4
$\begingroup$

This is a problem for homework and I could use a little help.

Define three polynomials $$ f_1(t, x) = t^2 + x^2 - 2, $$ $$ f_2(t, x) = tx - 1, $$ $$ f_3(t, x) = t^3 + 5tx^2 + 1. $$ I'm trying to write $1$ as a linear combination of $f_1, f_2, f_3$ with polynomial coefficients in order to show that $(f_1, f_2, f_3) = \mathbb{C}[x, t]$.

So far I have reduced the powers of $x$ and $t$ to $1$ by writing $$ f_3 - tf_1 - 4xf_2 = 4x + 2t + 1, $$ but I'm not sure how to get rid of $x$ and $t$ here. Any thoughts would be greatly appreciated!

$\endgroup$
0
4
$\begingroup$

No sane person would do that count: you can easily use Nullstellensatz to show that $(f_1,f_2,f_3)=\mathbb{C}[x,t]$.

Anyway, I'm insane: we have $$f_4=x(f_3-tf_1-4xf_2)-2f_2=4x^2+x+2$$ and $$f_5=x^2f_1-(tx+1)f_2=x^4-2x^2+1$$ Hence we have reduced the problem to one variable. $$x^2f_4-4f_5=x^3+10x^2-4$$ $$(4x^2-x)f_4-16f_5=4(x^2f_4-4f_5)-xf_4=39x^2-2x-16$$ $$(-4x^2+x+10)f_4+16f_5=x^2+12x+36$$ $$47f_6=4(-4x^2+x+10)f_4+64f_5-f_4=(-16x^2+4x+39)f_4+64f_5=47x+142$$ $$f_6=x+\frac{142}{47}$$ $$f_4(x)-f_4\left( -\frac{142}{47} \right)=\left(x+\frac{142}{47}\right) \left(4x-4 \frac{142}{47}+1\right)$$ $$1= f_4\left( -\frac{142}{47} \right)^{-1}\left[f_4 -\left(4x-4 \frac{142}{47}+1\right) f_6\right]$$ Now, if you want, you may substitute the expressions of $f_4,f_5,f_6$ in terms of $t_1,t_2,t_3$.

$\endgroup$
4
  • 1
    $\begingroup$ Only because he assumed the OP's partial result was correct. It should read $f_3-tf_1-4xf_2=4x-2t+1$ instead. $\endgroup$ – dafinguzman Feb 22 '14 at 9:36
  • 1
    $\begingroup$ @dafinguzman I think it is $4x + 2t + 1$. wolframalpha.com/input/?i=%28t%5E3+%2B+5*tx%5E2+%2B+1%29+-+t*%28t%5E2+%2B+x%5E2+-+2%29+-4*x*%28tx+-+1%29 $\endgroup$ – tylerc0816 Feb 23 '14 at 19:20
  • 1
    $\begingroup$ Edited using the right partial result. @user121097 this is euclidean algorithm, apart from the last step which is a trick. $\endgroup$ – Giulio Bresciani Feb 24 '14 at 0:26
  • $\begingroup$ @tylerc0816 did you really need the computation or you only wanted to proof that the generated ideal was all the ring? $\endgroup$ – Giulio Bresciani Feb 24 '14 at 0:29
2
$\begingroup$

You may use $$f_1,f_2,f_3,\\tf_1, txf_1,t^2f_1, \\ xf_2,x^2f_2,t^2f_2, \\ xf_3, \text{ and }\, tf_3.$$

This will give you $11$ expressions of the form $$g_i(x,y)=l_i(t^2,x^2,tx,t^3,tx^2,x,t,tx^3,xt^3,t^4,t^2x^2,1),$$ where $l_i$ is a linear function. These actually are functions of $12$ arguments, but what you need is just to get rid of the terms which are not constant, and there are $11$ of those. Now, write the coefficients of each $g_i$ (not considering the constant) as a column, forming a $11\times11$ matrix $A$. What you need is to solve the linear system $A\lambda=0$ in order to find the real coefficients $\lambda_i$ which will get rid of all the nonconstant terms.


Edit.

If I made no mistakes, the matrix is

[ 1  0  0  0  0  0  0 -1  0  0 -2]  
[ 1  0  0  0  0 -1  0  0  0  0  0]  
[ 0  1  0  0  0  0  0  0 -2  0  0]  
[ 0  0  1  0  1  0  0  0  0  0  0]  
[ 0  0  5  1  1  0  0  0  0  0  0]  
[ 0  0  0 -1  0  0  1  0  0  0  0]  
[ 0  0  0  0 -2  0  0  0  0  1  0]  
[ 0  0  0  0  0  1  1  0  1  0  0]  
[ 0  0  0  0  0  0  1  1  1  0  0]  
[ 0  0  0  0  0  0  0  0  0  1  1]  
[ 0  0  0  0  0  0  0  0  0  5  1]. 

Edit.

However, I didn't get the result I expected. The only nontrivial solution up to scalar gets rid of the constant term too. $$f_1-2f_2+x^2f_2+t^2f_2-txf_1=0.$$

So this is of no use. Maybe some modification of this procedure can give you the result you are looking for.

$\endgroup$
2
  • 2
    $\begingroup$ You essentialy wrote $f_1f_2-f_2f_1=0$: I don't understand why this should help. You need the constant term, otherwise this is not even an answer. $\endgroup$ – Giulio Bresciani Feb 21 '14 at 15:13
  • 2
    $\begingroup$ @GiulioBresciani I know it is not an answer, I think I made it clear when I said that I "didn't get the result I expected", but I wanted to provide the matrix anyway, maybe it can be of some use. Maybe another vector in the null space can be the correct answer. $\endgroup$ – dafinguzman Feb 21 '14 at 18:37
0
$\begingroup$

Building off of Giulio Bresciani's answer, once you get the problem down to one variable you can simply morph the problem into one of determining if the columns of a matrix are independent.

Multiplying the equations Giulio gave by $x$ until they are both degree 5 gives:

$$x^4+2x^2+1$$ $$x^5+2x^3+x$$ $$4x^2+x+2$$ $$4x^3+x^2+2x$$ $$4x^4+x^3+2x^2$$ $$4x^5+x^4+2x^3$$

We want some linear combination of these so that all that is left is a constant coefficient of 1, so we want some combination of these such that all the variable coefficients add to 0. Note that there are 5 variables here: $x,x^2,x^3,x^4,x^5$, but 6 equations. If you write it out as a matrix, it isn't hard to use something like matlab to find one column of the matrix in terms of the others. This gives exactly the coefficients needed to add all these equations together and eliminate the variables.

Since there are only 6 equations, its unlikely the constants will also be eliminated in this sum, so you can just divide by whatever constant comes out to make the sum equal to 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.