# Galois Groups and a detailed example using the Quintic equation

I have been asking ChatGPT to explain the relation between Galois Groups and the impossibility of solving quintic equations using radicals. It is amazingly coherent. But I can’t seem to get it to be really explicit with an example.

Could you share your prompts/responses? I’ve been looking to prove my chops in prompt engineering!

I would be glad to do so. But I cannot seem to find where it went. Is there a good way to search through my chats with this bot?

Are you on mobile? If there isn’t a column of your previous conversations on the left, there should be a hamburger icon (≡) in the top-left that should display them

That particular bunch of chat seems to have disappeared??!! Other chats are marked with topics, sort of, in the column on the left. I am on my home computer, not mobile.
I asked if it could search for `Galois’ and it said it could not do that??!!

I suspect it was a long request and maybe it dropped the whole business?

As in the following ` I’ve realized that if you do short and specific requests, the AI will work very well, otherwise, if you do bigger requests, the AI may not work too good, giving some “network errors” and closing the body stream, resulting in a performance failure’ from somebody called homersimpson

In fact I took a screenshot. I wonder if i can include it here

can you see it

Cheers

John

That is bizarre. I could imagine it dropped a post, but not a whole thread. Did you regenerate the response? You may need to navigate to it if so

I tried a different but similar question and here is the result:
Can you provide an example of a quintic equation where by Galois theory there can be no radical solution, and explain the group theory in clear terms?

Sure! Here is an example of a quintic equation that does not have a radical solution:

x^5 + x + 1 = 0

Galois theory tells us that if the Galois group of a polynomial is solvable, then the polynomial can be solved by radicals. The Galois group of a polynomial is a group consisting of all permutations of the roots of the polynomial that can be obtained by repeatedly applying the field operations of addition, subtraction, multiplication, and division.

In this case, the Galois group of the quintic polynomial x^5 + x + 1 is the group S5, which is the group of all permutations of 5 elements. It can be shown that S5 is not a solvable group, which means that the quintic equation x^5 + x + 1 does not have a radical solution.

To see this, let’s suppose for the sake of contradiction that the equation x^5 + x + 1 does have a radical solution. This means that we can solve the equation by repeatedly applying the field operations of addition, subtraction, multiplication, and division to the coefficients of the polynomial.

Since the equation has degree 5, the solution must involve the roots of the polynomial. Let’s call the roots r1, r2, r3, r4, and r5. We can then write the solution as a function of the roots:

f(r1, r2, r3, r4, r5) = 0

Now let’s consider the permutation (r1 r2 r3 r4 r5) of the roots. This permutation can be obtained by applying a series of field operations to the coefficients of the polynomial. Since the polynomial has a radical solution, we can find a series of field operations that will transform the coefficients of the polynomial into the coefficients of the polynomial with the roots permuted.

In other words, there exists a series of field operations g(x) such that:

g(x^5 + x + 1) = (x - r2)(x - r3)(x - r4)(x - r5)(x - r1)

But this is a contradiction, because the left hand side is a polynomial with rational coefficients, while the right hand side is a polynomial with irrational coefficients (since the roots r1, r2, r3, r4, and r5 are not rational).

This contradiction shows that the quintic equation x^5 + x + 1 does not have a radical solution, as claimed.

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How do we know that the roots are not rational?

The roots of the quintic equation x^5 + x + 1 are not rational because the polynomial has no rational roots.

To see this, let’s suppose for the sake of contradiction that one of the roots, say r1, is a rational number. This means that r1 can be written in the form a/b, where a and b are integers and b is not equal to 0.

Since r1 is a root of the polynomial, it satisfies the equation x^5 + x + 1 = 0. Substituting r1 = a/b into this equation, we get:

(a/b)^5 + a/b + 1 = 0

Multiplying both sides of this equation by b^5, we get:

a^5 + ab^4 + b^5 = 0

This is a polynomial equation with integer coefficients, and it has the rational number a/b as a root. By the rational root theorem, the possible rational roots of this equation are of the form a/b, where a is a divisor of b^5 and b is a divisor of a.

Since a is a divisor of b^5, it must be of the form a = k * b^n, where k is an integer and 0 <= n <= 5. Substituting this into the equation a^5 + ab^4 + b^5 = 0, we get:

k^5 * b^5 + k * b^4 + b^5 = 0

This equation simplifies to:

(k^5 + k) * b^5 + b^4 = 0

Since b is not equal to 0, the equation b^4 = 0 has no solutions. Therefore, the only way that this equation can hold is if (k^5 + k) = 0.

The only integer solutions of this equation are k = 0 and k = -1. But these values do not work, because they give us the roots 0 and -1/b, which are not rational numbers (since b is not equal to 0).

Therefore, we have reached a contradiction, which means that our assumption that r1 is a rational number was false. This means that all of the roots of the quintic equation x^5 + x + 1 are irrational, as claimed.