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I'm new to this subject and not brilliant at it at all! All of the tutorials I've found online have just been a bit too overhwelming and I have an exam tomorrow so is there any way somebody could explain as simply as they possibly could!

An example I have in my paper is:

$$\text{Let }K =\{a+b\sqrt3\mid a, b\in\mathbb{Q}\}$$ Find the minimal polynomial of $ \sqrt3+\sqrt5$ over $\mathbb{Q}$

Is there any way somebody could explain step by step! Thank you so much

verret
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Lindsay
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  • over $\mathbb Q$ or $K$? – J. W. Tanner May 14 '20 at 18:31
  • @J.W.Tanner over Q it says :) – Lindsay May 14 '20 at 18:32
  • @DietrichBurde or I could just focus on these lovely and patient people that have helped me understand the subject just in time for tomorrow, but thank you for your unwelcome concern – Lindsay May 14 '20 at 18:54
  • @DietrichBurde I'm sorry if you thought you were helping, I don't mean to be rude I'm just very aware I'm not the most intelligent person so to come to a platform for help and have someone tell me to focus on learning instead of posting, and that 'I don't even know what something is' is upsetting. Thank you anyway – Lindsay May 14 '20 at 19:02
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    @DietrichBurde there's so much arguing in the comments of these posts I didn't know maths people hated each other so much lol – Lindsay May 14 '20 at 19:20

2 Answers2

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The conjugates of $\sqrt{3}$ are $\sqrt{3}$ and $-\sqrt{3}$. Same for $\sqrt{5}$ giving you $\sqrt{5}$ and $-\sqrt{5}$.

The minimal polynomial will have a factor of $(x-(\sqrt{3}+\sqrt{5})$ (in $\mathbb{C}[x]$). If you apply the (extension of the) automorphism that sends $\sqrt{3}$ to $-\sqrt{3}$ (and fixes $\sqrt{5}$), you will get a factor of $(x-(-\sqrt{3}+\sqrt{5}))$. If you apply the one that sends $\sqrt{5}$ to $-\sqrt{5}$ and fixes $\sqrt{3}$, you will get a factor $(x-(\sqrt{3}-\sqrt{5}))$. And if you take the composition of those two automorphisms, you’ll get a factor of $(x-(-\sqrt{3}-\sqrt{5}))$.

All of those must be factors of the minimal polynomial, because the minimal polynomial has coefficients in $\mathbb{Q}$, and so is fixed by the automorphisms. That gives you at least four linear factors. Check to see if you already have a polynomial with rational coefficients.

That there are automorphisms doing what is described (conjugating one root but not the other) follows because $5$ is not a square in $K$, which immediately yields that your element lives in an extension of degree $2$ over $K$ (and hence of degree $4$ over $\mathbb{Q}$).

Arturo Magidin
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  • @EricTowers: You don’t need to assume that, nor am I assumning that ahead of time. The factors I found were found applying automorphisms we know exist. If a degree two polynomial sufficed, then the automorphisms would yield repeated factors, which they plainly do not. (If you get repeated factors, then of course you only need to take one of them, you don’t need to take them multiple times). In other words, it’s not an “assumption”, it’s what drops out of the computations. – Arturo Magidin May 14 '20 at 18:46
  • @EricTowers: No, because there is no automorphism that sends $\sqrt{3}$ to $-\sqrt{3}$ but fixes $\sqrt{12}$. So, no, what I write would not apply to that situation., – Arturo Magidin May 14 '20 at 18:49
  • @EricTowers: The “fact” you claim I assumed I did not assume; what is being used is that $5$ is not a square in $\mathbb{Q}(\sqrt{3})$ and that $3$ is not a square in $\mathbb{Q}(\sqrt{5})$. That is: that it is the extension $\mathbb{Q}(\sqrt{3},\sqrt{5})$ that is of degree (at least) $4$. – Arturo Magidin May 14 '20 at 18:51
  • @EricTowers: You are free to say there is something that needs more explanation, but if you are going to claim that then at least identify something that is actually being used, not something else you think is being used. And don’t assert an argument “would hold” where it doesn’t. That is also pretty confusing for your putative future reader. – Arturo Magidin May 14 '20 at 18:54
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First, you should find an "upper bound" - find a simple polynomial which has $t=\sqrt3+\sqrt5$ as a root. Note that: $$x=\sqrt3+\sqrt5\Rightarrow x^2=8+2\sqrt3 \sqrt5\Rightarrow (x^2-8)^2=60$$ And so we have the polynomial: $$p(x)=x^4-16x^2+4$$

The minimal polynomial of $t$ must divide this polynomial. Due to the dimension theorem, the minimal polynomial has to be either of degree $1,2$ or $4$, since its degree must divide the degree of $p(x)$ (in the last case where $deg(p)=4$ this means that $p(x)$ IS the minimal polynomial).

The degree can't be $1$ since $t$ is clearly irrational. It also cannot be $2$ since if $x^2+bx+c$ has $t$ as a root, then: $$(\sqrt3+\sqrt5)^2+b(\sqrt3+\sqrt5)+c=0$$ $$\Rightarrow8+2\sqrt3 \sqrt5+b\sqrt3+b\sqrt5+c=0$$ But there is not $b,c\in\mathbb{Q}$ which can satisfy this, since $b,c$ are rational. Naturally we must have $c=-8$, but then: $$b=-\frac{2\sqrt3 \sqrt5}{\sqrt3+\sqrt5}\notin \mathbb{Q}$$

We are thus left to conclude that the degree of the minimal polynomial is $4$, and so $p(x)$ is the minimal polynomial. This is the general strategy I know of finding a minimal polynomial for simple cases - guess a polynomial that has $t$ as a root, and try to see if there can be a smaller polynomial.

J. W. Tanner
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GSofer
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