# Can two different roots of an irreducible polynomial generate the same extension?

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.

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As Qiaochu notes in his comment to my answer, you really need to specify that you are working inside a fixed algebraic closure of $K$ for the question of equality of the two extensions to make sense. – Arturo Magidin Dec 12 '10 at 23:49
This is always true if the degree of $f$ is exactly 2. – g.castro Feb 26 '12 at 13:22

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.

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For the question to really make sense we need to fix an algebraic closure of $K$ and work always inside of it; otherwise, we can even make two constructions of $K(a)$ that would not be "equal" (e.g., constructing $\mathbb{Q}(i)$ as the set of pairs $(a,b)\in\mathbb{Q}^2$ with suitable addition/multiplication; or as $\mathbb{Q}[x]/(x^2+1)$; or as the set of formal symbols $a+bi$ with $a,b\in\mathbb{Q}$, etc.) Added: See Qiaochu's answer for a much better explanation of this issue.

Given that, yes, it is possible: for a trivial example, take $K$ to be a finite field with $q$ elements. Then both $K(a)$ and $K(b)$ are extensions of degree $\deg(f)$ over $K$; if $\deg(f)=n$, then this is a field with $q^n$ elements, and since there is a unique field of $q^n$ elements sitting inside the algebraic closure of $K$, $K(a)=K(b)$. In fact, in this case you always have that situation.

For a nonfinite example in a field of characteristic $0$, take either $\mathbb{Q}(\sqrt{d})$: the irreducible polynomial is $x^2-d$, which has roots $\sqrt{d}$ and $-\sqrt{d}$; naturally, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(-\sqrt{d})$ (always inside the same fixed algebraic closure)

For another example, take $f(x)$ to the cyclotomic polynomial $\Phi_n(x)$ with $K=\mathbb{Q}$; the roots are precisely the primitive $n$-roots of unity, and all the extensions are equal to $\mathbb{Q}(e^{2\pi i/n})$ (again, always sitting inside the same fixed algebraic closure of $\mathbb{Q}$).

Added: More generally, any simple extension $K(a)$ which is Galois over $K$ will have the property, since the irreducible polynomial of $a$ necessarily splits in $K(a)$. That means that if $f(x)$ is the irreducible polynomial of $a$, then for any root $r$ of $f(x)$ in the fixed algebraic closure of $K$ containing $K(a)$ we have $r\in K(a)$, so $K(r)\subseteq K(a)$; but applying the usual isomorphism we get that $f(x)$ must also split in $K(r)$, so $a\in K(r)$ hence $K(a)\subseteq K(r)$, giving equality. In particular, this holds for any finite Galois extension of a perfect field. This includes all the examples I gave above.

Conversely, I claim that if $f(x)$ is an irreducible polynomial over $K$ with degree at least $2$, and for any two roots $a$ and $b$ of $f(x)$ in a fixed algebraic closure of $K$ we have that $K(a) = K(b)$, then $K(a)$ is the splitting field of $f(x)$ (and so if $f(x)$ is separable, $K(a)$ is a finite Galois extension of $K$). Indeed, this is trivial, since for any root $b$ of $f(x)$ we have $b\in K(b)=K(a)$, so $K(a)$ contains all the roots of $f(x)$; since $K(a)$ is the least field containing $a$, $K(a)$ is contained in the splitting field of $f(x)$ over $K$, thus showing $K(a)$ is the splitting field.

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Arturo, I think it is somewhat imprecise to speak of "equality" of fields without specifying a field containing both. I suppose you're working in a fixed algebraic closure of K? – Qiaochu Yuan Dec 12 '10 at 23:41
@Qiaochu: Yes; I thought it was clear when I said "sitting inside the algebraic closure of $K$" in the first paragraph, but I'll make it even more explicit; in fact, I almost added that as a comment to the question! Thanks. – Arturo Magidin Dec 12 '10 at 23:45