Condition for pairwise different zeroes in a polynomial K[X] where K is a field

[xenonforlife]

the polynomial K[X] = X³ + aX + b where K is field

For this polynomial to have 3 pairwise different zeroes in K prove that 4a³ + 27b² is non zero.

How do I proceed?

 

[daon2]

What is "pairwise different"? Do you just mean distinct roots?

 

[xenonforlife]

Yes...exactly that

 

[daon2]

Assume that two roots are identical. Then \(\displaystyle f(x)=(x-\alpha)^2(x-\beta)\) in some extension \(\displaystyle E/K[x]\). Note I do not exclude the possibility that \(\displaystyle \alpha=\beta\)...

Then \(\displaystyle f(x) = x^3-(2\alpha+\beta)x^2+(2\alpha\beta+\alpha^2)x - \alpha^2\beta \in K[x]\)

So:

\(\displaystyle \alpha^2\beta = -b\)
\(\displaystyle 2\alpha\beta+\alpha^2 = a\)
\(\displaystyle 2\alpha+\beta = 0\)

Solving this and playing a bit will show that \(\displaystyle 4a^3+27b^2 = 0\)

 

[xenonforlife]

Assume that two roots are identical. Then \(\displaystyle f(x)=(x-\alpha)^2(x-\beta)\) in some extension \(\displaystyle E/K[x]\). Note I do not exclude the possibility that \(\displaystyle \alpha=\beta\)...

Then \(\displaystyle f(x) = x^3-(2\alpha+\beta)x^2+(2\alpha\beta+\alpha^2)x - \alpha^2\beta \in K[x]\)

So:

\(\displaystyle \alpha^2\beta = -b\)
\(\displaystyle 2\alpha\beta+\alpha^2 = a\)
\(\displaystyle 2\alpha+\beta = 0\)

Solving this and playing a bit will show that \(\displaystyle 4a^3+27b^2 = 0\)

Superb...thank you so so much...now I think I should have thought about that...What I did is I did not reach to the \(\displaystyle 4a^3+27b^2 = 0\) with the help of these three equations...rather I calculated the value of \(\displaystyle 4a^3+27b^2 = 0\) and showed that indeed it is coming out to be equal to 0...many thanks for your help.

 

[SlipEternal]

Out of curiosity, is it possible the question is worded differently? I agree that it is a sufficient condition that \(\displaystyle 4a^3+27b^2 \ne 0\), but it may not a necessary one. If you assume that \(\displaystyle 4a^3+27b^2=0\), and assume that the three roots are \(\displaystyle r_1, r_2, r_3\) are distinct, do you arrive at any contradictions? I didn't seem to when I tried plugging it into Mathematica, although I did not attempt to plug the results I got into the original equation (ran out of time). Still, it appeared that three distinct roots were possible.

 

[daon2]

Out of curiosity, is it possible the question is worded differently? I agree that it is a sufficient condition that \(\displaystyle 4a^3+27b^2 \ne 0\), but it may not a necessary one. If you assume that \(\displaystyle 4a^3+27b^2=0\), and assume that the three roots are \(\displaystyle r_1, r_2, r_3\) are distinct, do you arrive at any contradictions? I didn't seem to when I tried plugging it into Mathematica, although I did not attempt to plug the results I got into the original equation (ran out of time). Still, it appeared that three distinct roots were possible.

Yes, it is true both ways. If the equation is satisfied, then:

\(\displaystyle a =-3(\frac{b^2}{4})^{1/3}\)

Then use a fact from calculus: \(\displaystyle r\) is a multiple root of \(\displaystyle f\) if and only if \(\displaystyle f(r)=f'(r)=0\)

There are two scenarios when solving for the roots, but the correct value of \(\displaystyle x\) turns out to be \(\displaystyle x = (\frac{b^2}{4})^{1/6}\). It is both a root of \(\displaystyle f'\) and \(\displaystyle f\).

edit: This actually might require that b>=0, and that taking the other root of f' would switch the sign of the root, but still work when b<0