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I have been reading about fields and I would like assistance with the following:

Let $k$ be an arbitrary field and $f(x)$ be an irreducible polynomial in $k[x]$. Then:

Prop: There exists a field $K$ containing $k$ and an element $\alpha\in K$ such that $f(\alpha)=0$.

I want to show that given a field $k$ and a polynomial $f(x)\in k[x]$ there is a field $K \supset k$ such that $[K:k]$ is finite and $f(x)=(x-\alpha_{1})(x-\alpha_{2})\cdots (x-\alpha_{n})$ in $K[x]$. Thanks.

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So you don't need to prove the Proposition? then just apply it repeatedly to build a tower of fields. If you have constructed field $F_n$, and $f(x)$ still does not split completely, extend to provide a root for one of the factors of $f(x)$ irreducible over $F_n$. –  André Nicolas Feb 16 '12 at 5:41
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Nothing to do with "finite fields" (which are fields that are finite as sets); perhaps "finite extensions of fields"? –  Arturo Magidin Feb 16 '12 at 5:44
    
@Andre : OP is also asking for $[K : k]$ to be finite ; this means he is asking for something stronger than the proposition, hence this proposition is almost irrelevant here... –  Patrick Da Silva Feb 16 '12 at 5:45
    
@Arturo : The number-theory/finite fields tags are both unappropriate here. I edited them. –  Patrick Da Silva Feb 16 '12 at 5:45
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@Arturo: Agreed...but also nothing really to do with number theory either, right? This is pure field theory. –  Pete L. Clark Feb 16 '12 at 5:46
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3 Answers

up vote 4 down vote accepted

Induction on the degree of $f$ (which must be assumed to be monic and nonconstant given your factorization, though you did not say so). The result is trivially true if the degree of $f$ is $1$, by taking $K=k$.

Assuming the result holds for polynomials of degree strictly less than $n$, let $f$ be an arbitrary polynomial of degree $n$. Since $k$ is a field, $k[x]$ is a UFD so we can factor $f$ into irreducibles. Let $g(x)$ be an irreducible factor of $f$. By the proposition you quote, there is an extension $K_1$ of $k$ that has a root $\alpha$ for $g$; moreover, we can take $K$ to be of degree $\deg(g)$ over $k$. In $K[x]$ we have $x-\alpha | g | f$, so we can write $f(x) = (x-\alpha)f_2(x)$, with $\deg(f_2)\lt n$. By the induction hypothesis, there is an extension of $L$ of $K$, finite over $K$, where $f_2$ splits; hence $f$ splits in $L$, and $[L:k] = [L:K][K:k]$ is finite.

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It may be slightly cleaner to prove the result with the weaker hypothesis that $f$ is nonconstant. Then $f$ factors as a nontrivial product of irreducible polynomials. If all of these have degree one, you're done. Otherwise, choose any irreducible factor $f_i$ of degree $d_i > 1$, and let $K_1 = k[x]/(f_i(x))$. This is a degee $d_i$ field extension such that -- in a wonderful, tautological way -- $f_i$ has a root over $K_1$.

Well, this doesn't prove the result, but unmistakable progress has been made. Repeat this process until all the irreducible factors have degree one, and use the fact that if $K = K_n \supset \ldots \supset K_1 \supset K_0 = k$ and $[K_{i+1}:K_i]$ is finite for all $i$, then $[K_n:K_0] = [K:k]$ is finite.

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Use finite induction :

Let $n = \deg f$. If $\alpha_1 \in k$, fine. If not, there exists an extension $K_1$ such that $\exists \alpha_1 \in K_1$ with $f(\alpha_1) = 0$, hence $f(x) = (x-\alpha_1) g_1(x)$ in $K_1[x]$. Now $g_1$ is a polynomial in $K_1[x]$ and has degree $n-1$, so this is fine, and our induction hypothesis goes well since for degree $1$ polynomials, since $ax+b = a(x-(-a^{-1}b))$ always has its root in its field of coefficients.

Obviously $[K_1 : k]$ can be chosen to be a finite degree extension ; you can take $K_1 = k(\alpha_1)$. After taking the $n^{th}$ extension if necessary, $[K_n : k] = [K_n : K_{n-1} ] \dots [K_1 : k] < \infty$.

Hope that helps,

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