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Prove that the $q(x)=X^2+3$ is irreducible over $\Q[\sqrt{2}]$. Is my proof correct?

Proof. Since it is a polynomial of second degree it factors iff it has $2$ polynomials of degree $1$. Since we are looking for a factorization over $F[x]$ where $F$ is a field.The factorization is unique.

Now I know factorization is always possible over $\Bbb C$. So $(X^2-i\sqrt{3})(X^2+i\sqrt{3})$.!!! And this factorization is unique. And it cannot be factored in any other way to any subfields of $\Bbb C$!!. (I use this fact by intuition don't know why???). So only job left is to prove that $$i\sqrt{3}$$ does not belong to $\Q[\sqrt{2}]$. Say for contradiction it does then $$i\sqrt{3}=a+b\sqrt{2}$$ ......$$\sqrt{2} =\frac{a^2+2b^2+3}{ab} $$ where a,b are integers so we have a contradiction. Does my proof miss a step where I should prove something that I have taken for granted? Is it bulletproof and complete? Also. Another way to go is that this polynomial is equal to the minimal polynomial of the extension $\Q[i\sqrt{3}]:\Q[\sqrt{2}]$. But where I stuck is that $$X^2+3$$ is irreducible over $\Q$ and $ i\sqrt{3}$is a root so $\Q[i\sqrt{3}]:\Q$ is of degree 2 but that doesn't make sense since $$[\Q[i\sqrt{3}]:\Q]=[\Q[i\sqrt{3}]:\Q[\sqrt{2}] ][\Q[\sqrt{2}]:\Q]$$ so $2=2 *1$ which means $[\Q[i\sqrt{3}]=\Q[\sqrt{2}]$ which is a contradiction. Am I correct? Any help on how to do a proof with the second method?

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    $\begingroup$ If $Q$ is $\mathbb{Q}$, then as you said in the first line, if it is reducible, then it has a root. Now notice that $x^2+3\ge 3$ for all choices of $x\in\mathbb{R}\supset \mathbb{Q}[\sqrt{2}]$. $\endgroup$ – b00n heT May 31 '16 at 15:30
  • $\begingroup$ trying to avoid as many "granted steps as much" I mean Your using the fact that it is a subfield and also that $x^2.>=0$ for all x in R which isnt obvious. $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:45
  • $\begingroup$ I am sorry to tell you, but must be most definitely drunk. You are overcomplicating things and claiming that I am using hard math, although I am not even using the irrationality of $\sqrt{2}$... $\endgroup$ – b00n heT May 31 '16 at 15:52
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A monic quadratic polynomial $f(x)$ over a field $K$ is reducible if and only if it can be written as $(x-a)(x-b)$ with $a,b\in K$. Now take $f(x)=x^2+3$ and $K=\mathbb{Q}[\sqrt{2}]$. We have $a=x_1+y_1\sqrt{2}$ and $b=x_2+y_2\sqrt{2}$. Compute $(x-a)(x-b)$ and compare it with $x^2+3$.

Remark: The degree $[L:K]$ is only defined for $K\subseteq L$. This is not the case with your two fields.

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  • $\begingroup$ I thought of this as well at first, but I think my way is even faster (see comment) $\endgroup$ – b00n heT May 31 '16 at 15:31
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    $\begingroup$ Perhaps your mean "reducible" where you write "irreducible", especially since you immediately reduce it to factors. $\endgroup$ – Eric Towers May 31 '16 at 15:32
  • $\begingroup$ WHat about my second method? Where am i wrong $Q$ is a subfield of $Q[i\sqrt{3}]$ $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:36
  • $\begingroup$ OOOOOH $Q[\sqrt{2}]$ is not a subfield of $Q[i\sqrt{3}]$ Didnt notice that.!!! $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:38
  • $\begingroup$ Is my method Correct? $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:56
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Simply: $ \mathbb Q(\sqrt{2})=\{a+b\sqrt{2}| a,b\in \mathbb Q\}$. So, if $t\in \mathbb Q(\sqrt{2})$, then $q(t)=t^2+3\geq 3$. If a second degree polynom has no root, it is irreducible.

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  • $\begingroup$ Have to prove that this is true for all t. $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:40
  • $\begingroup$ The square of real number is greater or equal to zero $\endgroup$ – guestDiego May 31 '16 at 15:47
  • $\begingroup$ Trying to use as less as possible "granted" results. $\endgroup$ – Manolis Lyviakis May 31 '16 at 15:49
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    $\begingroup$ Ok! So let's avoid completely the short argument. Let $ t=a+b\sqrt{2}$. Then: $q(t)=(a^2+2b^2+3)+2\sqrt{2}ab$ This expression can't be zero because $\sqrt{2}$ is not rational. $\endgroup$ – guestDiego May 31 '16 at 15:52
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    $\begingroup$ That is also correct: you can factor a polynom on $\mathbb C$ through its roots (so the factorization is unique if you use monic polynomails: the roots determine it completely). Now $ \mathbb Q(\sqrt{2})$ is a subfield of $\mathbb C$, so the game is done. $\endgroup$ – guestDiego May 31 '16 at 16:04

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