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Find polynomials in S, then find basis for ideal (S)

rapid

New member
Jan 31, 2012
13
Hi There,

I posted this question over at MHF to no avail, I'm not really sure what the ruling is on this kind of thing, I know this site was setup when MHF was down for a long time but you seem to still be active and a lot of clever people are still here so hopefully you don't mind taking a look at this for me :)

I have a couple of example questions that I'm trying to get my head around, a bit of guidance would be fabulous.


\(S:=\{f\in\mathcal{Q}[X,Y]\mid f(X,Y)=f(Y,X) \mbox{ and } \deg(f)\geq 0\}\)


1a: Give two polynomials that belong to \(S\).
1b: Find a finite basis of the ideal \((S)\) of \(\mathcal{Q}[X,Y]\) and justify your answer.


I then have the question where the questions are the same but based on this
\(S:=\{f\in\mathcal{Q}[X,Y]\mid f(X,Y)=-f(Y,X)\}.\)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Hi There,

I posted this question over at MHF to no avail, I'm not really sure what the ruling is on this kind of thing, I know this site was setup when MHF was down for a long time but you seem to still be active and a lot of clever people are still here so hopefully you don't mind taking a look at this for me :)

I have a couple of example questions that I'm trying to get my head around, a bit of guidance would be fabulous.


\(S:=\{f\in\mathcal{Q}[X,Y]\mid f(X,Y)=f(Y,X) \mbox{ and } \deg(f)\geq 0\}\)


1a: Give two polynomials that belong to \(S\).
Do you understand what Q[X,Y] is? It is the set of all polynomials in variables X and Y with rational coefficients. Examples are X+ Y, [tex]3X^2+ 2Y[/tex], and [tex]X^2+ XY+ Y^2[tex]
To be in S requires that it be symmetric- that is that swapping X and Y does not change the polynomial. X+ Y and [tex]X^2+ XY+ Y^2[/tex] are in S but [tex]3X^2+ 2Y[/tex] is not.

1b: Find a finite basis of the ideal \((S)\) of \(\mathcal{Q}[X,Y]\) and justify your answer.


I then have the question where the questions are the same but based on this
\(S:=\{f\in\mathcal{Q}[X,Y]\mid f(X,Y)=-f(Y,X)\}.\)
 

rapid

New member
Jan 31, 2012
13
Yeh, I thought that would be the case, thanks for confirming. What about part b however, a finite basis?

Also with the second question where \(f(X,Y)=-f(Y,X)\) i'm honestly struggling to think of any polynomials, other than \(0\), that fit because the minus makes it more tricky.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,716
Also with the second question where \(f(X,Y)=-f(Y,X)\) i'm honestly struggling to think of any polynomials, other than \(0\), that fit because the minus makes it more tricky.
How about $X-Y$ ?