Splitting fields and algebraic closure

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $\left(f_i\left(x\right)\right), f_i\left(x\right)\in 𝔽\left[x\right],$ 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 $$f\left(x\right)=f_1\left(x\right)\cdots f_n\left(x\right)\in 𝔽\left[x\right].$$

Define an ordering on pairs $\left(𝔼,\iota \right)$ where $\iota :𝔽\to 𝔼$ is a field homomorphism by $$\left(𝔼,\iota \right)\le \left(𝕂,\phi \right)if\; there\; is\; a\; field\; homomorphism\psi :𝔼\to 𝕂 \; such\; that\; \phi =\psi \circ \iota .$$ Then, every increasing sequence $$\left(𝔽,{\mathrm{id}}_{𝔽}\right)<\left({𝔼}_1,{\iota }_1\right)<\left({𝔼}_2,{\iota }_2\right)<\cdots$$ has a maximal element $$\left(𝔼,\iota \right)defined\; by𝔼=\bigcup {𝔼}_i \; and\; \iota \left(x\right)={\iota }_k\left(x\right),x\in {𝔼}_k.$$ 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 $$\stackrel{\_}{ℂ\left(x\right)}\cong ℂ$$ 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 $\left(f_i\right(x\left)\right)$ be a family of polynomials in $𝔽\left[x\right]$. The splitting field of the family $\left(f_i\right(x\left)\right)$ exists.

		Proof.
	For each polynomial $f_i\left(x\right)=a_d{x}^d+\cdots +a_{1}x+a_0$ in the family $\left(f_i\right(x\left)\right)$ let $$A_i=𝔽 [x_{i,1},...,x_{i,d}] /I_i,where\; I_i \; is\; the\; ideal\; generated\; by\; e_r\left(x_{i,1},...,x_{i,d}\right)-a_r.$$ Let $$A=\underset{i}{⨂}{A}_{i}and\; let𝔼=A/𝔪,$$ 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 $$f_i\left(x\right)=\prod _r(x-{\alpha }_{i,r})in\; 𝔼\left[x\right].$$ $\square$

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 $$\stackrel{\_}{𝔽}=\frac{𝔽\left[x_p\right]}{𝔪}.$$ 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 $x_p+𝔪)$ 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 $$1=\sum _q{u}_{q}q\left(x_q\right),with\; u_q\in 𝔽\left[x_p\right].$$ 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 $$\begin{array}{rrcl}\phi :& 𝔽\left[x_p\right]& \to & 𝔼\\ & \xi & \mapsto & \xi ,for\xi \in 𝔽,\\ & x_q& \mapsto & {\alpha }_q,\\ & x_p& \mapsto & 0,\end{array}$$ 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

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References

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