# Abstract Nonsense

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

Point of Post: This is a continuation of this post.

Theorem(Isomorphism Extension Theorem): Let $k,k'$ be two fields and $\sigma:k\to k'$ a ring isomorphism. Then, if $\{f_j(x)\}\subseteq k[x]$ is a set of polynomials we can consider the set $\{\sigma(f_j(x))\}\subseteq k'[x]$ where $\sigma(f_j(x))$ is the polynomial obtained by applying $\sigma$ to the coefficients of $f_j$. Let $K$ be a splitting field for $\{f_j(x)\}$ and $K'$ be a splitting field for $\{\sigma(f_j(x))\}$. Then, there is an isomorphism $\varphi:K\to K'$ which is an extension of $\sigma$. Furthermore, if $\alpha\in K$ is some distinguished element we can take $\varphi$ to be such that $\varphi(\alpha)=\alpha'$ where $\alpha'$ is any root of $\sigma(m_{\alpha,k})$.

Proof: Basically we can create monomorphisms $L\to K'$ where $L$ is some “small” subextension of $K/k$ (namely we can just specify where we want the generators to go), the problem is in extending all the way to $K$. Of course,  phrased this way it’s clear that Zorn’s lemma is going to be our lemma of choice for this proof [relevant]. Ok, so let’s formally define

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$\mathcal{S}=\left\{(L,\varphi):L\text{ is a subextension of }K/k\text{ and }\varphi\text{ is an extension of }\sigma\right\}$

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We define a partial ordering structure $\leqslant$ on $\mathcal{S}$ by demanding that $(L,\varphi)\leqslant (L,\varphi')$ if and only if $L\subseteq L'$ and $\varphi'$ is an extension of $\varphi$. Of course, $\mathcal{S}$ is not empty since $(k,\sigma)\in\mathcal{S}$. Now, let’s show that every chain in $\mathcal{S}$ has a maximal element. Indeed, let $\{(L_c,\varphi_c)\}_{c\in C}$ be a chain. Define then $\displaystyle L=\bigcup_{c\in C}L_c$ and $\varphi:L\to K'$ by defining $\varphi(x)=\varphi_c(x)$ for any $c\in C$ with $x\in L_c$ (by the requirements on the $\varphi_c$ and the fact that we have a chain this is well-defined). Clearly then $(L,\varphi)\in\mathcal{S}$ and $(L_c,\varphi_c)\leqslant (L,\varphi)$ for al $c\in C$. Thus, we see that every chain in $\mathcal{S}$ has an upper bound and so our best friend Zorn hands us some maximal element $(M,\varphi)$ of $\mathcal{S}$. Our job now is to prove (we hope) that $M$ must actually be $K$ and that $\varphi(M)=K'$[there is, of course, no need to prove that $\varphi$ is injective since it’s a ring map out of a field].  To prove that $M=K$ we merely note that if this wasn’t true then necessarily there is some $f_j$ such that $f_j$ does not split in $M$ (since otherwise $M$ contains all the roots of all the $f_j$ and so must be equal to $K$). Thus, we can find some root $\alpha$ of $f_j$ not conatined in $M$. Then, by the lemma we can extend $\varphi:M\to \varphi(M)$ to $\varphi'M(\alpha)\to \varphi(M)(\alpha)$ which contradicts the maximality of $(M,\varphi)$.  Thus, we see that $M=K$. Now, to see that $\varphi(M)=\varphi(K)=K'$ we merely note that $\varphi(K)$ is a splitting field for $\{f_j^\sigma\}$ and thus must necessarily be equal to $K'$. $\blacksquare$

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Ok, cool so we have proven two things up to this point: splitting fields exist for a single polynomial, and splitting fields for an arbitrary set of polynomials are unique up to isomorphism. Well, there seems to be somewhat of a discrepancy for these two statements. Namely, we have a theorem telling us a fact about the splitting field for arbitrarily polynomials yet, a priori, we only know that the splitting field of finitely polynomials exists. Strange, right? This begs the questions as to whether or not the splitting field of arbitrarily polynomials even exists. Well, for a finite set of polynomials $\{f_1,\cdots,f_n\}$ we actually have nothing to wonder about since, as one can easily deduce, a splitting field for $f_1\cdots f_n$ is a splitting field for $\{f_1,\cdots,f_n\}$. But, what about infinite polynomials? We surely can’t just take their product. Indeed, for the infinitary case we are going to need slightly more sophisticated machinery.

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To create splitting fields for arbitrarily many polynomials in $F[x]$ it suffices to find a field containing all the roots for all the polynomials in $F[x]$. Indeed, suppose that we have constructed such a field, let’s call it $A$, then to find a the splitting field for $S\subseteq F[x]$ we merely let $R$ be the union of all the sets of roots in $A$ of all the elements in $S$. We can easily see then that $F(R)\subseteq A$ is a splitting field for $S$.

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To create such a field, let’s consider the field $F$ and for each nonconstant $f(x)\in F[x]$ let $t_f$ be an indeterminate. Consider then the polynomial ring $R=F[\{t_f:f\in F[x]\}]$. Let then $\mathfrak{a}$ be the ideal generated by the $f(t_f)$ for each nonconstant $f\in F[x]$. What we’d like to do is put $\mathfrak{a}$ inside some maximal ideal $\mathfrak{m}$ (using Krull’s theorem), but for this we need to know that $\mathfrak{a}\subsetneq\mathfrak{m}$. To prove that this is impossible we note that if it’s true we could find $g_1,\cdots,g_n\in R$ and $t_{f_1},\cdots,t_{f_n}$ such that

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$g_1f_1(t_{f_1})+\cdots+g_nf_n(t_{f_n})=1\quad\mathbf{(1)}$

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Ok, but now here’s the cool part, let’s rename the variables $t_{f_i}=t_i$ for $i\in[n]$ and let $t_{n+1},\cdots,t_m$ be the other variables that occur in the $g_1,\cdots,g_n$ so that we can rewrite $\mathbf{(1)}$ as

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$g_1(t_1,\cdots,t_m)f_1(t_1)+\cdots+g_n(t_1,\cdots,t_m)f_n(t_n)=1\quad\mathbf{(2)}$

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That said, we can find some extension $k/F$ such that $k$ contains roots $\alpha_i$ for $f_i$. Plugging this into $\mathbf{(2)}$ gives $0=1$ which is evidently ridiculous. Thus, we see that $\mathfrak{a}\subsetneq R$ and so we can put $\mathfrak{a}\subseteq\mathfrak{m}$ where $\mathfrak{m}$ is maximal. Consider then $k_1=R/\mathfrak{m}$. Clearly then $k_1$ is an extension of $F$ which contains. We may repeat this process for $k_1$ to get another field $k_2$, and so on. Define then $A=F\cup k_1\cup k_2\cup k_3\cup\cdots$. Clearly then $A$ is a field containing $F$, and since any polynomial in $A[x]$ has coefficients in some $K_r$ which has a root in $K_{r+1}$. Thus, every polynomial in $A[x]$ has a root in $A$. But, this clearly implies that every polynomial $f(x)\in A[x]$ has all its roots in $A$ since once we have a root, say $\alpha$, we can write $f(x)=(x-\alpha)h(x)$ for some $h(x)$. But, then $h(x)$ has a root in $A$, so we can write $f(x)=(x-\alpha)(x-\beta)c(x)$, etc. Thus, we know, in particular, every polynomial $f(x)\in F[x]\subseteq A[x]$ has a root in $A$, and considering our previous discussion we can finally conclude that :

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Theorem: Let $F$ be a field and $S\subseteq F[x]$. Then, $S$ has a splitting field.

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We call a splitting field for $F[x]$ (all the polynomials) an algebraic closure of $F$ and denote it $\overline{F}$–note that while this notation may seem ambiguous, it really isn’t since (as we have proven) all algebraic closures for $F$ must necessarily be isomorphic.

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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 5, 2012 -

## 3 Comments »

1. […] of the field) generate the field we may conclude that really is a splitting field for . Using the fact that splitting fields are unique up to isomorphism we may finally state the full classification of […]

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2. […] an irreducible polynomial in and suppose that with . Let be another root of . We then apply the isomorphism extension theorem to lift to a map with . Since the image of this map is we may conclude that […]

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3. […] polynomials. We say that a polynomial is separable if it has distinct roots in (the algebraic closure of ). For example, is separable since it has distinct roots in . If a polynomial is not separable […]

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