I am looking at the solutions to a problem that asks me to show

The only subfields of $\mathbb{Q}(i,\sqrt{5})$ are $\mathbb{Q},\mathbb{Q}(i),\mathbb{Q}(\sqrt{5}),\mathbb{Q}(i\sqrt{5}),\mathbb{Q}(i,\sqrt{5})$.

I found the exact same problem here

Show that the only subfields of $\mathbb{Q}(i, \sqrt{5})$ is $\mathbb{Q}, \mathbb{Q}(i),\mathbb{Q}(\sqrt{5}), \mathbb{Q}(i \sqrt{5})$ and itself?

But the answer given is something I was not familiar with; where it says the "quadratic field has form $\mathbb{Q}(\sqrt{D})$"

Though I think it links to my question. The solution reasons that

(It uses the tower law and figures that the extension from $\mathbb{Q}$ must be 4, so the only possbilities are having $1\cdot 4$ or $2\cdot2$.It then assumes that there is a intermediate field $K$ and since $[K:\mathbb{Q}]=2$, it says there is a quadratic such that $\alpha$ where $K=\mathbb{Q}(\alpha)$ satisfies. I assume this is the minimal polynomial of $\alpha$. After all this, it says) Now, by shifting we may assume $\alpha^2 \in \mathbb{Q}$.

If I can understand why $\alpha=\sqrt{D}$ for a quadratic extension, as in this case, I am good. But what does "shifting" mean? Shift what to what? Is this some mathematical jargon?

If someone would explain it to me, that would be appreciated very much

  • $\begingroup$ Hint: Let $a,b\in\mathbb{Q}$ and $D\in\mathbb{Z}$. $\mathbb{Q}(a+b\sqrt{D})=\mathbb{Q}(\sqrt{D})$. $\endgroup$ May 13, 2016 at 11:10

1 Answer 1


Hint: Let $a,b\in\mathbb{Q}$ and $D\in\mathbb{Z}$. $\mathbb{Q}(a+b\sqrt{D})=\mathbb{Q}(\sqrt{D})$. The reason is that the sets are the same. More precisely, $\mathbb{Q}(a+b\sqrt{D})$ is the set of all linear combinations of the form $$ c_1+c_2(a+b\sqrt{D}) $$ while $\mathbb{Q}(\sqrt{D})$ is the set of all linear combinations of the form $$ d_1+d_2\sqrt{D}. $$ Here $c_1,c_2,d_1,d_2\in\mathbb{Q}$.

In your setting, you have $[K:\mathbb{Q}]$ and $K=\mathbb{Q}(\alpha)$. Therefore, $\alpha$ satisfies a degree $2$ polynomial with coefficients in $\mathbb{Q}$, let's write this as $a_0x^2+a_1x+a_2$ with $a_0>0$. Using the quadratic formula, we can write $\alpha=a+b\sqrt{D}$ as above. On the other hand, if we let $y=x-\frac{a_1}{2a_0}$ (compare to completing the square), then this polynomial becomes $a_0y^2+\left(\frac{a_1^2}{4a_0^2}-\frac{a_1^2}{2a_0}+a_2\right)$. If $\beta$ is a root of this polynomial, then $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$. We can even go further by using the substitution $z=y/\sqrt{a_0}$.

  • $\begingroup$ Right, I see that $\mathbb{Q}(a+b\sqrt{D})=\mathbb{Q}(\sqrt{D})$ since by multplication or addition, $\sqrt{D}$ becomes a multiple of itself or $D$. So we only need $\sqrt{D}$ to express all elements in that field. So, then the degree of extension is, as the basis is $\{1, \sqrt{D}\}$, $2$ but the above in my question seems to reason for the other way round/ i.e. degree extension is $2$. thus $\mathbb{Q}(\sqrt{D})$ instead of $\mathbb{Q}(\sqrt{D})$ so degree $2$(which is what you've kindly hinted). $\endgroup$
    – Melba1993
    May 13, 2016 at 11:17
  • $\begingroup$ @Melba1993 If you have an extension $\Bbb Q(\alpha)$ of degree $2$, that means that $\alpha^2$ may be represented as a linear combination of $1$ and $\alpha$, say $\alpha^2 = a + b\alpha$ (because every element in the field may be represented this way). This means that $\alpha$ is a root of the equation $x^2 - bx - a = 0$, which means that $$\alpha = \frac{b\pm\sqrt{b^2 + 4a}}{2}$$for some fixed choice of $\pm$. Therefore $\alpha \in \Bbb Q(\sqrt{b^2 + 4a})$. Also, $$\sqrt{b^2 + 4a} = \pm (2\alpha - b) \in \Bbb Q(\alpha)$$ which means that $\Bbb Q(\alpha) = \Bbb Q(\sqrt{b^2 + 4a})$. $\endgroup$
    – Arthur
    May 13, 2016 at 11:56

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