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If it is known that $\mathbb{Q}$ is a PID, and that $x^2-2$ is irreducible in $\mathbb{Q}[x]$ (and thus maximal in $\mathbb{Q}[x]$), then how can we determine from this that $\mathbb{Q}[\sqrt{2}]$ is a field? I think there is a theorem that I do not know or have forgotten about in order to arrive at this conclusion. Would definitely appreciate some insight.

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    Explicitly find the inverse for an arbitrary nonzero element. – 2017-02-27
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    Thanks, I did that. I was nevertheless wondering how the desired conclusion could be drawn from the facts above. – 2017-02-27

3 Answers 3

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I believe you are thinking about results of simple algebraic extensions. $F[x]/(p(x))$ will be a field containing a root to $p(x)$ as long as $p(x)$ is irreducible, thus maximal. $R/M$ is a field when $M$ is maximal. So you are thinking about $\mathbb{Q}[X]/(X^2-2)$

http://sierra.nmsu.edu/morandi/OldWebPages/Math331Spring2003/Chapter5.pdf

This seems a decent writeup with plenty of examples to refresh you.

It is like the construction of $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$

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You might have learnt how to rationalize the denominator in expressions of the form $\displaystyle \frac{a+b\sqrt n}{c+d\sqrt n}$. This procedure, when the numerator is taken as $1\ ( a=1, b=0)$ proves that $\mathbf{Q}[\sqrt n]$ is a field.

In case you are not aware of this procedure it simply means muliply the numerator and the denominator by the 'conjugate' of the denominator, ie by $c-d\sqrt n$.

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    Thanks, I certainly realize that. I was nevertheless wondering how the desired conclusion could be drawn from the facts listed in my question, since the original question that I was trying to solve assumed I had to do just that - draw from those facts. – 2017-02-27
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    When you quotient by the ideal generated by $x^2-2$, the quotient ring will have only linear expressions in $x$. So we need to show all these linear expressions have their inverses also expressible as linear expressions. – 2017-02-27
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    $\mathbb{Q}[x]/\langle x^2-1 \rangle=f(x) + \langle x^2-1 \rangle$, so I don't see how the members of this coset would be linear expressions. – 2017-02-27
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    $x^2 + \langle x^2-1 \rangle = 1 + \langle x^2-1 \rangle$, and $x^3 + \langle x^2-1 \rangle = x + \langle x^2-1 \rangle$, and $x^4 + \langle x^2-1 \rangle = 1 + \langle x^2-1 \rangle$, and so on. So if you have an expression: $c_0+c_1x+c_2x^2+\cdots+c_nx^n + \langle x^2 - 1 \rangle$, then you can show that each term with degree $\geq 2$ is equivalent modulo this ideal to a term with degree $0$ or $1$. – 2017-02-27
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Here is the theorem :

Let $a$ be algebraic over the field $F$ and let $F(a)$ be the field generated by $a$ over $F$. Then $F(a) \cong F[x]/(m_a(x))$ where $m_a(x)$ is the minimal irreducible monic polynomial having $a$ as a root.

By this theorem you can directly arrive at your desired conclusion.

EDIT 1:

  1. $a$ is called algebraic over the field $F$ if $a$ is a root of some polynomial in $F[x]$.
  2. Also minimal irreducible monic polynomial having $a$ as a root, has smallest degree amongst all those polynomials having $a$ as a root. We denote it by $m_a(x)$.
  3. You can plug in $a=\sqrt 2$ and $m_a(x)=x^2-2$. Only thing you want to prove is that $x^2-2$ is the smallest degree polynomial with $\sqrt 2$ as a root. (Hint: Use contradiction with degree $0$ and degree $1$ polynomials.)

EDIT 2: (As per suggestion by Don Antonio in the comments)

Theorem :- $F(a)=F[a] \iff a \;\text{is algebraic over} \; F$.

Proof - Let $a$ be algebraic over $F$ and $p(x) \in F[x]$ be the minimal irreducible polynomial having $a$ as a root. Let $f(x) \in F[x]$ be such that $f(a) \neq 0$. Then $p(x) \not| f(x)$ and $p(x),f(x)$ are relatively prime. Hence there exist polynomials $g(x)$ and $h(x) \in F[x]$ such that $g(x)p(x)+h(x)f(x)=1 \Rightarrow g(a)(0)+h(a)f(a)=1 \Rightarrow h(a)f(a)=1$. That is $f(a)$ is invertible in $F[a]$. This means $F[a]$ is field. Since any field containing $F$ and $a$ also contains polynomials in $a$ hence $F[a] \subseteq F(a)$. But $F(a)$ is by definition the smallest field containing $F$ and $a$. Hence $F(a) \subseteq F[a]$. It follows $F(a)=F[a]$.

Conversely, assume $F(a)=F[a]$. Suppose $a$ is transcendental over $F$. Since $F(a)$ is smallest field containing $F$ and $a$, it also contains $\frac 1a$. But $\frac 1a \notin F[a]$. Hence we have a contradiction. and $a$ must be algebraic over $F$.

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    That's, but that seems too advanced for me yet (haven't covered some of the concepts which this theorem refers to). – 2017-02-27
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    Are you unfamiliar with the terms 'algebraic' and 'minimal irreducible monic polynomial'? – 2017-02-27
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    I'm not sure this answers or even helps the OP: he wants $\;F[a]=\;$ the polynomial ring with $\;a\;$ as unknown, so that he has to prove that in fact $\;F[a]=F(a)\iff a\;$ is algebraic over $\;F\;$ . You simply begin with **the field** $\;F(a)\;$ ... – 2017-02-27
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    @DonAntonio Oh okay. I didn't think it's necessary to add proof for $F[a]=F(a) \iff \; \text{a is algebraic over} F$. Also I thought that OP could learn something new by starting from here and then searching on his/her own further. But if you feel that I should include the proof then I will definitely do so. Thanks for suggestion. – 2017-02-27
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    @VikrantDesai It's just that **that exactly** is the OP's question: why is $\;\Bbb Q[a]\;$ a field, *not* $\;\Bbb Q(a)\;$ , which is trivially a field *by definition* . – 2017-02-27
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    @DonAntonio Thank you very much for this. Sometimes we get too much used to some things that we don't think to mention the details. EDIT 2 is in order. :) – 2017-02-27
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    @DonAntonio Please have a look at my answer now. If any further criticism, then let me know. – 2017-02-27
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    @VikrantDesai Very nice indeed. +1 Perhaps only to explain a little more why "But\frac1a\notin F[a]\;$...", but I think this bit can safely be let to the OP to complete. – 2017-02-27
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    @DonAntonio Yeah I left that part for the asker. :D Thanks again! – 2017-02-27