Abstract Nonsense

Crushing one theorem at a time

Normal Extensions


Point of Post: In this post we shall discuss normality for extensions.

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Motivation

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This post is where we start admitting to a god-given truth–not all field extensions are created equal. Obviously this could be interpreted as saying not all extensions are isomorphic, but this is too strong. Instead of going head-long at the isomorphism problem for extensions I’d rather first discuss some simple properties that extensions have that are isomorphism invariants. Namely, properties that an extension does or does not enjoy, that shall enable us to tell extensions apart.

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In this post we discuss one of these invariants, the notion of normality. Intuitively normality is the failure for a given extension to be a splitting field of some polynomial. For example, we have discussed before the field \mathbb{Q}(\sqrt[3]{2}). This is a perfectly fine extension, but it is NOT a splitting field. Why? Because splitting fields L of some polynomial f_0(x)\in F[x] have the very nice property that if you give me an irreducible f(x)\in F[x] that has a root in L then f(x) has all its roots in L. We see then that \mathbb{Q}(\sqrt[3]{2}) is definitely not a splitting field since x^3-2\in\mathbb{Q}[x] is irreducible, has the root \sqrt[3]{2} contained in itself, yet does not have the other two roots of x^3-2 contained in itself.

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Of course, there are other descriptions of normal extensions besides the two given above (being a splitting field/having the all-or-none roots policy for irreducible polynomials), but this is the property to keep in mind. Normality is one-half of the equation to defining Galois extensions which, in our very simple look at field theory, are the “perfect” (not in the technical sense) extensions.

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Normal Extensions

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Let F be a field and k/F and extension. We call k/F (or just k, if F is clear) normal if it has the property that every time f(x)\in F[x] is irreducible and has a root in k then it has all of it’s roots in k.

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Let’s look at some examples. I claim for example that \mathbb{Q}(\sqrt{2}) is a normal extension of \mathbb{Q}. Indeed, it’s easy to see that the only irreducible polynomials f(x)\in\mathbb{Q}[x] with a root in \mathbb{Q}(\sqrt{2}) are polynomials that have a+b\sqrt{2} as a root. But, it’s easy to see that such polynomials necessarily have all of their roots in \mathbb{Q}(\sqrt{2}) since one can show that a-b\sqrt{2} is necessarily a root and so (x-(a+b\sqrt{2})(x-(a-b\sqrt{2}))\mid f(x), but since (x-(a+b\sqrt{2}))(x-(a-b\sqrt{2}))\in\mathbb{Q}[x] it follows from irreducibility that f=c(x-(a+b\sqrt{2}))(x-(a-b\sqrt{2})) and so f does have all of it’s roots in \mathbb{Q}(\sqrt{2}).

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As we have pointed out above, \mathbb{Q}(\sqrt[3]{2}) is not normal though.

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We now give several characterizations of normality the most fundamental of which though is \mathbf{(1)}\Leftrightarrow\mathbf{(2)}.

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Theorem: Let k/F be an algebraic extension. Then, the following are equivalent:

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\begin{aligned}&\mathbf{(1)}\quad k\text{ is normal}\\ &\mathbf{(2)}\quad k\text{ is the splitting field for a set }S\subseteq F[x]\\ &\mathbf{(3)}\quad \text{Every embedding }\sigma:k\to\overline{F}\text{ is an automorphism of }k\end{aligned}

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Proof: 

\mathbf{(1)}\implies\mathbf{(2)}: We claim that k is the splitting field for S=\left\{\text{min}(\alpha,F):\alpha\in k\right\}. Indeed, clearly k contains all of the roots of all polynomials in S since each polynomial is irreducible over F and has a root in k. That said, it’s evidently true that k=F(\text{roots of polynomials in }S) and thus k is the splitting field as desired.

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\mathbf{(2)}\implies\mathbf{(3)}: It suffices to show that \sigma(k)=k. To see that \sigma(k)\subseteq k merely note that, by definition, k=F(R) for some set R of roots of irreducible polynomials f(x)\in F[x]. But, since \sigma must necessarily send each root of an irreducible polynomial to another such root we see that \sigma(R)=R and thus \sigma(k)=\sigma(F(R))=F(\sigma(R))=F(R)=k.

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\mathbf{(3)}\implies\mathbf{(1)}: Let f(x) be an irreducible polynomial in F[x] and suppose that f(\alpha)=0 with \alpha\in k. Let \beta be another root of f(x). We then apply the isomorphism extension theorem to lift to a map k\to \overline{F} with \alpha\mapsto\beta. Since the image of this map is k we may conclude that \beta\in k.

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\blacksquare

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I’d like to end this post with the following observation:

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Theorem: Let L/k/F be a tower of algebraic fields. If L/F is normal then L/k is normal.

Proof: Since k/F is algebraic it’s trivial that \overline{k}=\overline{F} and thus the conclusion follows from the previous theorem, part three. \blacksquare

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It is NOT true that k/F needs to be normal. For example, consider the tower \mathbb{Q}(\sqrt[3](2),\omega)/\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q} where \omega is a primitive 3^{\text{rd}}-root of unity. Then, \mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q} is normal since the top field is the splitting field of x^3-2 (as I leave to you to verify) but \mathbb{Q}(\sqrt[3]{2})/\mathbb{Q} is not normal since it contains a root of x^3-2 but not any of the other two.

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.print

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April 30, 2012 - Posted by | Algebra, Field Theory | , , ,

1 Comment »

  1. […] time we discussed normal extensions which were the nice kind of extensions which (among several other definitions) are just the […]

    Pingback by Separable Extensions (Pt. I) « Abstract Nonsense | May 4, 2012 | Reply


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