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To prove that $X^4-23$ is irreducible in $\mathbb{Q}[X]$ we can do the following:

We use the Eisenstein criterion with $a=23$ to see that it is irreducible in $\mathbb{Z}[X]$, and then we conclude that it must be irreducible in $\mathbb{Q}[X]$ because $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$.

I was trying to see if a similar argument was possible for $X^4-\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})[X]$ using the fact that $\mathbb{Q}(\sqrt{2})=\mathbb{Q}[\sqrt{2}]$ (this equality commonly known in field theory) is the field of fractions of $\mathbb{Z}[\sqrt{2}]$.

The purpose of all this is to verify that $X^4-\sqrt{2}=Irr(\sqrt[8]{2},\mathbb{Q}(\sqrt{2}))$.

IAG
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  • How about using brute force? Since your target is of degree 4, if it is reducible, you must factorize it into deg 2 $\times$ deg 2. Then you calculate the coefficients and draw a contradiction. – SouthChinaSeaPupil Apr 06 '24 at 12:17
  • @WilliamZheng True, but I like to look and learn a not brute way too, or to find one if it is possible : ) – IAG Apr 06 '24 at 12:46

4 Answers4

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Note that $f(x)=x^8-2$ is irreducible over $\mathbb{Q}$ by Eisenstein. Also, $x^8-2=(x^4-\sqrt{2})(x^4+\sqrt{2})$. So if $\alpha\in\mathbb{C}$ is any root $x^4-\sqrt{2}$ then $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and so:

$8=[\mathbb{Q}(\alpha):\mathbb{Q}]\leq [\mathbb{Q}(\alpha,\sqrt{2}):\mathbb{Q}]=[\mathbb{Q}(\alpha,\sqrt{2}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]\leq$

$\leq 2\cdot [\mathbb{Q}(\alpha,\sqrt{2}):\mathbb{Q}(\sqrt{2})]$.

Thus, $x^4-\sqrt{2}$ must be the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$, as otherwise the extension degree $[\mathbb{Q}(\alpha,\sqrt{2}):\mathbb{Q}(\sqrt{2})]$ would be smaller than $4$, a contradiction. In particular $x^4-\sqrt{2}$ is irreducible over $\mathbb{Q}(\sqrt{2})$.

Mark
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  • This is a beautiful argument. How did you come to it? Do you know another example in which you would use this technique so I can grasp it in a more general way? Thanks – IAG Apr 06 '24 at 13:27
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    @IAG Since it's in general difficult to work over fields like $\mathbb{Q}(\sqrt{2})$, it makes sense to try to reduce this to a problem over $\mathbb{Q}$ where we have arguments like the Eisenstein criterion. How to do it, that depends on the problem. Here I just saw immediately that the polynomial is a factor of $x^8-2$, which is irreducible over $\mathbb{Q}$. But in general, it is often useful to prove irreducibility by using extension degree, which can show that this is the minimal polynomial of its root. (like I did here) – Mark Apr 06 '24 at 13:42
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The non-trivial factors of $\ X^4-\sqrt{2}\ $ are $\ X-2^\frac{1}{8},$$\,X+2^\frac{1}{8},$$\,X-i2^\frac{1}{8},$$\,X+i2^\frac{1}{8},$$\,X^2-2^\frac{1}{4},$$\,X^2+2^\frac{1}{4},$$\,X^2-(1+i)X+i2^\frac{1}{4},$$\,X^2-(1-i)2^\frac{1}{8}X-i2^\frac{1}{4},$$\,X^2+(1-i)2^\frac{1}{8}X-i2^\frac{1}{4},$$\,X^2+(1-i)2^\frac{1}{8}X+i2^\frac{1}{4},$$\,X^3+2^\frac{1}{8}X^2+2^\frac{1}{4}X+2^\frac{3}{8},$$\,X^3-2^\frac{1}{8}X^2+2^\frac{1}{4}X-2^\frac{3}{8},$$\,X^3+i2^\frac{1}{8}X^2-2^\frac{1}{4}X-i2^\frac{3}{8}\ ,$ and $ X^3-i2^\frac{1}{8}X^2-2^\frac{1}{4}X+i2^\frac{3}{8}\ ,$ none of which belong to $\ \Bbb{Q}\big(\sqrt{2}\big)[X]\ .$

lonza leggiera
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  • This is nice as a general way to see it, brute force works if one is enough patient. : ) – IAG Apr 06 '24 at 13:31
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More generally, the polynomial $X^n - \sqrt{2}$ is irreducible over $\mathbb{Q}(\sqrt{2})$ for any $n \ge 1$. Indeed, the ring $\mathbb{Z}[\sqrt{2}]$ is a UFD (in fact, a Euclidean domain) so applying Eisenstein's criterion with respect to the prime $\sqrt{2}$ gives us irreducibility over $\mathbb{Z}[\sqrt{2}]$. The polynomial remains irreducible over the field of fractions $\mathbb{Q}(\sqrt{2})$ by Gauss lemma.

Daniel Arreola
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Just for beginners to whom it may interest.

►If $f(X)=X^4-\sqrt2$ has an unitary linear factor $X-(a+b\sqrt2)$ in the field $\Bbb Q(\sqrt2)$ then $(a+b\sqrt2)^4-\sqrt2=0$ which gives $a^4+4b^4+12a^2b^2+(4a^3b+8ab^3)\sqrt2=\sqrt2\Rightarrow a=b=0$. Thus $f(X)$ has neither linear factor nor cubic factor.

►► By division in general we have . $$\frac{X^4-\sqrt2}{X^2+AX+B}=X^2-AX+(A^2-B)-\frac{A(A^2-b)X+B(A^2-B)+\sqrt2} {X^2+AX+B}$$ so if $X^2+AX+B$ is a divisor of $f(X)$, the residue should be equal to $0$ so $$A(A^2-B)X+B(A^2-B)+\sqrt2=0$$ and because it is a polynomial we have the coefficients are null $$A(A^2-B)=B(A^2-B)+\sqrt2=0$$ which implies a contradiction straightforward deduced.

Ataulfo
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