TITLE: Can two different roots of an irreducible polynomial generate the same extension?
QUESTION [10 upvotes]: Let $K$ be a field and $f(x)$ be an irreducible polynomial over $K$. Suppose, $f(x)$ has degree at least $2$. Is it possible that if $a,b$ are two roots of $f(x)$ with $a\neq b$, then $K(a)=K(b)$. Note I need equality, not isomorphism.

REPLY [11 votes]: I think it's worth elaborating on this distinction between equality and isomorphism.  The problem occurs if the extension you're considering isn't normal and can therefore embed into an algebraic closure in more than one way.  For example, the abstract field $\mathbb{Q}[x]/(x^3 - 2)$ embeds in three ways into $\mathbb{C}$, corresponding to the three roots of $x^3 - 2$.  So it doesn't make sense to ask whether $\mathbb{Q}[x]/(x^3 - 2)$ is "equal to" $\mathbb{Q}[y]/(y^3 - 2)$ without specifying an embedding into a larger field.  
This is a fairly subtle point which I don't think is addressed particularly well in introductions to Galois theory (at least the ones I've seen).  There are three categories one might work in when studying fields (where $K$ is a fixed field):

The category of fields.  Here it is "evil" (really, impossible) to speak of the "equality" of two fields and one can only speak of isomorphism.
The category of (say, algebraic) field extensions $K \to L$.  (The morphisms are morphisms $\phi : L \to L'$ making the obvious triangle commute.)  Here it is still "evil" to speak of the "equality" of two extensions, and one can only speak of isomorphism, but the isomorphism type of an extension $K \to L$ is not determined by the isomorphism type of $L$ (that is, $L$ can be an extension of $K$ in more than one way).  
The category of subfields of $\bar{K}$ containing $K$ (for a fixed embedding $K \to \bar{K}$).  This is the category in which it makes sense to take intersections and composita of fields; you can't do either of these constructions in the above category.  Here, at last, one can speak of equality (as subfields of $\bar{K}$), and it is not the same as isomorphism of $K$-extensions, which is in turn not the same as isomorphism of abstract fields.

People sometimes don't specify which of the above categories they're working in, and until you do this you can't be precise: there are three different notions of equality or isomorphism in the last category.