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# primitive element of biquadratic field

###### Theorem.

Let $m$ and $n$ be distinct squarefree integers, neither of which is equal to $1$. Then the biquadratic field $\mathbb{Q}(\sqrt{m},\sqrt{n})$ is equal to $\mathbb{Q}(\sqrt{m}+\sqrt{n})$.

In other words, $\sqrt{m}+\sqrt{n}$ is a primitive element of $\mathbb{Q}(\sqrt{m},\sqrt{n})$.

###### Proof.

We clearly have $\mathbb{Q}(\sqrt{m}+\sqrt{n})\subseteq\mathbb{Q}(\sqrt{m},\sqrt{n})$. For the reverse inclusion, it is equivalent to show that $\sqrt{m}+\sqrt{n}$ does not belong to any of the quadratic subfields of $\mathbb{Q}(\sqrt{m},\sqrt{n})$, which are $\mathbb{Q}(\sqrt{m})$, $\mathbb{Q}(\sqrt{n})$, and $\mathbb{Q}(\sqrt{mn})$.

Suppose that $\sqrt{m}+\sqrt{n}\in\mathbb{Q}(\sqrt{m})$. Then $\sqrt{n}\in\mathbb{Q}(\sqrt{m})$. Thus, $\mathbb{Q}(\sqrt{n})=\mathbb{Q}(\sqrt{m})$, which is proven to be false here. By a similar argument, $\sqrt{m}+\sqrt{n}\notin\mathbb{Q}(\sqrt{n})$.

Suppose that $\sqrt{m}+\sqrt{n}\in\mathbb{Q}(\sqrt{mn})$. Let $a,b,c,d\in\mathbb{Z}$ with $\gcd(a,b)=\gcd(c,d)=1$, $b\neq 0$, and $d\neq 0$ such that

$\displaystyle\sqrt{m}+\sqrt{n}=\frac{a}{b}+\frac{c}{d}\sqrt{mn}.$ | (1) |

$\displaystyle bd\sqrt{m}+bd\sqrt{n}$ | $\displaystyle=ad+bc\sqrt{mn}$ | ||

$\displaystyle bd\sqrt{m}+bd\sqrt{n}-ad$ | $\displaystyle=bc\sqrt{mn}$ | ||

$\displaystyle(bd\sqrt{m}+bd\sqrt{n}-ad)^{2}$ | $\displaystyle=(bc\sqrt{mn})^{2}$ | ||

$\displaystyle b^{2}d^{2}m+2b^{2}d^{2}\sqrt{mn}-2abd^{2}\sqrt{m}+b^{2}d^{2}n-2% abd^{2}\sqrt{n}+a^{2}d^{2}$ | $\displaystyle=b^{2}c^{2}mn$ | ||

$\displaystyle b^{2}d^{2}m+b^{2}d^{2}n+a^{2}d^{2}-b^{2}c^{2}mn$ | $\displaystyle=2abd^{2}(\sqrt{m}+\sqrt{n})-2b^{2}d^{2}\sqrt{mn}$ |

Now, we use equation (1) to eliminate the $\sqrt{m}+\sqrt{n}$ and obtain

$\displaystyle b^{2}d^{2}m+b^{2}d^{2}n+a^{2}d^{2}-b^{2}c^{2}mn$ | $\displaystyle=2abd^{2}\left(\frac{a}{b}+\frac{c}{d}\sqrt{mn}\right)-2b^{2}d^{2% }\sqrt{mn}.$ |

Now, we perform some more basic algebraic manipulations.

$\displaystyle b^{2}d^{2}m+b^{2}d^{2}n+a^{2}d^{2}-b^{2}c^{2}mn$ | $\displaystyle=2a^{2}d^{2}+2abcd\sqrt{mn}-2b^{2}d^{2}\sqrt{mn}$ | ||

$\displaystyle b^{2}d^{2}m+b^{2}d^{2}n-a^{2}d^{2}-b^{2}c^{2}mn$ | $\displaystyle=2bd(ac-bd)\sqrt{mn}$ |

Since $\sqrt{mn}\notin\mathbb{Q}$, $b\neq 0$, and $d\neq 0$, we must have $ac-bd=0$. Thus, $\frac{c}{d}=\frac{b}{a}$. (Note that we have $a\neq 0$ since $ac=bd\neq 0$.) Using this in equation (1), we obtain

$\sqrt{m}+\sqrt{n}=\frac{a}{b}+\frac{b}{a}\sqrt{mn}.$ |

Now we perform similar calculations as before.

$\displaystyle ab\sqrt{m}+ab\sqrt{n}$ | $\displaystyle=a^{2}+b^{2}\sqrt{mn}$ | ||

$\displaystyle ab\sqrt{m}+ab\sqrt{n}-a^{2}$ | $\displaystyle=b^{2}\sqrt{mn}$ | ||

$\displaystyle(ab\sqrt{m}+ab\sqrt{n}-a^{2})^{2}$ | $\displaystyle=(b^{2}\sqrt{mn})^{2}$ | ||

$\displaystyle a^{2}b^{2}m+2a^{2}b^{2}\sqrt{mn}-2a^{3}b\sqrt{m}+a^{2}b^{2}n-2a^% {3}b\sqrt{n}+a^{4}$ | $\displaystyle=b^{4}mn$ | ||

$\displaystyle a^{2}b^{2}m+a^{2}b^{2}n+a^{4}-b^{4}mn$ | $\displaystyle=2a^{3}b(\sqrt{m}+\sqrt{n})-2a^{2}b^{2}\sqrt{mn}$ | ||

$\displaystyle a^{2}b^{2}m+a^{2}b^{2}n+a^{4}-b^{4}mn$ | $\displaystyle=2a^{3}b\left(\frac{a}{b}+\frac{b}{a}\sqrt{mn}\right)-2a^{2}b^{2}% \sqrt{mn}$ | ||

$\displaystyle a^{2}b^{2}m+a^{2}b^{2}n+a^{4}-b^{4}mn$ | $\displaystyle=2a^{4}+2a^{2}b^{2}\sqrt{mn}-2a^{2}b^{2}\sqrt{mn}$ | ||

$\displaystyle a^{2}b^{2}m+a^{2}b^{2}n-a^{4}-b^{4}mn$ | $\displaystyle=0$ |

Since $b^{2}$ divides $a^{4}$ and $\gcd(a,b)=1$, we must have $b^{2}=1$. Plugging into the equation above yields

$a^{2}m+a^{2}n-a^{4}-mn=0.$ |

Now for yet some more algebraic manipulations.

$\begin{array}[]{rl}a^{2}m-a^{4}-mn+a^{2}n&=0\\ a^{2}(m-a^{2})-n(m-a^{2})&=0\\ (m-a^{2})(a^{2}-n)&=0\end{array}$

Thus, $m=a^{2}$ or $n=a^{2}$, a contradiction. It follows that $\mathbb{Q}(\sqrt{m}+\sqrt{n})=\mathbb{Q}(\sqrt{m},\sqrt{n})$. ∎

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## Comments

## oddity of equations in html mode

In this entry, in html mode, the last set of centered equations are shifted quite far to the right. In page images mode, it displays the way I want it to (i.e., no shift to the right). Any ideas how to make the html look better without wrecking how the entry looks in page images mode?

Warren

## Re: oddity of equations in html mode

There must be something weird going on. Look at the stub entry PrescisiveAbstraction that I just now used as a sandbox. Also, the system seems really slow tonight.

Jon Awbrey

## Re: oddity of equations in html mode

Try \begin{equation} ... \end{equation}

## Re: oddity of equations in html mode

Thanks for the suggestion Chi. Unfortunately, it did not work. In case anyone wants to know, this is the workaround I used:

\begin{center}

$\begin{array}{rl}

...

\end{array}$

\end{center}

I used to use the array and center environments together a lot, but got out of the habit since the align environment centers and converts to math mode without the use of dollar signs; moreover, the align environment has an extra benefit in that I do not have to add displaystyles as often. Unfortunately, due to the weird shift, array is better in this particular scenario.

Warren

## slower than a snail (was oddity of equations in html mode)

> Also, the system seems really slow tonight.

Jon Awbrey said this yesterday. I noticed a problem with the speed yesterday morning, and the system seems to be even slower now. Does anyone have any idea what is going on???

## Re: slower than a snail (was oddity of equations in html mod...

I noticed the slowdown too.. but not sure why.

## Re: slower than a snail (was oddity of equations in html mod...

I see that our source pages are now in color -- maybe the little colorizing gremlins have only one box of crayons. JA

## unlikely culprit (was slower than a snail)

At least on my end, it seems that the culprit that is slowing down the system is the Google ads! Now, when I am going to a page on PM, most of it appears almost instantly, but the page is still loading, and the only thing missing is the Google ad (the upper-right hand corner is blank).

## Re: unlikely culprit (was slower than a snail)

It may be that Google has to search rather far and wide to Taylor its ads to our special needs. For some odd reason, whose critique I omit, I always seem to get the ones for Fool-Proof Weight-Loss Rules and Degrees in Public Relations. JA