# What is the meaning of “algebraically indistinguishable”

I heard the term couple of times (in Field theory class and book), for example: The different roots of $p(x)=x^3-2$ are "algebraically indistinguishable".

I understand the meaning intuitively, but what does it really mean ?

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It means that there is no polynomial with coefficients in the ground field for which one is a root and the other isn't. For instance, $\pm i$ over $\Bbb R$. – Arthur Jun 10 '14 at 17:02
Emphasizing @Arthur's comment, you should turn to model theory. Taking a structure $\mathfrak M$ over some first-order language $\mathcal L$, two elements $a,b$ are indistinguishable if they have same type. – Pece Jun 10 '14 at 17:10

Suppose you look at the polynomial $x^3 -2 \in \Bbb{Q}[x]$. Then you should know that

$$\Bbb{Q}[x]/(x^3 - 2) \cong \Bbb{Q}(\sqrt[3]{2})\cong \Bbb{Q}(\omega\sqrt[3]{2}) \cong \Bbb{Q}(\omega^2 \sqrt[3]{2})$$

where $\omega = e^{2 \pi i/3}$. If you view each each of the three right most fields as sitting inside the complex numbers they are not the same object, because one of them is entirely contained within the reals while the other two contain non-real numbers. However if we view these as purely algebraic objects not sitting inside of anything, they are isomorphic in the sense of the following theorem:

Theorem: Let $F \subseteq E_1$ and $F \subseteq E_2$ be two field extensions and let $f \in F[x]$ be irreducible. Suppose $\alpha_i \in E_i$ with $f(\alpha_i) = 0$. Then $F[\alpha_1]$ and $F[\alpha_2]$ are $F$ - isomorphic via an isomorphism that carries $\alpha_1$ to $\alpha_2$.

By an $F$ - isomorphism we mean a map $\theta: E_1 \to E_2$ that extends the identity map on $F$.

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Said elementarily, it means that you cannot distinguish field-theoretically (over $\mathbb Q$) between the roots, i.e. any polynomial equality $\rm\:f(x) = g(x),\: f,g \in \mathbb Q[x]\:$ that holds true for one root also holds true for the other root. Therefore the roots are indistinguishable in the language of fields (over $\mathbb Q$).

This can be proved simply without field theory. Namely, since $\rm\:p(x)\:$ is irreducible, it is the least degree polynomial $\in \mathbb Q(x)\:$ with root $\alpha = \sqrt[3]{2}.\:$ Now $\rm\:f(\alpha) = g(\alpha)\iff f(\alpha)-g(\alpha) = 0,\:$ and the polynomials $\rm\:h(x)\in\mathbb Q(x)\:$ such that $\rm\:h(\alpha) = 0\:$ are precisely the multiples of the minimal polynomial $\rm\:p(x),\:$ for else $\rm\:gcd(p(x),h(x))\:$ would be a proper divisor of $\rm\:p(x)\:$ having $\:\alpha\:$ as root (since $\rm\:gcd(p(x),h(x)) = a(x)p(x)+b(x)h(x)\:$ by Bezout). Therefore $\rm\:h(\alpha) = 0\:$ $\Rightarrow$ $\rm\:h(x) = p(x)\hat h(x),\:$ for some $\rm\: \hat h(x)\in \mathbb Q(x),\:$ so every root of $\rm\:p\:$ is also a root of $\rm\: h = f - g.\:$ Therefore if $\rm\:f(\alpha) = g(\alpha)\:$ is true for one root of $\rm\:p\:$ then it is true for all roots of $\rm\:p.$

When one learns field theory the reason for this becomes clearer. Namely, the "conjugation" map sending one root to another $\:\alpha\to \alpha'\:$ is a ring homomorphism over $\mathbb Q$ (i.e. it preserves sums and products, and it is the identity map on $\mathbb Q$). Thus $\rm\:f(\alpha) = g(\alpha)\:$ maps to $\rm\:f(\alpha') = g(\alpha')\:$ because hom's preserve sums and products so also polynomials (and the coefficients are $\in \mathbb Q$ so are fixed).

Notice in particular that it is the definition of a (ring) homomorphism (structure preserving map)which serves to specify what we consider to be the "essential" algebraic properties of ring structure. It specifies precisely what aspects of the structure we are interested in, and excludes those that we are not. Thus while we have many different representations of $\mathbb C$ there are all isomorphic are fields, whether the elements are represented by pairs, matrices, polynomials, etc. They all have the same multiplication and addition tables, and that is the only structure we desire to capture in the abstraction of a ring, viz. how the elements are related to each other under the operations of the structure. Any other structure such as representational details (e.g. inner structure of the elements, say as a matrix) cannot be distinguished in the language of rings.

Returning to our example, the field obtained by adjoining one root of $\rm\:p\:$ to $\mathbb Q$ is isomorphic as a field to that obtained by adjoining any other root. Any other properties of the roots (size, real vs. complex, etc) cannot be distinguished purely field-theoretically (over $\mathbb Q$).

Hopefully this helps convey a more precise meaning to the informal term algebraically indistinguishable. The phrase is also used in analogous ways for other algebraic structures.

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It means that for any pair of roots $p$, $q$ of $x^3-2$, there is a field isomorphism $F$ from $\Bbb{Q}[x]/\langle x^3-2\rangle$ to itself which has $F(p)=q$ and $F(q)=p$.

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What do you mean by $\mathbb{Z}/\langle x^3-2 \rangle$? – Aaron Mazel-Gee May 22 '12 at 15:56
I meant one takes the ring of polynomials $\Bbb Z[x]$ and forms the quotient with the ideal generated by $x^3-2$. I accidentally left out the "$[x]$", but I have put it in where it belongs now. Thanks for pointing this out. – MJD May 22 '12 at 16:05

Which of the two square roots of $−1$ should we call $i$ and which should we call $−i$? Whichever one we call $i$ we will show one unit above $0$ when we draw a picture. Does it make any sense to say there's a difference between choosing one of them for that role rather than the other? If not, then in that sense they're indistinguishable.

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