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I have this false prove which I can't find the problem in. Trying to find the field extension degree $[E:\mathbb{Q}]$ where $E$ is the splitting field of the polynomial $f(x)=x^3-5$ over $\mathbb{Q}$.

My false proof is: The roots of $f$ in E are $a,b,c$, we don't need to state them precisely at the moment. Since $f$ is irreducible over $\mathbb{Q}$, $f(x)=irr(a,\mathbb{Q})=irr(b,\mathbb{Q})=irr(c,\mathbb{Q})$, and hence $\mathbb{Q}/(x^3-5) \cong \mathbb{Q}(a) \cong \mathbb{Q}(b) \cong \mathbb{Q}(c)$, therefore $b,c\in \mathbb{Q}(a)$, therefore $E=\mathbb{Q}(a,b,c)=\mathbb{Q}(a)$, therefore $[E:\mathbb{Q}]=3$.

This contradicts several other things, which are the reason I believe this prove is false. Can you spot the not?

Thanks, G.

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btw, when writing a post, how do I start a new line without making an empty line in between? when I just press "enter" it stays in the same line... :) – cruvadom Jan 26 '13 at 16:20
I don't know exactly what your typesetting complaint is. Either you start a new paragraph (with a blank line in the input) which is displayed with a bit of vertical whitespace (given that paragraphs have no indentation on their first line, they would be hard to recognise without this whitespace), or you don't start a new paragraph and then you get nothing in the output (but you can make your input start a new line for your personal convenience). – Marc van Leeuwen Jan 26 '13 at 16:25
"but you can make your input start a new line for your personal convenience" how do I do that? – cruvadom Jan 26 '13 at 16:29
You already know how to do that, you started "The roots", "Since" and "G." on fresh lines, and it still shows in the input (what you see when you het "edit"). I think you are asking the same for the output (visible to everyone), but you can't get that without vertical whitespace. – Marc van Leeuwen Jan 26 '13 at 16:43
up vote 5 down vote accepted

The error is in "$\mathbb{Q}(a) \cong \mathbb{Q}(b) \cong \mathbb{Q}(c)$ therefore $b,c\in \mathbb{Q}(a)$". The three fields you list are isomorphic but distinct subfields of $E$, which has degree $6$ over $\Bbb Q$. Indeed one can see that $\frac ab,\frac bc, \frac ac$ are all roots of $X^2+X+1$, which is an irreducible quadratic polynomial over $\Bbb Q$, whence neither of these roots can lie in any of the degree $3$ extensions listed, so each of those extensions can contain only one of $a,b,c$.

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same comment as below - just because I believe you might also have something interesting to say about it: consider this example: the splitting field of $x^2+x+1$ over $\mathbb{Q}$. there are two complex roots "$a$" and "$b$"$=a^2$, and the splitting field $E=\mathbb{Q}(a)=\mathbb{Q}(b)$. so this situation can happen. – cruvadom Jan 26 '13 at 16:59
@cruvadom: Yes, in general a field that contains one root of a quadratic polynomial contains them all. This is specific for degree $2$ though, and somehow related to the identity $2!=2$. Or to the following truism: if a polynomial in $K[X]$ of degree $d>0$ is divisible by a product of $d-1$ factors of degree $1$ in $K[X]$, then it can actually be written as a product of $d$ factors of degree $1$ in $K[X]$ (applied for $d=2$). – Marc van Leeuwen Jan 26 '13 at 17:41

As others said, your mistake is to assume that isomorphism of fields implies equality. To make this absolutely clear consider the following. Exactly one of the roots of $x^3-5$ is real. Therefore exactly one of those three fields consists of real numbers only - the other two do not. Hence they cannot be equal (even though they are isomorphic).

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consider this example: the splitting field of $x^2+x+1$ over $\mathbb{Q}$. there are two complex roots "$a$" and "$b$"$=a^2$, and the splitting field $E=\mathbb{Q}(a)=\mathbb{Q}(b)$. so this situation can happen. – cruvadom Jan 26 '13 at 16:56
@cruvadom: Correct, but there the equation $$x^2+x+1=(x-a)(x-b)=$$ dictates a relation $a+b=1$ (in addition to the $b=a^2$ that you mentioned). And the field equality $\mathbb{Q}(a)=\mathbb{Q}(b)$ is a consequence of these. When the irreducible polynomial has three roots, this may or may not happen. It did not happen with the polynomial in your OP. But it does happen e.g. with the polynomial $$(x-2\cos\frac{2\pi}7)(x-2\cos\frac{4\pi}7)(x-2\cos\frac{8\pi}7)=x^3+x^2-2x+1$$‌​(unless I made a mistake), because $\cos 2\alpha=2\cos^2\alpha-1$ is polynomially related with $\cos\alpha$. – Jyrki Lahtonen Jan 26 '13 at 20:18

Your error is with the misunderstanding of what an isomorphism is and what it does. The two groups $\mathbb{Z}/2\mathbb{Z}$ and the group of permutations of $\{1,2\}$ are isomorphic but they have completely different elements (one consists of classes of integers, the other permutations).

Just because the three fields you have are isomorphic it doesnt mean they are equal.

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Thanks, I sure know the difference between equality and isomorphism. my misconception was that the equivalent of $b$ under the isomorphism, lies somewhere inside $\mathbb{Q}(a)$ – cruvadom Jan 26 '13 at 16:43
Yes, so $b$ doesn't necessarily lie in $\mathbb{Q}(a)$ since isomorphism isn't equality. – fretty Jan 27 '13 at 19:15

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