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One of the definitions of a Galois extension is that $E/K$ is Galois iff $E$ is the splitting field of some separable polynomial $f(x) \in K[x]$, yes?

I want to understand why the following is true:

Let $f(x) \in \mathbb{Q}[x]$ be a polynomial and let $F$ be a splitting field of $f(x)$. Then $F/\mathbb{Q}$ is Galois.

My question is: How do we know that $f(x)$ is a separable polynomial? Can you please explain this part?

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This "definition" only applies to finite extensions, though. An infinite Galois extension cannot be the splitting field of a single polynomial. –  Arturo Magidin May 12 '11 at 1:49
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2 Answers 2

up vote 6 down vote accepted

We don't. What we do know is that every irreducible polynomial over a field of characteristic zero is separable, and $f$ is the product of its irreducible factors, so its splitting field is a compositum of separable extensions, hence separable. (In other words, the splitting field of $f$ is the splitting field of a separable polynomial, but that polynomial is not $f$; it's the product of the distinct irreducible factors of $f$.)

This isn't a particularly transparent definition of Galois extension, though. The definition I prefer is that $F/K$ is Galois if $F^{\text{Aut}(F/K)} = K$. Among other things, it has the benefit of working for infinite extensions.

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thanks! exactly what I needed. By the way, there are many Galois theory books, any particular favorite of yours? –  user6495 May 12 '11 at 1:57
    
@user6495: not particularly. I like Keith Conrad's notes on Galois theory (math.uconn.edu/~kconrad/blurbs) and I also like Szamuely's Galois Groups and Fundamental Groups (renyi.hu/~szamuely/fg.pdf), although the approach taken there may be slightly intimidating. –  Qiaochu Yuan May 12 '11 at 2:03
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That is because there is a theorem that states that in every perfect field ($\mathbb Q$ and $\mathbb F_p$ are examples), every irreducible polynomial is separable (i.e. all their roots have multiplicity 1). Your polynomial f(x) in your assertion is not necessarily separable, but if you write $$ f(x) = (p_1(x))^{a_1} (p_2(x))^{a_2}... (p_n(x))^{a_n}, \quad a_i \ge 1 $$ with $p_i(x)$ being the irreducible factors of $f$, then F is also the splitting field for $$ g(x) = p_1(x) p_2(x) ... p_n(x). $$ since both of these polynomials ($f$ and $g$) have exactly the same roots. The polynomial g is separable since it is a product of irreducible polynomials (if you use the Dummit & Foote, which you should if you don't, I recommend ; I am speaking of Corollary 34 page 547.) so that every root of $g$ has multiplicity 1.

Therefore, the splitting field of $f$ being Galois over $\mathbb Q$ doesn't mean that $f$ is separable, but rather that some polynomial (the $g$ one) divides $f$ and has the same roots as $f$, with $g$ being separable and having the same splitting field as $f$.

I hope this helped.

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thank you very much! –  user6495 May 12 '11 at 1:58
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