# Field isomorphism

#### Fantini

MHB Math Helper
The following question appeared in my last Rings and Fields exam.

Let $\alpha \in \mathbb{R}$ be a root of $x^3 -2$. Let $K = \mathbb{Q}(\alpha) \subseteq \mathbb{R}$ and $\sigma: K \to K$ a field isomorphism. Prove that $\sigma (x) = x$ for all $x \in K$.

My attempt is as follows: since this is an homomorphism from the field to itself we have that $\sigma (0) = 0$ and $\sigma (1) = 1$. Using this I showed that $\sigma (n) = n$ for all $n \in \mathbb{Z}$ and subsequently that $\sigma (q) = q$ for all $q \in \mathbb{Q}$. Since this is an isomorphism, this means it is surjective and injective, and the only missing piece is $\alpha$, therefore we must have $\sigma (\alpha) = \alpha$. Concluding, we have that $\sigma (x) = x$ for all $x \in K$.

Even if my answer is wrong, do you think this should be awarded zero points at all?

Thank you for all help. P.S.: There were more passages in my solution, such as $\sigma (0) = \sigma (0+0) = \sigma (0) + \sigma (0)$, thus $\sigma (0) = 0$, and similar ones for $\sigma (1) = 1$, $\sigma(n) = n$ and $\sigma (q) = q$. I don't think it is necessary to repeat them all here, specially since I've just written one of them here. If you find it necessary, though, I will. Thanks.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
The following question appeared in my last Rings and Fields exam.

Let $\alpha \in \mathbb{R}$ be a root of $x^3 -2$. Let $K = \mathbb{Q}(\alpha) \subseteq \mathbb{R}$ and $\sigma: K \to K$ a field isomorphism. Prove that $\sigma (x) = x$ for all $x \in K$.

My attempt is as follows: since this is an homomorphism from the field to itself we have that $\sigma (0) = 0$ and $\sigma (1) = 1$. Using this I showed that $\sigma (n) = n$ for all $n \in \mathbb{Z}$ and subsequently that $\sigma (q) = q$ for all $q \in \mathbb{Q}$. Since this is an isomorphism, this means it is surjective and injective, and the only missing piece is $\alpha$, therefore we must have $\sigma (\alpha) = \alpha$. Concluding, we have that $\sigma (x) = x$ for all $x \in K$.

Even if my answer is wrong, do you think this should be awarded zero points at all?

Thank you for all help. P.S.: There were more passages in my solution, such as $\sigma (0) = \sigma (0+0) = \sigma (0) + \sigma (0)$, thus $\sigma (0) = 0$, and similar ones for $\sigma (1) = 1$, $\sigma(n) = n$ and $\sigma (q) = q$. I don't think it is necessary to repeat them all here, specially since I've just written one of them here. If you find it necessary, though, I will. Thanks.
As a last resort you could do the following:
Let $\sigma(\alpha)=a+b\alpha+c\alpha^2$, for some $a,b,c\in \mathbb{Q}$.
Then $(\sigma(\alpha))^3=\sigma(\alpha^3)=\sigma(2)=2 \Rightarrow (a+b\alpha+c\alpha^2)^3=2$. This last equality will boil down to three equations whose solutions must be $a=0,b=1,c=0$ for the desired result.
I'll try to come up with a more elegant solution.

#### Fantini

MHB Math Helper
Thank you caffeine. But the question is: regardless of your solution, is mine wrong? That's the point. I didn't solve it using your way, precisely the one he wanted, but I need to know if my reasoning is incorrect. If it isn't, I want my points in the test!

Thanks for your input. #### caffeinemachine

##### Well-known member
MHB Math Scholar
The following question appeared in my last Rings and Fields exam.

Let $\alpha \in \mathbb{R}$ be a root of $x^3 -2$. Let $K = \mathbb{Q}(\alpha) \subseteq \mathbb{R}$ and $\sigma: K \to K$ a field isomorphism. Prove that $\sigma (x) = x$ for all $x \in K$.

My attempt is as follows: since this is an homomorphism from the field to itself we have that $\sigma (0) = 0$ and $\sigma (1) = 1$. Using this I showed that $\sigma (n) = n$ for all $n \in \mathbb{Z}$ and subsequently that $\sigma (q) = q$ for all $q \in \mathbb{Q}$. Since this is an isomorphism, this means it is surjective and injective, and the only missing piece is $\alpha$, therefore we must have $\sigma (\alpha) = \alpha$. Concluding, we have that $\sigma (x) = x$ for all $x \in K$.

Even if my answer is wrong, do you think this should be awarded zero points at all?

Thank you for all help. P.S.: There were more passages in my solution, such as $\sigma (0) = \sigma (0+0) = \sigma (0) + \sigma (0)$, thus $\sigma (0) = 0$, and similar ones for $\sigma (1) = 1$, $\sigma(n) = n$ and $\sigma (q) = q$. I don't think it is necessary to repeat them all here, specially since I've just written one of them here. If you find it necessary, though, I will. Thanks.
But you haven't shown that $\sigma(\alpha)=\alpha$. And that is the most important part.
Since this is an isomorphism, this means it is surjective and injective, and the only missing piece is $\alpha$, therefore we must have $\sigma (\alpha) = \alpha$.
This does not prove that $\sigma(\alpha)=\alpha$.
I can't say how many credits your solution would fetch. (I am not even a math major )

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#### Fantini

MHB Math Helper
Okay. Thank you! Edit: Why does it not prove that $\sigma (\alpha) = \alpha$? Since $\sigma$ is surjective we have that there is some $k \in K$ such that $\sigma (k) = \alpha$. Suppose $k \in \mathbb{Q}$. We already showed that $\sigma (k) = k$ for that case, but $\alpha \notin \mathbb{Q}$, therefore it can't be rational. If it can't be rational, the only possibility, since $\sigma$ is injective, is that $\sigma (\alpha) = \alpha$.

In other words, I don't mind being wrong. Mistakes is where I learn most. But I need know WHERE exactly I am going wrong. What is the problem with my reasoning? Am I using faulty assumptions? Is doing $(a+b \alpha + c \alpha^2)^3$ the only way? Thanks.

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#### Deveno

##### Well-known member
MHB Math Scholar
all we are given to go on for $\alpha$ is that:

$\alpha^3 = 2$ and $\alpha \in \Bbb R$

since $\sigma$ fixes $\Bbb Q$ taking $\sigma$ of both sides gives:

$(\sigma(\alpha))^3 = 2$

now that doesn't seem to help much, but at least it does tell us $\sigma(\alpha)$ is a root of $x^3 - 2$.

let's expand our investigation of $x^3 - 2$ a bit. suppose $\omega$ is a root of:

$x^2 + x + 1 = \dfrac{x^3 - 1}{x - 1}$.

a little algebra tells us that:

$x^3 - 2 = (x - \alpha)(x - \alpha\omega)(x - \alpha\omega^2)$

examining the discriminant of $x^2 + x + 1$ which is $1^2 - 4(1)(1) = -3 < 0$ tells us that $\omega,\omega^2$ and thus $\alpha\omega,\alpha\omega^2$ are not real.

since we are given that $\sigma(K) = K$, it follows that $\sigma(\alpha) \in \Bbb R$.

since $\sigma(\alpha) = \alpha,\alpha\omega,\text{or }\alpha\omega^2$

and only one of these is real, it must be that $\sigma(\alpha) = \alpha$.

#### Fantini

MHB Math Helper
Nice solution. What is wrong with mine? Also, given my arguments, do you think I am right to claim at least half the question?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Okay. Thank you! Edit: Why does it not prove that $\sigma (\alpha) = \alpha$? Since $\sigma$ is surjective we have that there is some $k \in K$ such that $\sigma (k) = \alpha$. Suppose $k \in \mathbb{Q}$. We already showed that $\sigma (k) = k$ for that case, but $\alpha \notin \mathbb{Q}$, therefore it can't be rational. If it can't be rational, the only possibility, since $\sigma$ is injective, is that $\sigma (\alpha) = \alpha$.

In other words, I don't mind being wrong. Mistakes is where I learn most. But I need know WHERE exactly I am going wrong. What is the problem with my reasoning? Am I using faulty assumptions? Is doing $(a+b \alpha + c \alpha^2)^3$ the only way? Thanks.
$\sigma(k)=\alpha \Rightarrow k \not \in \mathbb{Q}$. This is good. But this doesn't imply that $k=\alpha$. $k$ could be $\alpha^2$ or $1+\alpha$ or $1+2\alpha+\alpha^2$ or anything like that. So here's where the proof needs work. Denevo has provided a nice solution above.

#### Deveno

##### Well-known member
MHB Math Scholar
Nice solution. What is wrong with mine? Also, given my arguments, do you think I am right to claim at least half the question?
it is not, in general true, that if E = F(a) that if an automorphism of E fixes F it must fix a as well (since it's the "only thing left").

to see this, look at C = R(i).

the automorphism x+yi --> x-yi fixes R, but does NOT fix i.

you MUST prove that the image of $\alpha$ is $\alpha$ (and not something which perhaps might work as well like $-\alpha$ or $1/\alpha$ which is the case for SOME extension fields).

any automorphism of an extension of Q necessarily fixes Q (because Q is the prime field of characteristic 0). this is basic, and is not the "real meat" of the problem.

#### Fantini

Good explanation. I haven't acquired a firm grip with this subject yet, but your examples and details are helping a good deal. Thanks! 