Is there a formula for the roots of a Quintic Equation?

I can get my head around this so someone explain it please.

$(1)$ From Galois theory it is known there is no formula to solve a general quintic equation.

But it is known a general quintic can be solved for the 5 roots exactly. Back in 1858 Hermite and Kronecker independently showed the quintic can be exactly solved for (using elliptic modular function). Also I think they're maybe other solution for the quintic which means a formula for each of the 5 roots.

So why is the claim in Galois theory that there is no formula to solve it? I know I am missing something here because the above $(1)$ is an established result.

So what is the value in saying, using Galois theory we do not have a formula for the 5 roots? Since for practical purposes we can actually find the 5 roots each time using say for example the formula based on elliptic modular functions.

• Galois Theory tells us that the roots of a general quintic can not be obtained by the use of radicals alone (that is, with the usual operations of arithmetic and taking $n$-th roots for various $n$). That does not mean that the roots can't be expressed using other types of functions. – Geoff Robinson Feb 1 '16 at 14:39
• @GeoffRobinson But what is the point in knowing that if you can solve for each of the five roots (exactly and numerically) using the Hermite method? – Developer Feb 1 '16 at 14:44
• @Developer, the value lies in the difference between $1.4142$ and $\sqrt2$ – Mariano Suárez-Álvarez Feb 1 '16 at 20:41
• @Mariano Suárez-Alvarez word play on the word "value"for a numerical value compared to a algebraic number. If that what you mean then. It depends on the situation if all that is required is a numerical value (or an expression for infinite precision like $sqrt(2)$) for each of the five roots then you can find them using the Hermite metthod-it doesnt depend on the fact that Galois proved that the Quintic is not solvable by his method (in other words that fact here has no value) – Developer Feb 2 '16 at 2:39
• The «infinite precision» is irrelevant, @Developer. – Mariano Suárez-Álvarez Feb 2 '16 at 3:02

Galois-theory only says that there is no general formula to solve a quintic equation in terms of radicals. That is, there is no formula only using the arithmetic operations "sum, multiplication etc. and taking the $n$-th root".

For instance for the polynomial $x^5 - 4x + 2$ it is known that it has a root that is not expressible in the above mentioned operations (as its Galois-group is $S_5$). (Edit: Another example is $x^5 + x + 1$ which also has Galois-group $S_5$. If you Wolframalpha this polynomial you see nicely how four of its roots can be expressed by radicals, but the fifth can't.)

The solution you mean is the solution using Bring radicals - wikipedia-article here: https://en.wikipedia.org/wiki/Bring_radical -, which is not a contradiction, as it is not expressed in form of radicals (in the sense of $n$-th roots of something).

• So what is the value in knowing the Galois result above, if you can solve for each of the five roots (exactly or numerically) with a formula which is not restricted to nth roots and arithmetic operations? – Developer Feb 1 '16 at 14:55
• solving things in radicals is an algebraic viewpoint. taking the $n$-th root of something can still be made sense of in any field. note that the solution of the problem is more generally: for any field $k$, a polynomial in $k[X]$ is solvable in radicals if and only if its Galois-group is solvable. in a general field you can't use analytic methods like you do with the real numbers (which is done with the Bring radical). – Steven Feb 1 '16 at 20:04
• I cheated however a bit in the above comment. the result which was historically important was indeed the special case of the rational numbers. but these kind of questions are more of the style: "what is the point in constructing things only with ruler and compass?" or "why care about Fermat's Last Theorem?" or "why would anyone care about classical music like Schönberg's that can't be used in movies?" or "why do painters paint if we have photography?" – Steven Feb 1 '16 at 20:08
• $x^5+x+1=(x^2+x+1)(x^3-x^2+1)$? – CO2 Jun 24 '18 at 17:07

The problem of solving algebraic equations algebraically is one of the oldest problems humans have considered —thousands of years old— arising from everything, going from the subdivision of inheritance to riddles by mithological figures to the construction of actual buildings. The fact that we know exactly when we can do it and, when it is possible, that we can in fact carry out the construction of solutions is one of the greatest achievements of mankind.

That's the value of it.

No one cares about infinite precision.

• Jacobi's «for the honor of the human spirit» comes to mind. – Mariano Suárez-Álvarez Feb 2 '16 at 3:01
• And the reason no one cares about infinite precision is (I looked up the proof just now) because the proof basically it goes like this. The Galois group solvability implies (indirectly) there is a sequence of field extension that imply that there are algebra operations so the root finally pops out at the end. However if the group isnt solvable "you cant get to the end". – Developer Feb 2 '16 at 3:16
• No, infinite precision is useless. – Mariano Suárez-Álvarez Feb 2 '16 at 3:18
• In other words infinite precsion isnt used by the group theory or abstract algebra in the proof outlined above; youre just stating the obvious. – Developer Feb 2 '16 at 3:18
• Huh? ${}{}{}{}$ – Mariano Suárez-Álvarez Feb 2 '16 at 3:22