Question

Question in comments

Answer 1

So I am in a modern algebra class and I am working on a problem.

I have the set of all polynomials in two variables \(x\) and \(y\) with \(0\) constant term. So if \(p(x,y)\) is one such polynomials that is not the zero polynomial, we can write it as \[p(x,y)=\sum_{i=1,j=1}^{m,n}\alpha_{i,j}x^iy^j\text{ where } \alpha_{0,0}=0\]

I want to show that if I take some polynomial \(r(x,y)\) (not necessarily with \(0\) constant term) then \(r(x,y)p(x,y)=0\) implies \(r(x,y)=0\).

Suppose \(r(x,y)p(x,y)=0\), then \(\sum_{i=1,j=1}^{m,n}r(x,y)\alpha_{i,j}x^iy^j=0\).


Here is where the problem is, I said that this implies \(r(x,y)\alpha_{i,j}=0\) for all \(i\in \{1,2,3,...,m\}, j\in\{1,2,3,...,n\}\). My teacher said this was wrong. How am I wrong?

Answer 2

@SolomonZelman this looks very interesting

Answer 3

your soluton looks ok to me

Answer 4

Well there is more to it, but this is everything up until the part I was told is wrong. But I don't see the problem...

Answer 5

Summing to 0 isn't the same as saying all terms are going to be zero

Answer 6

?

ax^2+bx+c=0 if and only if a=b=c=0

Answer 7

When you went from
\[\Large \sum_{i=1,j=1}^{m,n}r(x,y)\alpha_{i,j}x^iy^j=0\]
to
\[\Large r(x,y)\alpha_{i,j}=0\]
why did you drop the \(\Large x^iy^j\) term? I think that will create at least 2 pairs of like terms (which means you'll have summing going on, eg: -2x^3y^2 and 5x^3y^2)

Answer 8

so polynomials of xy are like \(2x^2y^+3y^2+2x\) the only way to get 0 is if the coefficients are 0.

so \(ax^2+by+cx^2y^2\) is the zero polynomial if \(a=0,b=0,c=0\)

Answer 9

x and y are indeterminates.

Answer 10

It's possible you made a mistake in notation a step before that? Maybe you should have an intermediate step showing how r(x,y) is formed? But that would make things even more messy

Answer 11

Or maybe your teacher has a specific way s/he wants you to show?

Answer 12

There is one notation mistake I can see but not that important, my index on the sum should start at 0 on both i and j

All I am saying here is if I am given some polynomial like \(2x^2y^2+3x\)
Then the only polynomial I can multiply that by to get the 0 polynomial is the zero polynomial.

So like \((ax^4+by^2)(2x^2y^2+3x)=0\) implies \((ax^4+by^2)2x^2y^2+(ax^4+by^2)3x=0\) only if \((ax^4+by^2)2=0, \text{ and } (ax^4+by^2)3=0\)

Answer 13

I see what you mean by your second part

Answer 14

So it's possible that the indexing is the only issue. I can't see any other problems to be honest.

Answer 15

If m=2, n=2 I should get

\(x^0y^0+x^1y^0+x^2y^0+x^0y+x^1y+x^2y+x^0y^2+x^1y^2+x^2y^2\) and the coefficients.