Unverified Commit 64f94bdb authored by velleto's avatar velleto
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(MR): Accepted helf's MR.

parents a141dbd6 6beb43fb
......@@ -61,7 +61,7 @@ We can easily prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is 4
\implies 3
& = x^2 + 2y^2 + 2\sqrt{2}\underset{= 0}{\underbrace{xy}}
\end{align*}
We know that $xy = 0$ as otherwise the RHS contains irrationals whilst the LHS does not. If, however $xy = 0$, then either $x = 0 \implies y = \sqrt{\frac{3}{2}} \notin \mathbb{Q}$ or $y = 0 \implies x = \sqrt{3} \notin \mathbb{Q}$. Both contradictions. We can do the same for $\mathbb{Q}(\sqrt{2}) \not\subseteq \mathbb{Q}(\sqrt{3})$. Hence we can state that, $\mathbb{Q}(\sqrt{3}, \sqrt{2})$ is an extension over $\mathbb{Q}(\sqrt{2}$, which in turn is an extension over $\mathbb{Q}$, to which we can apply the dimension formula to arrive at our result. $\blacksquare$
We know that $xy = 0$ as otherwise the RHS contains irrationals whilst the LHS does not. If, however $xy = 0$, then either $x = 0 \implies y = \sqrt{\frac{3}{2}} \notin \mathbb{Q}$ or $y = 0 \implies x = \sqrt{3} \notin \mathbb{Q}$. Both contradictions. We can do the same for $\mathbb{Q}(\sqrt{2}) \not\subseteq \mathbb{Q}(\sqrt{3})$. Hence we can state that, $\mathbb{Q}(\sqrt{3}, \sqrt{2})$ is an extension over $\mathbb{Q}(\sqrt{2})$, which in turn is an extension over $\mathbb{Q}$, to which we can apply the dimension formula to arrive at our result. $\blacksquare$
We now want to use this information to think about the autmorphism group of the extension of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}$. Let's go via the extension of the adjoin of only $\sqrt{2}$:
......@@ -120,13 +120,13 @@ In summary, we have exactly two automorphisms. As we have found two, and there a
### Finding a tight bound for the dimension formula
Suppose $E$ is a field. Let $\Sigma = \{ \sigma_1, \ldots, \sigma_n \}$ be distinct automorphisms on $E$. Let $H$ be the field fixed by these automorphisms. Lecture 2 told us that $\mathrm{dim}(E/F) \geq |\Sigma|$. What if $\Sigma$ is a **group**?
Suppose $E$ is a field. Let $\Sigma = \{ \sigma_1, \ldots, \sigma_n \}$ be distinct automorphisms on $E$. Let $H$ be the field fixed by these automorphisms. Lecture 2 told us that $\mathrm{dim}(E/H) \geq |\Sigma|$. What if $\Sigma$ is a **group**?
*Remark* We usually write this so that $\sigma_1$ is the identity.
*Remark.* We usually write this so that $\sigma_1$ is the identity.
Obviously, as $\Sigma$ is a group, its closed under group operations and inverses. We now want to show equality between the size of the automorphism group and the dimension of the extension
**Theorem.** (Lecture 3) *Let $\Sigma, E, F$ be as above. Then we have*
**Theorem.** (Lecture 3) *Let $\Sigma, E, H$ be as above. Then we have*
\begin{align*}
\mathrm{dim}(E/H) = |\Sigma| = n
\end{align*}
......@@ -134,20 +134,20 @@ Obviously, as $\Sigma$ is a group, its closed under group operations and inverse
We already know that the dimension is greater or equal to the size of the group. When is the dimension stricly greater than the size of the group? We will assume this to be the case, and look for a contradiction.
*Proof. (by contradiction)*
Assume $r = \mathrm{dim}(E/H) > n$. Then we have a basis $\{w_1, \ldots, w_r\} \in E$. We now define the matrix $M \in \mathrm{Mat}_{r \times n} (E)$
Assume $r = \mathrm{dim}(E/H) > n$. Then we have a basis $\{w_1, \ldots, w_r\} \in E$. We now define the matrix $M \in \mathrm{Mat}_{n \times r} (E)$
\begin{align*}
M_{ij} = \sigma _i (w_j)
\end{align*}
As $r > n$, $M$ has a non-trivial kernel. Let $0 \neq a \in E$ be such an element.
As $r > n$, $M$ has a non-trivial kernel. Let $0 \neq a \in E^r$ be such an element.
We now claim
(i) **Not all $a_i$ are $0$.** By assumption.
(ii) **Not all $a_i$ are in $H$.** Let $\sigma _1 = \mathrm{id}$. Then the first row of $M$ is just $(w_1, \ldots, w_r)$, and thus the first entry of $Ma$ would be $\sum _i a_i w_i = 0$ - a trivial representation of $0$, but ${w_1, \ldots, w_r}$ form a basis, a contradiction.
(ii) **Not all $a_i$ are in $H$.** Let $\sigma _1 = \mathrm{id}$. Then the first row of $M$ is just $(w_1, \ldots, w_r)$, and thus the first entry of $Ma$ would be $\sum _i a_i w_i = 0$ - a non trivial representation of $0$, but ${w_1, \ldots, w_r}$ form a basis, a contradiction.
(iii) **It can't be that only one $a_i$ is non-zero.** If this were the case (with $a_i$ non-zero), $Ma = a_i \cdot Me_i = a_i \cdot M^{(i)} \neq 0$.
Let's look for elements of the non-trivial kernel. We can characterise this by counting the number of zero entries in the vector. From point (iii), we know that this number can't be 1. Let $s \geq 2$ be the minimal number of non-zero coefficients of any non-zero kernel vector. We can reorder the basis so that the first $s$ entries of $a$ are non-zero. We will get a contradiction by finding a solution with even less non-zero elements.
We now write $a = (a_i, \ldots, a_{s-1}, 1, 0, \ldots, 0)^{\mathrm{T}}$. We can force a $1$ in the $s$'th entry by dividing by $a_s$. We can now consider the equations that arise from $Ma = 0$. Up until this point, we haven't used the group structure of the $\sigma$'s. We remind that not all $a_i$ are not in $H$. Let's claim that $a_1$ is not in $H$.
We now write $a = (a_1, \ldots, a_{s-1}, 1, 0, \ldots, 0)^{\mathrm{T}}$. We can force a $1$ in the $s$'th entry by dividing by $a_s$. We can now consider the equations that arise from $Ma = 0$. Up until this point, we haven't used the group structure of the $\sigma$'s. We remind that not all $a_i$ are not in $H$. Let's claim that $a_1$ is not in $H$.
Now the idea is to take the $a_1 \notin H$. What does it mean for $a_1 \notin H$? Well, we know that it is not fixed by some $\sigma \in \Sigma$. Let $\sigma_k$ be the automorphism that moves it. As the set of equations $Ma = 0$ are all in $E$, we can apply $\sigma _k$. Let's do exactly that:
\begin{align*}
......
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