Splitting field

 ( Field ) 

In abstract algebra, a splitting field of a polynomial with in a field is the smallest field extension of that field over which the polynomial splits, i.e., decomposes into linear factors.

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Definition A splitting field of a polynomial p( X) over a field K is a field extension L of K over which p factors into linear factors

$p(X)\; =\; c\; \backslash prod\_\{i=1\}^\{\backslash deg\; p\}\; (X\; -\; a\_i)$

where $c\; \backslash in\; K$ and for each $i$ we have $X\; -\; a\_i\; \backslash in\; LX$ with a_i not necessarily distinct and such that the roots a_i generate L over K. The extension L is then an extension of minimal degree over K in which p splits. It can be shown that such splitting fields exist and are unique up to isomorphism. The amount of freedom in that isomorphism is known as the Galois group of p (if we assume it is separable).

A splitting field of a set P of polynomials is the smallest field over which each of the polynomials in P splits.

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Properties An extension L that is a splitting field for a set of polynomials p( X) over K is called a normal extension of K.

Given an algebraically closed field A containing K, there is a unique splitting field L of p between K and A, generated by the roots of p. If K is a field extension of the , the existence is immediate. On the other hand, the existence of algebraic closures in general is often proved by 'passing to the limit' from the splitting field result, which therefore requires an independent proof to avoid circular reasoning.

Given a separable extension K′ of K, a Galois closure L of K′ is a type of splitting field, and also a Galois extension of K containing K′ that is minimal, in an obvious sense. Such a Galois closure should contain a splitting field for all the polynomials p over K that are minimal polynomials over K of elements of K′.

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Constructing splitting fields

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Motivation Finding roots of polynomials has been an important problem since the time of the ancient Greeks. Some polynomials, however, such as over , the , have no roots. By constructing the splitting field for such a polynomial one can find the roots of the polynomial in the new field.

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The construction Let F be a field and p( X) be a polynomial in the polynomial ring F X of degree n. The general process for constructing K, the splitting field of p( X) over F, is to construct a chain of fields $F=K\_0\; \backslash subseteq\; K\_1\; \backslash subseteq\; \backslash cdots\; \backslash subseteq\; K\_\{r-1\}\; \backslash subseteq\; K\_r=K$ such that K_i is an extension of K _i−1 containing a new root of p( X). Since p( X) has at most n roots the construction will require at most n extensions. The steps for constructing K_i are given as follows:

• Factorize p( X) over K_i into irreducible factors $f\_1(X)f\_2(X)\; \backslash cdots\; f\_k(X)$.
• Choose any nonlinear irreducible factor f( X).
• Construct the field extension K _i+1 of K_i as the quotient ring K _i+1 = K _i X / ( f( X)) where ( f( X)) denotes the ideal in K _i X generated by f( X).
• Repeat the process for K _i+1 until p( X) completely factors.

The irreducible factor f( X) used in the quotient construction may be chosen arbitrarily. Although different choices of factors may lead to different subfield sequences, the resulting splitting fields will be isomorphic.

Since f( X) is irreducible, ( f( X)) is a maximal ideal of K _i X and K _i X / ( f( X)) is, in fact, a field, the residue field for that maximal ideal. Moreover, if we let $\backslash pi\; :\; K\_iX\; \backslash to\; K\_iX/(f(X))$ be the natural projection of the ring onto its quotient then

$f(\backslash pi(X))\; =\; \backslash pi(f(X))\; =\; f(X)\backslash \; \backslash bmod\backslash \; f(X)\; =\; 0$

so π( X) is a root of f( X) and of p( X).

The degree of a single extension $K\_\{i+1\}$ is equal to the degree of the irreducible factor f( X). The degree of the extension K is given by $K\_r\; \backslash cdots\; K\_2\; K\_1$ and is at most n!.

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The field K_iX/(f(X)) As mentioned above, the quotient ring K _i+1 = K _i X/( f( X)) is a field when f( X) is irreducible. Its elements are of the form

$c\_\{n-1\}\backslash alpha^\{n-1\}\; +\; c\_\{n-2\}\backslash alpha^\{n-2\}\; +\; \backslash cdots\; +\; c\_1\backslash alpha\; +\; c\_0$

where the c_j are in K_i and α = π( X). (If one considers K _i+1 as a vector space over K_i then the powers α^j for form a basis.)

The elements of K _i+1 can be considered as polynomials in α of degree less than n. Addition in K _i+1 is given by the rules for polynomial addition, and multiplication is given by polynomial multiplication modulo f( X). That is, for g( α) and h( α) in K _i+1 their product is g( α) h( α) = r(α) where r( X) is the remainder of g( X) h( X) when divided by f( X) in K _i X.

The remainder r( X) can be computed through polynomial long division; however there is also a straightforward reduction rule that can be used to compute r( α) = g( α) h( α) directly. First let

$f(X)\; =\; X^n\; +\; b\_\{n-1\}\; X^\{n-1\}\; +\; \backslash cdots\; +\; b\_1\; X\; +\; b\_0.$

The polynomial is over a field so one can take f( X) to be monic polynomial without loss of generality. Now α is a root of f( X), so

$\backslash alpha^n\; =\; -(b\_\{n-1\}\; \backslash alpha^\{n-1\}\; +\; \backslash cdots\; +\; b\_1\; \backslash alpha\; +\; b\_0).$

If the product g( α) h( α) has a term α ^m with it can be reduced as follows:

$\backslash alpha^n\backslash alpha^\{m-n\}\; =\; -(b\_\{n-1\}\; \backslash alpha^\{n-1\}\; +\; \backslash cdots\; +\; b\_1\; \backslash alpha\; +\; b\_0)\; \backslash alpha^\{m-n\}\; =\; -(b\_\{n-1\}\; \backslash alpha^\{m-1\}\; +\; \backslash cdots\; +\; b\_1\; \backslash alpha^\{m-n+1\}\; +\; b\_0\; \backslash alpha^\{m-n\})$.

As an example of the reduction rule, take K_i = Q X , the ring of polynomials with rational number coefficients, and take f (X ) = X ⁷ − 2. Let $g(\backslash alpha)\; =\; \backslash alpha^5\; +\; \backslash alpha^2$ and h (α ) = α ³ +1 be two elements of QX/( X⁷ − 2). The reduction rule given by f( X) is α⁷ = 2 so

$g(\backslash alpha)h(\backslash alpha)\; =\; (\backslash alpha^5\; +\; \backslash alpha^2)(\backslash alpha^3\; +\; 1)\; =\; \backslash alpha^8\; +\; 2\; \backslash alpha^5\; +\; \backslash alpha^2\; =\; (\backslash alpha^7)\backslash alpha\; +\; 2\backslash alpha^5\; +\; \backslash alpha^2\; =\; 2\; \backslash alpha^5\; +\; \backslash alpha^2\; +\; 2\backslash alpha.$

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Examples

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The complex numbers Consider the polynomial ring R x, and the irreducible polynomial The quotient ring is given by the congruence As a result, the elements (or equivalence classes) of are of the form where a and b belong to R. To see this, note that since it follows that , , , etc.; and so, for example

The addition and multiplication operations are given by firstly using ordinary polynomial addition and multiplication, but then reducing modulo , i.e. using the fact that , , , , etc. Thus:

$(a\_1\; +\; b\_1x)\; +\; (a\_2\; +\; b\_2x)\; =\; (a\_1\; +\; a\_2)\; +\; (b\_1\; +\; b\_2)x,$

$(a\_1\; +\; b\_1x)(a\_2\; +\; b\_2x)\; =\; a\_1a\_2\; +\; (a\_1b\_2\; +\; b\_1a\_2)x\; +\; (b\_1b\_2)x^2\; \backslash equiv\; (a\_1a\_2\; -\; b\_1b\_2)\; +\; (a\_1b\_2\; +\; b\_1a\_2)x\; \backslash ,\; .$

If we identify with ( a, b) then we see that addition and multiplication are given by

$(a\_1,b\_1)\; +\; (a\_2,b\_2)\; =\; (a\_1\; +\; a\_2,b\_1\; +\; b\_2),$

$(a\_1,b\_1)\backslash cdot\; (a\_2,b\_2)\; =\; (a\_1a\_2\; -\; b\_1b\_2,a\_1b\_2\; +\; b\_1a\_2).$

We claim that, as a field, the quotient ring is isomorphic to the , C. A general complex number is of the form , where a and b are real numbers and Addition and multiplication are given by

$(a\_1\; +\; b\_1\; i)\; +\; (a\_2\; +\; b\_2\; i)\; =\; (a\_1\; +\; a\_2)\; +\; i(b\_1\; +\; b\_2),$

$(a\_1\; +\; b\_1\; i)\; \backslash cdot\; (a\_2\; +\; b\_2\; i)\; =\; (a\_1a\_2\; -\; b\_1b\_2)\; +\; i(a\_1b\_2\; +\; a\_2b\_1).$

If we identify with ( a, b) then we see that addition and multiplication are given by

$(a\_1,b\_1)\; +\; (a\_2,b\_2)\; =\; (a\_1\; +\; a\_2,b\_1\; +\; b\_2),$

$(a\_1,b\_1)\backslash cdot\; (a\_2,b\_2)\; =\; (a\_1a\_2\; -\; b\_1b\_2,a\_1b\_2\; +\; b\_1a\_2).$

The previous calculations show that addition and multiplication behave the same way in and C. In fact, we see that the map between and C given by is a homomorphism with respect to addition and multiplication. It is also obvious that the map is both injective and surjective; meaning that is a bijective homomorphism, i.e., an ring isomorphism. It follows that, as claimed:

In 1847, Cauchy used this approach to define the complex numbers.

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Cubic example Let be the rational number field and . Each root of equals $\backslash sqrt3\{2\}$ times a cube root of unity. Therefore, if we denote the cube roots of unity by

$\backslash omega\_1\; =\; 1,\backslash ,$

$\backslash omega\_2\; =\; -\backslash frac\{1\}\{2\}\; +\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}\; i,$

$\backslash omega\_3\; =\; -\backslash frac\{1\}\{2\}\; -\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}\; i.$

any field containing two distinct roots of will contain the quotient between two distinct cube roots of unity. Such a quotient is a primitive cube root of unity—either $\backslash omega\_2$ or $\backslash omega\_3=1/\backslash omega\_2$. It follows that a splitting field of will contain ω₂, as well as the real cube root of 2; conversely, any extension of containing these elements contains all the roots of . Thus

$L\; =\; \backslash mathbf\{Q\}(\backslash sqrt3\{2\},\; \backslash omega\_2)\; =\; \backslash \{\; a\; +\; b\backslash sqrt3\{2\}\; +\; c\{\backslash sqrt3\{2\}\}^2\; +\; d\backslash omega\_2\; +\; e\backslash sqrt3\{2\}\backslash omega\_2\; +\; f\{\backslash sqrt3\{2\}\}^2\; \backslash omega\_2\; \backslash mid\; a,b,c,d,e,f\; \backslash in\; \backslash mathbf\{Q\}\; \backslash \}$

Note that applying the construction process outlined in the previous section to this example, one begins with $K\_0\; =\; \backslash mathbf\{Q\}$ and constructs the field $K\_1\; =\; \backslash mathbf\{Q\}X\; /\; (X^3\; -\; 2)$. This field is not the splitting field, but contains one (any) root. However, the polynomial $Y^3\; -\; 2$ is not irreducible over $K\_1$ and in fact:

$Y^3\; -2\; =\; (Y\; -\; X)(Y^2\; +\; XY\; +\; X^2).$

Note that $X$ is not an indeterminate, and is in fact an element of $K\_1$. Now, continuing the process, we obtain $K\_2\; =\; K\_1Y\; /\; (Y^2\; +\; XY\; +\; X^2)$, which is indeed the splitting field and is spanned by the $\backslash mathbf\{Q\}$-basis $\backslash \{1,\; X,\; X^2,\; Y,\; XY,\; X^2\; Y\backslash \}$. Notice that if we compare this with $L$ from above we can identify $X\; =\; \backslash sqrt3\{2\}$ and $Y\; =\; \backslash omega\_2$.

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Other examples

• The splitting field of x^q − x over F _p is the unique finite field F _q for q = pⁿ. Sometimes this field is denoted by GF( q).

• The splitting field of x² + 1 over F₇ is F₄₉; the polynomial has no roots in F₇, i.e., −1 is not a square there, because 7 is not congruent to 1 modulo 4.Instead of applying this characterization of odd prime number moduli for which −1 is a square, one could just check that the set of squares in F₇ is the set of classes of 0, 1, 4, and 2, which does not include the class of −1 ≡ 6.

• The splitting field of x² − 1 over F₇ is F₇ since x² − 1 = ( x + 1)( x − 1) already splits into linear factors.

• We calculate the splitting field of f( x) = x³ + x + 1 over F₂. It is easy to verify that f( x) has no roots in F₂; hence f( x) is irreducible in F₂ x. Put r = x + ( f( x)) in F₂ x/( f( x)) so F₂( r) is a field and x³ + x + 1 = ( x + r)( x² + ax + b) in F₂( r) x. Note that we can write + for − since the characteristic is two. Comparing coefficients shows that a = r and b = 1 + r². The elements of F₂( r) can be listed as c + dr + er², where c, d, e are in F₂. There are eight elements: 0, 1, r, 1 + r, r², 1 + r², r + r² and 1 + r + r². Substituting these in x² + rx + 1 + r² we reach ( r²)² + r( r²) + 1 + r² = r⁴ + r³ + 1 + r² = 0, therefore x³ + x + 1 = ( x + r)( x + r²)( x + ( r + r²)) for r in F₂ x/( f( x)); E = F₂( r) is a splitting field of x³ + x + 1 over F₂.

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Notes

• Dummit, David S., and Foote, Richard M. (1999). Abstract Algebra (2nd ed.). New York: John Wiley & Sons, Inc. .

Categories: Field

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