# Abstract Nonsense

## Splitting Fields and Algebraic Closures (Pt. I)

Point of Post: In this post we discuss the notion of splitting fields for sets of polynomials as well as the algebraic closure of a given field.

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Splitting Fields and Algebraic Closures

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Up until this point we have committed a bit of an atrocity–well, sort of. Continually we have been considering some field $F$ and some extension $k$ and saying “well, if $f(x)\in F[x]$ is some polynomial and $f(\alpha)=0$ with $\alpha\in k$ then we can consider the field $F(\alpha)$ generated by $F$ and $\alpha$ in $k$“. My point being that to consider things like $F(\alpha)$, which fits with our original goal to consider minimal fields over which we can discuss the roots of a polynomial, we actually have to start with/find a field containing a root. Unfortunately, up until this point we have failed to do this. We have done a lot of math assuming we already have such a field, but alas, none are to be found. In this post we remedy this by actually showing that for any polynomial $f\in F[x]$ we can find a field extension $k/F$ such that $f(\alpha)=0$ for some $\alpha\in k$ (i.e. such that it has a root in $k$) but that we can factor $f=(x-\alpha_1)\cdots(x-\alpha_n)$ (i.e. that it has all of its roots in $k$). It turns out that this is not too bad, but lurking quietly in the background is a much more interesting, and deep, question. Namely, the way in which we shall construct such a field for a given $f$ shall be simple-minded, but at the price of this simplicity comes a lack of “cohesion”. Namely, we shall have a difficult time trying to compare results about a polynomial $f$ to a polynomial $g$ by looking at the respective fields we constructed since, in a sense, they don’t related well together. What would be nice is if we didn’t have to look for these fields containing the roots of $f,g,h,\cdots$ separately, if instead we could think about all of these fields living inside some bigger (realistically huge) field. In other words, can we find a an extension $k/F$ such that $k$ contains all of the roots of all $f\in F[x]$? This is a somewhat sensitive question construction wise. In a perfect world we’d just take the “union” of all the fields we get by examining the polynomials individually. Alas, being the semi-rigorous minded students that we are, this just won’t do. We shall construct the algebraic closure by a sneaky trick first thought up by Artin.

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Splitting Fields and Algebraic Closures

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As we stated above our first goal is to, given a field $F$ and polynomial $f(x)\in F[x]$, construct some extension $k/F$ such that $f(x)$ factors completely into linear factors, or in the lingo, such that $f(x)$ splits. If $k/F$ is an extension such that $f(x)$ splits and $k$ is generated over $F$ by the roots of $f$ we call $k$splitting field for $f$. Thus, the content of the first sentence could be rephrased as the desire to prove every polynomial in $f(x)\in F[x]$ has a splitting field (it’s obvious the statements are equivalent).

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Remark: We define the splitting field for a set of polynomials $\{f_j(x)\}\subseteq F[x]$ similarly.

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We go about find a splitting field for $f$ by baby steps. Namely, instead of trying to find a field which contains all the roots of $f$ we first show that we can find a field which contains a root of $f$. Of course, we need only consider monic irreducible polynomials since the general case will follow by factoring given polynomials into their irreducibles, etc. So, let’s consider the case when $f(x)\in F[x]$ is monic and irreducible. The idea behind this construction is, as was said in the motivation, simple. In fact, from what we have already proven we actually have no real choice in the matter. Namely, suppose for a second that $K$ is some extension of $f$ containing a root $\alpha\in K$. Then, of course, there is no reason to consider the entirety of $K$ and instead to only consider $F(\alpha)\subseteq K$. But, note then that $\alpha$ is necessarily algebraic since $f(\alpha)=0$. Thus, we know that $F(\alpha)\cong F[x]/(m_{\alpha,F})$ where, as usual, $m_{\alpha,F}$ is the minimal polynomial of $\alpha$ over $F$. But, check it out, we know that $f$ is a monic irreducible polynomial annihilating $\alpha$. Thus, by first principles we know that $m_{\alpha,F}$ is, secretly, just $f$. Thus, we know that $F(\alpha)=F[x]/(f)$. Thus, our only hope for such an $F(\alpha)$ and thus such a $K$ to exist is that $F[x]/(f)$ is our desired field.

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Ok, well, one part of this statement is immediately clear–namely, since $f$ is irreducible we know that $F[x]/(f)$ is actually a field, so we can check that part off. That said, what about the important part–the existence of a root of $f$ in $F[x]/(f)$? Luckily, the fact that whatever this mysterious root $\alpha$ is to be it has the property that $F(\alpha)\cong F[x]/(f)$ comes to our rescue again. Namely, the isomorphism $F[x]/(f)\xrightarrow{\approx}F(\alpha)$ is the result of applying the first isomorphism theorem to the natural epimorphism $F[x]\to F(\alpha)$ taking $x\mapsto \alpha$. In particular, for all intents and purposes not only are we forced to concede that the only true identity of $F(\alpha)$ is $F[x]/(f)$ but the only true identity of $\alpha$ is $x+(f)$. Thus, the moment of truth. For everything to work we need to check that $f$ applied to $x+(f)$ is zero. By a well-constructed miracle this is a tautology since $f(x+(f))=f(x)+(f)=0$. Yay! Thus we see that there is justice in the world and, in fact, $F[x]/(f)$ is an extension of $F$ containing a root of $f$.

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Ok, cool, so we’ve found a root of $f$ hiding in some extension of $F$, now we have to find the rest. Of course though, this easy, the first step’s always the hardest. Indeed, let $F[x]/(f)$ and $\alpha$ be equal to $F_1$ and $x+(f)$ respectively. The fact that $f$ has $\alpha$ as a root in $F_1$ allows us to factor $f$ as $f=(x-\alpha)g_1$ for some $g_1\in F_1[x]$ with $\deg g_1=\deg f-1$. Now we can take an irreducible, monic factor $h$ of $g_1=hu$ and apply the same method to get a new field $F_2$ such that $h=g_2(x-\alpha_2)$ so that $f=(x-\alpha_1)(x-\alpha_2)ug_2$. Repeating this process $n$ times will give us the desired field $K$ as desired.

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Note that not only do we have such a $K$ but we have control on it’s degree. Indeed, if $\deg f=n$ we can begin by noting that $F_1$ is $n$ dimensional, and then $F_2$ is an $n-1$ dimensional $F_1$-space, and then $F_2$ is an $n-2$ dimensional $F_2$-space, etc. we can actually conclude that $K$ actually has the property that $[K:F]\leqslant n!$.

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Now, once we have our $K$ actually getting a splitting field for $f$ is utterly trivial. Indeed, if $\alpha_1,\cdots,\alpha_n$ are the roots of $f$ in $K$ then we merely consider $F(\alpha_1,\cdots,\alpha_n)\subseteq K$. Clearly this is a field in which $f$ splits and is, by definition, generated by the roots of $f$. Thus, putting it all together we get the following theorem:

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Theorem: Let $F$ be a field and $f\in F[x]$ with $\deg f=n$. Then, there exists a splitting field $K$ for $f$ with $[K:F]\leqslant n!$.

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Of course, we can repeat this process finitely man times to produce a splitting field for any finite set $\{f_1,\cdots,f_n\}\subseteq F[x]$.

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The first obvious question considering our construction is whether or not splitting fields are (up to isomorphism) unique. I mean, we have surely constructed a splitting field, but whose to say that we couldn’t find another field, constructed entirely differently, which is also a splitting field for $f$ yet is not isomorphic to it? Of course, as all is right in the mathematical world, this is not the case.

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Indeed following [1] we prove the uniqueness of splitting fields in a piecewise manner. Namely, we shall start small and show that “small” extensions of a field [by appending roots] are unique up to isomorphism and then use transfinite techniques to take this finitary fact to an infinitary one. So, first comes first:

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Lemma: Let $k,k'$ be two fields and $f(x)\in k[x]$ irreducible. Define then $f^\sigma(x)\in k'[x]$ to be the polynomial obtained by applying $\sigma$ to the coefficients of $f$. Then, if $\alpha$ is a root of $f$ in some extension $K$ of $k$ and $\alpha'$ a root of $f^\sigma$ in some extension $K'$ of $k'$. There then exists an isomorphism $\varphi:k(\alpha)\to k(\alpha')$ extending $\sigma$ which takes $\alpha$ to $\alpha'$.

Proof: Since $f$ is irreducible and $f(\alpha)=0$ we know that $(f)=(m_{\alpha,k})$ and so $\kappa:k(\alpha)\cong k[x]/(m_{\alpha,k})\cong k[x]/(f)$ with $x+(f)\mapsto \alpha$, and similarly we have an isomorphism $\tau:k'(\alpha)\cong k'[x]/(f^\sigma)$ with $x+(f^\sigma)\mapsto \alpha$. So, now it’s fairly obvious that $\nu:k[x]/(f)\to k[x]/(f')$ defined by $g+(f)\mapsto g^\sigma+(f^\sigma)$ is an isomorphism extending $\sigma$. Clearly then $\tau^{-1}\circ\nu\circ\kappa:k(\alpha)\to k'(\alpha')$ is an isomorphism extending $\sigma$ and with $\alpha\mapsto \alpha'$. $\blacksquare$

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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.p