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I'm trying to solve the following problem.

Let $\mathbb{Q}(\sqrt{2}) = \{a + b \sqrt{2} : a,b \in \mathbb{Q}\}$. Prove that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{R}$.

The hint given in the problem is that it's enough to check that addition, negation, multiplication, and inversion are well-defined in $\mathbb{Q}(\sqrt{2})$.

I'm not sure I fully understand exactly what I need to prove. Surely $\mathbb{Q}(\sqrt{2})$ is a subset of $\mathbb{R}$, and I'm attempting to endow it with the inherited operations from $\mathbb{R}$, so the addition is the same addition on $\mathbb{R}$ restricted to $\mathbb{Q}(\sqrt{2})$, and similarly for the multiplication. But I'm not sure I fully understand, given the context, what a subfield is. Is it simply a subset that admits the structure of a field? In that case, I'd need to take this subset with the two given operations, and check each of the field axioms. Distributivity, commutativity, and associativity are almost given for free from the inherited operation, which leaves the axioms the problem statement says I should check.

Is this the idea?

JohnT
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  • "Is this the idea?" Yes. You don't have to check the properties that are "inherited" from $\mathbb R$. Addition, multiplication are still commutative and associative. What's more $1$ and $0$ are still the multiplicative and additive identities. But you have more elements. You don't the that every element has an additive inverse (negation) or that every non-negative element has a multiplicative inverse. So you must show those. But otherwise all other properties are done. – fleablood Jan 15 '22 at 23:13
  • It might not be a bad idea for the experience and understanding to go through the definition of a field axiom by axiom and state why it holds. We it comes to, for example. "$+$ is associative" to state "for all $q,r,s\in \mathbb Q[\sqrt 2]$ then$q,r,s\in \mathbb R$ as $\mathbb Q[\sqrt 2]$ is a subset of $\mathbb R$. As $+$ is associative in $\mathbb R$ we have $q+(r+s)=(q+r)+s$. Therefore $+$ is associative in $\mathbb Q[\sqrt]$." This may be tedious and get silly but it will prove that $\mathbb Q$ is a field. – fleablood Jan 15 '22 at 23:19
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    Oh.... you do have to prove addition and multiplication are closed That if $r,s\in \mathbb Q[\sqrt 2]$ then $r + s$ and $rs$ are both in $\mathbb Q[\sqrt 2]$ as well. – fleablood Jan 15 '22 at 23:20
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    "But I'm not sure I fully understand, given the context, what a subfield is. Is it simply a subset that admits the structure of a field?" Well, the superset is a field with elements and operations. A subfield will have the same operations and a subset of the elements. The elements must have the attributes of a field with the operations. An example of something that isn't a subfield would be ${1,5,\pi}$ it still has the operations and the multiplicative identity and the ops are assoc, commut but... c'mon. They aren't closed ($1+5=6$ is not in the set and all sorts of other wrong stuff. – fleablood Jan 15 '22 at 23:48
  • Very helpful, thank you! – JohnT Jan 16 '22 at 00:18

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