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Let $ f=X^{4}+aX^{3}+bX^{2}+cX+d \in K[X] $ where $ K[X] $ is a field with $ char(K) \neq 2,3 $ and let $ \alpha_{1}, \dots , \alpha_{4} $ be its roots( in an extension of $ K $ ). Define $$ C_{1}=(\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{3})^{2} $$ $$C_{2}=(\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4})^{2}$$ $$ C_{3}=(\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4})^{2} $$ The exercise at hand first asks to verify that the $ S_{4} $ action permutes the $ C_{i} $ and that the subgroup $ H=\{ (1),(12)(34),(13)(24),(14)(23) \} $ fixes the $ C_{i} $ and then to show that the following relations hold: $$ C_{1}+C_{2}+C_{3}=3a^{2}-8b $$ $$ C_{1}C_{2}+C_{2}C_{3}+C_{1}C_{3}=3a^{4}-16a^{2}b+16b^{2}+16ac-64bd $$ $$ C_{1}C_{2}C_{3}=(a^{3}-4ab+8c)^{2} $$ This were clearly just computational and I solved them quickly, having used Newton's identities for instance for the second task. I couldn't provide an answer though for the last question regarding this exercise which is to verify that, given the above information, one can solve in general the fourth degree equation.

What I did was to express $ \alpha_{1} $ in terms of the $ \sqrt{C_{i}}' $s and the coefficient $ a $ of $ f $. Then I think one could get the other roots by solving the system of equations in the square roots of the $ C_{i}' $s which if I am not wrong means solving a third degree equation. I am quite sure this is not the way to do it though.

In the literature, I have read about solving fourth degree equations in general, including the wikipedia article regarding this and though they seem to have a similar approach to the one this exercise provides, I wasn't able to deduce how to solve the general fourth degree equation using only the information provided by this exercise.

I would appreciate any help concerning this. Thank you!

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  • $\begingroup$ Expand $(x-C_1)(x-C_2)(x-C_3)$. What are the coefficients? $\endgroup$ – tehjh Mar 13 '17 at 15:31
  • $\begingroup$ Well, of course the coefficients will be the elementary symmetric polynomials in the $ C_{i} $'s which we now how to express them in terms of the coefficients of f, but how does one actually get the roots? $\endgroup$ – Raizen Mar 14 '17 at 12:49
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    $\begingroup$ Ah yes sorry I did not see that you had already arrived at a third degree equation, which is in fact right, because the solving of a cubic is well known: first depress it and then the most elegant way would be through factorizing $x^3+p^3+q^3-3xpq$ and looking at it as a polynomial in $x$ $\endgroup$ – tehjh Mar 14 '17 at 12:53

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