cong \mathbb{Q}[z_1,z_2]$$ To prove that $\ker (f)=\langle u,v\rangle$ you may want to look at Magdiragdag's answer.

To prove $\langle u-1,v,x,y\rangle$ is maximal...

Consider the evaluation map $f:\mathbb{Q}[u,v,x,y]\rightarrow \mathbb{Q}[u,v,x,y]$ $$f(u)=1, \text{ } f(v)=v, \text{ } f(x)=x, \text{ } f(y)=y$$ Then $\ker(f)=\langle u-1,v,x,y\rangle$. (For explanation see Magdiragdag's answer.) Then we it follows $$\frac{\mathbb{Q}[u,v,x,y]}{\langle u-1,v,x,y\rangle}\cong \mathbb{Q}$$

learning_math
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  • how do I prove that (u,v),(u,x) are prime ideals and (u-1,v,x,y) is a maximal ideal.I can't relate Magdiragdag'answer with this. –  Dec 27 '17 at 02:59