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Supposing you want to define an algebraic number as the root of a polynomial. The problem is the polynomial has more than one root. But each root is has different properties. (Or do they?)

If the roots are from irreducible polynomials, e.g. roots of $x^3-2=0$, does any operations in the field depend on the particular root you choose?

i.e. the polynomial $(x-3)(x^2-3)=0$ has two types of roots.

But in general if you multiplied two algebraic numbers which is the nth and mth root of polynomials. Could you say the result is an algebraic number which is the root of a bigger polynomial which is the pth root.

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  • $\begingroup$ If a polynomial is irreducible over the rationals, then its complex roots can be permuted by automorphisms of the complex field. So nothing in the field structure can specify just one root. In the other hand, if you allow complex conjugation in addition to the field operations, then you can distinguish one root $z$ of $x^3-2$ by the fact that $z^*=z$, but the other two roots remain indistinguishable. $\endgroup$ Jul 24, 2018 at 22:46
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    $\begingroup$ More generally, different real roots can be distinguished by means of a rational number between them, because the $<$ relation on reals can be defined in terms of field operations and complex conjugation: $a<b$ iff $b=a+z^*z$ for some non-zero $z$. $\endgroup$ Jul 24, 2018 at 22:49
  • $\begingroup$ The easiest way (for real or complex roots) is to show the polynomial it is a root of, together with enough decimal places to distinguish it from the other roots. $\endgroup$
    – GEdgar
    Jul 25, 2018 at 0:31

1 Answer 1

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If the roots are from irreducible polynomials, e.g. roots of $x^3−2=0$, does any operations in the field depend on the particular root you choose?

Which field? There are several relevant fields you may be having in mind.

If $F$ is a field and $p(x) \in F[x]$ an irreducible polynomial, the field $K=F[x]/\langle p(x) \rangle$ is an extension field of $F$ containing a root of $p$. There's also the splitting field $L$ of $F$ which extends $F$ and contains all the roots of $p$. And finally, when the base field $F$ is the rationals $\mathbb{Q}$, each of $K$ and $L$ can be realized as subfields of $\mathbb{C}$; in the case of $K$, this may be doable in more than one way, as is the case with your example $p(x)=x^3-2$.

The different embeddings of $K$ in $\mathbb{C}$ are all isomorphic, so no operations inside those fields "depend" in any way on which embedding you've chosen (meaning, which root of the complex roots you've chosen to bring in). The abstract field $K$ of course doesn't depend on any such choice, since it has a unique definition independent of any choices.

The field $L$ is different (in your case, it has degree $6$ over $\mathbb{Q}$, instead of $3$), but it, too, is uniquely determined. No choices involved.

But in general if you multiplied two algebraic numbers which is the nth and mth root of polynomials. Could you say the result is an algebraic number which is the root of a bigger polynomial which is the pth root.

The product of two algebraic numbers is an algebraic number, yes. There's no natural order on the roots, if that's what you mean, so we can't make a statement like "the 3rd root of this polynomial times the 5th root of that polynomial is the 12th root of this other polynomial."

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    $\begingroup$ If the minimal polynomial of both $\alpha$ and $\beta$ is $x^2 - 2$ then the minimal polynomial of $\alpha \beta$ can be either $x-2$ or $x+2$. Likewise, if $\alpha$ has minimal polynomial $x^2-2$ and $\beta$ has minimal polynomial $x^2-8$ then $\alpha+\beta$ can have minimal polynomial either $x^2-18$ or $x^2-2$. $\endgroup$ Jul 25, 2018 at 0:02
  • $\begingroup$ Fair point, I misspoke. I'll correct that. $\endgroup$
    – Alon Amit
    Jul 25, 2018 at 0:24
  • $\begingroup$ @Daniel Yes, that's kind of what I'm getting at. Because wouldn't that depend on which roots you picked? Or at least you would have to pair algebraic numbers in some way. $\endgroup$
    – zooby
    Jul 25, 2018 at 0:25
  • $\begingroup$ @zooby Algebraic numbers can be unambiguously added and multiplied once you think of them as belonging to one ambient field, such as the complex numbers. Again, there's no way to distinguish the roots of an irreducible polynomials before you've embedded them in this way. $\endgroup$
    – Alon Amit
    Jul 25, 2018 at 0:30

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