Splitting fields and algebraic closure

Last update: 02 February 2012

Splitting fields and algebraic closure

Let $𝔽$ be a field.

• A field $𝕂$ is algebraically closed if every nonconstant polynomial in $𝕂\left[x\right]$ has a root in $𝕂$.
• An extension $𝔼$ of $𝔽$ is a pair $\left(𝔼,\iota \right)$ where $𝔼$ is a field and $\iota :𝔽\to 𝔼$ is a field homomorphism.
• Let $𝔼$ be an extension of $𝔽$. An element $\alpha \in 𝔼$ is algebraic over $𝔽$ if $f\left(\alpha \right)=0$ for some polynomial $f\left(x\right)\in 𝔽\left[x\right].$
• An extension $𝔼$ of $𝔽$ is algebraic if every element of $𝔼$ is algebraic over $𝔽$.
• The splitting field of a family of polynomials is the smallest extension $𝔼$ of $𝔽$ such that all the ${f}_{i}\left(x\right)$ split in $𝔼\left[x\right]$, i.e. the extension of $𝔽$ such that each ${f}_{i}\left(x\right)$ splits in $𝔼\left[x\right]$ and $𝔼=𝔽\left({\alpha }_{i,r}\right),$ where ${\alpha }_{i,r}$ are the roots of ${f}_{i}\left(x\right)$.
• The algebraic closure of $𝔽$ is the splitting field $\stackrel{_}{𝔽}$ of (the family of polynomials) $𝔽\left[x\right]$.

Let $𝔽$ be a field and let $\left({f}_{i}\left(x\right)\right)$ be a family of polynomials in $𝔽\left[x\right]$.

1. The splitting field of the family $\left({f}_{i}\left(x\right)\right)$ exists and is unique.
2. The algebraic closure $\stackrel{_}{𝔽}$ exists and is unique.

(Fundamental theorem of algebra) The field $ℂ$ is the algebraic closure of $ℝ$.

Let $𝔽$ be a field.

1. The splitting field $𝔼$ of a single polynomial $f\left(x\right)$ is a finite extension since it has degree $\le \mathrm{deg}\left(f\right)$.
2. The splitting field $𝔼$ of a finite family $\left({f}_{1}\left(x\right),...,{f}_{n}\left(x\right)\right)$ of polynomials is actually the splitting field of the single polynomial $f(x) = f1(x) ⋯ fn(x) ∈𝔽[x].$
Define an ordering on pairs $\left(𝔼,\iota \right)$ where $\iota :𝔽\to 𝔼$ is a field homomorphism by Then, every increasing sequence $𝔽id𝔽< 𝔼1ι1< 𝔼2ι2<⋯$ has a maximal element If each $\left({𝔼}_{k},{\iota }_{k}\right)$ is an algebraic extension of $𝔽$ then $\left(𝔼,\iota \right)$ is an algebraic extension of $𝔽$.

Note: The ordering we have defined above is not necessarily well defined. For example, $ℂ$ is the field of complex numbers and if $x$ is trancendental over $ℂ$ then $ℂ(x)_ ≅ℂ$ as fields.

Let $𝔽$ be the algebraic closure of $𝔽$. The algebraic closure $\stackrel{_}{𝔽}$ of $𝔽$ is the smallest extension of $𝔽$ that contains every algebraic extension of $𝔽$.

Let $𝔽$ be a field and let $\left({f}_{i}\left(x\right)\right)$ be a family of polynomials in $𝔽\left[x\right]$. The splitting field of the family $\left({f}_{i}\left(x\right)\right)$ exists.

 Proof. For each polynomial ${f}_{i}\left(x\right)={a}_{d}{x}^{d}+\cdots +{a}_{1}x+{a}_{0}$ in the family $\left({f}_{i}\left(x\right)\right)$ let Let $A= ⨂ i Ai and let 𝔼=A/𝔪,$ where $𝔪$ is a maximal ideal of $A$. Then $𝔼=𝔽\left({\alpha }_{i,r}\right),$ where ${\alpha }_{i,r}$ is the image of ${x}_{i,r}$ in $𝔼$. Also $\square$

The algebraic closure $\stackrel{_}{𝔽}$ exists.

 Proof. Let ${x}_{p}$ be a set of variables indexed by the elements of $𝔽\left[x\right]$. Since the ideal $I= ⟨p\left({x}_{p}\right)|p\in 𝔽\left[x\right]⟩ \subseteq 𝔽{ [{x}_{p}] }_{p\in 𝔽\left[x\right]}$ is proper, and it is contained in a maximal ideal $𝔪$ of $𝔽\left[{x}_{p}\right]$. Let $𝔽_ = 𝔽[xp] 𝔪 .$ Then $\stackrel{_}{𝔽}$ is a field extension of $𝔽$ such that every polynomial $p\left(x\right)\in 𝔽\left[x\right]$ has a root (namely ${x}_{p}+𝔪\right)$ in $\stackrel{_}{𝔽}.$ Further, $\stackrel{_}{𝔽}$ is an algebraic extension of $𝔽$. For the proof of a.: If $I$ is not proper then there is a finite linear combination Let $𝔼$ be the extension of $𝔽$ which contains all the roots of the polynomials $q\in 𝔽\left[x\right]$ which appear in this sum. For each polynomial $q$ choose a root ${\alpha }_{q}\in 𝔼$. Then $φ: 𝔽[xp] → 𝔼 ξ ↦ ξ, for ξ∈𝔽, xq ↦ αq, xp ↦ 0,$ is a ring homomorphism and $1=\phi \left(1\right)=0$ in $𝔼$, a contradiction. So $I$ is a proper ideal. For the proof of b.: If $\alpha \in \stackrel{_}{𝔽}$ then $\alpha$ is an element of $𝔽\left(S\right)$ where $S$ is a finite subset of the ${\alpha }_{p}$. This is a finite extension of $𝔽$ and hence it is algebraic. So $\alpha$ is algebraic over $𝔽$. $\square$

The algebraic closure of $\stackrel{_}{𝔽}$ exists.

 Proof. Let $S$ be a set of cardinality greater than ${\aleph }_{0}\mathrm{Card}\left(𝔽\right).$ Then there is an injection of $𝔽$ into $S$ and so we may view $𝔽$ as a subset of $S$. Consider the class of fields $𝔼$ which as subsets of $S$ and which are algebraic extensions of $𝔽$. Then this set of fields is ordered and every linearly ordered collection of such fields has an upper bound. So this set of fields has a maximal element. $\square$

The algebraic closure $\stackrel{_}{𝔽}$ exists.

 Proof. Consider the directed system consisting of all finite extensions of $𝔽$. Let $\stackrel{_}{𝔽}$ be the direct limit of this directed system. $\square$

Notes and References

Where are these from?

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