State and prove a necessary and sufficient condition on $\alpha, \beta \in F$ so that $F(\sqrt{\alpha})=F(\sqrt{\beta})

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In summary, a necessary and sufficient condition means that the condition must be both required and enough for a particular statement or situation to be true. The square root of an element is a necessary and sufficient condition for the element to be algebraic over a field, making it a useful tool for constructing field extensions. The difference between a necessary and sufficient condition is that a necessary condition is not enough on its own to guarantee the truth of a statement, while a sufficient condition is. An example of $\alpha$ and $\beta$ that satisfy the necessary and sufficient condition in the statement is $\alpha=2$ and $\beta=8$ in the field of rational numbers. Proving that a condition is both necessary and sufficient involves demonstrating that it is necessary
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Chris L T521
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Here's this week's problem.

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Problem
: Let $F$ be a field of characteristic not equal to $2$. State and prove a necessary and sufficient condition on $\alpha, \beta \in F$ so that $F(\sqrt{\alpha})=F(\sqrt{\beta})$. Use this to determine whether $\mathbb{Q}(\sqrt{1-\sqrt{2}})=\mathbb{Q}(i, \sqrt{2}).$

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This week's problem was correctly answered by johng and Opalg. You can find Opalg's solution below.

[sp]The equation $F(\sqrt\alpha) = F(\sqrt\beta)$ must mean that $F(\sqrt\alpha)$ and $F(\sqrt\beta)$ should coincide when considered as subfields of $F(\sqrt\alpha,\sqrt\beta)$.

To avoid trivial cases, assume that neither $\alpha$ nor $\beta$ has a square root in $F$. If $F(\sqrt\alpha) = F(\sqrt\beta)$ then in particular $\sqrt\beta \in F(\sqrt\alpha)$, so there are elements $a,b\in F$ such that $\sqrt\beta = a+b\sqrt\alpha$. Square both sides to get $\beta = a^2 + b^2\alpha + 2ab\sqrt\alpha$. Since $\sqrt\alpha \notin F$, its coefficient $2ab$ must be $0$. But $b\ne0$ because that would imply $\sqrt\beta = a \in F$. Also $2\ne0$ (that's where we use the fact that $\mathrm{char}(F)\ne2$). Therefore $a=0$, so that $\sqrt\beta = b\sqrt\alpha$ and consequently $\beta = b^2\alpha.$

Conversely, if $\beta = b^2\alpha$ then $\sqrt\alpha$ and $\sqrt\beta$ are multiples of each other by an element of $F$, and therefore $F(\sqrt\alpha) = F(\sqrt\beta)$.

Putting it into words, the necessary and sufficient condition for $F(\sqrt\alpha) = F(\sqrt\beta)$ is that $\alpha/\beta$ should have a square root in $F$.

To determine whether $\mathbb{Q}(\sqrt{1-\sqrt2}) = \mathbb{Q}(i,\sqrt2)$, take $F = \mathbb{Q}(\sqrt2)$, $\alpha = 1-\sqrt2$ and $\beta = -1$. The necessary and sufficient condition then says that these two fields will be the same if and only if $\sqrt2-1$ has a square root in $ \mathbb{Q}(\sqrt2)$. But if $\sqrt2-1 = (x+y\sqrt2)^2 = x^2 + 2y^2 + 2xy\sqrt2$ (with $x,y\in \mathbb{Q}$) then $-1 = x^2+2y^2$ and $1 = 2xy$. The first of those equations clearly has no solution. Therefore those two fields are not the same.

Interesting corollary. The proof that $F(\sqrt\alpha) = F(\sqrt\beta)$ implies that $\alpha/\beta$ has a square root in $F$ only used the fact that $F(\sqrt\beta) \subseteq F(\sqrt\alpha)$. It follows that, provided that neither $\alpha$ nor $\beta$ has a square root in $F$, the condition $F(\sqrt\beta) \subseteq F(\sqrt\alpha)$ automatically implies the reverse inclusion $F(\sqrt\alpha) \subseteq F(\sqrt\beta)$.[/sp]
 

FAQ: State and prove a necessary and sufficient condition on $\alpha, \beta \in F$ so that $F(\sqrt{\alpha})=F(\sqrt{\beta})

What does it mean for two elements in a field to satisfy a necessary and sufficient condition?

A necessary and sufficient condition means that the condition must be both required and enough for a particular statement or situation to be true. In this case, the condition on $\alpha$ and $\beta$ in $F(\sqrt{\alpha})=F(\sqrt{\beta})$ must be both necessary and sufficient for the two elements to generate the same field extension.

What is the significance of $\sqrt{\alpha}$ and $\sqrt{\beta}$ in this statement?

The square root of an element in a field is important because it is a necessary and sufficient condition for the element to be algebraic over the field. In other words, the square root of an element allows us to extend the field by including all of its powers, making it an important tool for constructing field extensions.

What is the difference between a necessary and a sufficient condition?

A necessary condition is one that must be satisfied for a statement to be true, but it is not enough by itself to guarantee the truth of the statement. On the other hand, a sufficient condition is one that alone is enough to guarantee the truth of the statement. In this case, the condition on $\alpha$ and $\beta$ must be both necessary and sufficient for the two elements to generate the same field extension.

Can you provide an example of $\alpha$ and $\beta$ that satisfy the necessary and sufficient condition in the statement?

Yes, for example, if we take $\alpha=2$ and $\beta=8$ in the field of rational numbers $\mathbb{Q}$, then $F(\sqrt{2})=F(\sqrt{8})$ since both generate the same field extension $\mathbb{Q}(\sqrt{2})$.

How do we prove that a condition is both necessary and sufficient for the statement to be true?

In order to prove that a condition is both necessary and sufficient, we need to show that it is necessary by demonstrating that if the condition is not satisfied, then the statement is false. Additionally, we need to show that the condition is sufficient by demonstrating that if the condition is satisfied, then the statement is true. In this case, we need to show that if $\alpha$ and $\beta$ satisfy the given condition, then they generate the same field extension, and if they do not satisfy the condition, then they do not generate the same field extension.

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