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It might be too little to care, but there is a term that isn't really defined in a math text, and I was looking if someone could provide a more complete understanding of it. For context:

$\Bbb{Q}(a_0, ... , a_{n-1})$ is the field of the rationals adjoined the coefficients $a_0, ..., a_{n-1}$ of a polynomial.

$x_1, ... , x_n$ are the roots of this polynomial.


The text: "The goal of solution by radicals is to extend $ℚ(a_0, ... , a_{n-1})$ by adjoining radicals until a field containing the roots $x_1, ... , x_n$"

The author gives the example of the quadratic equation and how one can have $x_1, x_2$ in the field extension of $\Bbb{Q}(a_0, ... , a_{n-1})$ the polynomial by adjoining the root of $a_1^2-4a_0$.

My question, I think, is rather simple. What exactly do they mean by "radical"? As in, radical of what? Do they mean only radicals of the coefficients? Or are we also talking about radicals of the roots? Do these have to be radicals of only rational numbers?


I would truly appreciate any help/thoughts!

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    $\begingroup$ It takes 1 minute to write this question and 10 seconds to google the answer. $\endgroup$ – franz lemmermeyer Sep 14 '17 at 16:17
  • $\begingroup$ A radical is the symbol used to denote an $n^{th}$ root. $\endgroup$ – Yves Daoust Sep 14 '17 at 16:33
  • $\begingroup$ @Yves Daoust. It's not that I didn't know what a radical was haha. It's that I didn't know what they were getting the radical of. I appreciate the comment though. $\endgroup$ – Leo Sep 14 '17 at 16:47
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    $\begingroup$ @Leo: hem, from the title "What is a radical ?" and from the question "What exactly do they mean by "radical"?" I naively inferred that you wanted to know what a radical is, sorry. $\endgroup$ – Yves Daoust Sep 14 '17 at 16:53
  • $\begingroup$ @Leo $\mathbb{Q}(\zeta_n,\sqrt[5]{1+\zeta_n})/\mathbb{Q}$ is a radical extension (ie. adding $k$-th roots of unity and of other elements iteratively). They are interesting because their Galois closure is a tower of abelian extensions with easily understood Galois group (but the whole extension isn't abelian, it is only solvable) $\endgroup$ – reuns Sep 14 '17 at 22:17
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It's radicals as in $\sqrt x$, $\sqrt[3]x$, and so on. And radicals of what? Well, radicals of the coefficients (not the roots) and more generally of the polynomial expressions obtained from the coefficients. And you can apply the radicals again to these new numbers.

So, in the case of quadratic equations $x^2+ax+b=0$, first you compute $a^2-b$. Then you compute $\Delta$, which is any square root of that number. Then the roots will be $\dfrac{\pm\Delta-a}2$.

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  • $\begingroup$ You're missing a square root symbol. $\endgroup$ – Bernard Sep 14 '17 at 16:22
  • $\begingroup$ @Bernard Where? $\endgroup$ – José Carlos Santos Sep 14 '17 at 16:24
  • $\begingroup$ The formula for the roots. $\endgroup$ – Bernard Sep 14 '17 at 16:24
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    $\begingroup$ @Bernard Did you notice that $\Delta$ is already a square root? $\endgroup$ – José Carlos Santos Sep 14 '17 at 16:25
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    $\begingroup$ Oh! Sorry, I didn't notice. This isn't traditional. $\endgroup$ – Bernard Sep 14 '17 at 16:26

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