I am trying to prove Q(√2)={a+b√2|a,b∈Q}\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in \mathbb{Q}\}. I have proved that {a+b√2|a,b∈Q}\{a+b\sqrt{2}|a,b\in \mathbb{Q}\} is indeed a field with unit. However is this enough to show that this is indeed equivilant to Q(√2)\mathbb{Q}(\sqrt{2})?

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What can be said about the number (√2)n(\sqrt{2})^n? Hence, what can be said about P(√2)P(\sqrt{2}) if P(x):=a0+a1x+⋯+anxnP(x) := a_0 + a_1 x + \cdots + a_n x^n where the ai∈Qa_i \in \mathbb{Q}?

– user1892304

2 days ago

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How is Q(√2)\mathbb{Q}\left(\sqrt{2}\right) defined?

– Evan Aad

2 days ago

Where Q(√2)\mathbb{Q}(\sqrt{2}) is the field of Q\mathbb{Q} with all possible unions of √2\sqrt{2} under the operations.

– Daniel Houston

2 days ago

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3 Answers

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The minimal polynomial of √2\sqrt{2} over Q\mathbb{Q} is t2−2t^2-2, so that [Q(√2):Q]=2[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2. A basis of the 22-dimensional vector space

Q(√2)\mathbb{Q}(\sqrt{2}) over Q\mathbb{Q} is (1,√2)(1,\sqrt{2}), which is easy to see. Hence every element has the form a+b√2a+b\sqrt{2}. To verify that it is a field has been done on MSE already, see for example here.

That makes since, and I have already shown that this is a field.

– Daniel Houston

2 days ago

Q(√2)\mathbb{Q} (\sqrt{2}) is defined to be the intersection of every subfield of C\mathbb{C} that contains √2\sqrt{2}. As such it is a subfield of C\mathbb{C} and is contained in every subfield of C\mathbb{C} that contains √2\sqrt{2}. So if you have shown that {a+b√2:a,b∈Q}\{ a+b \sqrt{2} : a,b \in \mathbb{Q} \} is a field (which clearly has √2\sqrt{2} as one of its elements) you have shown Q(√2)⊆{a+b√2:a,b∈Q}\mathbb{Q} (\sqrt{2}) \subseteq \{ a+b \sqrt{2} : a,b \in \mathbb{Q} \}. To show the inclusion the other way you need to show why Q(√2)\mathbb{Q} (\sqrt{2}) must contain every rational as well as √2\sqrt{2}. Then since Q(√2)\mathbb{Q} (\sqrt{2}) is a subfield of C\mathbb{C} it must contain any sum and product of such elements.

You need to show that {a+b√2}⊆Q(√2)\{a+b\sqrt{2}\} \subseteq \mathbb{Q}(\sqrt{2}), and that Q(√2)⊆{a+b√2}\mathbb{Q}(\sqrt{2}) \subseteq \{a+b\sqrt{2}\}.

Then it follows that Q(√2)={a+b√2}\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}\}. So you don’t have to prove that {a+b√2}\{a+b\sqrt{2}\} has unit and inverse, if it equals a field then it definitely already has those.