# C. Algebraic Elements.

Let F be a field and E an extension field of F. If a is an ele­ment of E we may ask whether there are polynomials with coefficients in F which have a as root, a is galled algebraic with respect to F if tkere are such polynomials. NQW let a be algebraic and select among all polynomials in F which have a as root 0(1(3, f(x), of lowest degree.

We fliey assume that the highest coefficient of f(x) is 1. We con­tend that this f(x) is uniquely determined, that it is irreducible and that each polynomial in F with the foot a is divisible by f (x ). If, in­deed, g (* ) is a polynomial in F with g(a) = 0, we may divide g(x) = f(x)q(x) t r(x) where r(x) hft* a degree smaller than that of f(x). Substituting X = a we get r(p) = 0, NOW r(x) has to be identically 0 since otherwise r ) would h9V£ the root a gpcj be of lo\»er degree thap f (x ). 3o g ( x ) is divisible by f (x This also shows the uniqueness of f (x ). If f (}{ ) were not irreducible, one of the factors WQtild have to vanish for x = a contradicting again the choice of f ( y ),

where g(x) is a polynomial in F of degree less than n (n being the de­gree of f(x)). This set E^ is closed under addition and multiplication. The latter may be verified as follows:

are two polynomials of degree less than n we

put

Finally we see that the constants c , c ^ . . , c j are uniquely deter­mined by the element 0. Indeed two expressions for the same \$ would lead after subtracting to an equation for a of lower degree than n.

We remark that the internal structure of the set E does not depend on the nature of a but only on the irreducible f (x ). The knowledge of this polynomial enables us to perform the operations of addition and multiplication in our set E . We shall see very soon that E0 is a field; in fact, Eo is nothing but the field F(a). As soon as this is shown we have at once the degree, ( F (a) /F), determined as n, since the space F(a) is generated by the linearly independent 1, a, a2, • • • » a11"1. of a degree lower than n. This set forms a group under addition. We now introduce besides the ordinary multiplication

We shall now try to imitate the set EQ without having an exten­sion field E and an element a at our disposal. We shall assume only an irreducible polynomial as given.

We select a symbol £ and let Ej be the set of all formal polynomials

a new kind of multiplication of two elements g (£) and h (£) of E { denoted by g (£) x h (£). It is defined as the remainder r (£) of the ordinary product g (£) h (£) under division by f ( £ ). We first remark that any product of m terms gl( £), g2( ,,,, gm( £) is again the re­mainder of the ordinary product g ^ £) g2( £).., gm( £). This is true by definition for m = 2 and follows for every m by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials. This fact shows that our new product is associative and commutative and also that the new product g x g2( £) x , , , x g m(£) will coincide with the old product g £) g2( ,, gm( £) if the latter does not exceed n in degree. The distributive law for our multiplication is readily verified.

The set E , contains our field F and our multiplication in Ex has for F the meaning of the old multiplication. One of the polynomials of Ex is £ Multiplying it i-times with itself, clearly will just lead to as long as i < n. For i = n this is not any more the case since it leads to the remainder of the polynomial This remainder is We now give Up our old multiplication altogether and keep only the new one; we also change notation, using the point (or juxtaposition) as symbol for the new multiplication. Computing in this sense

involved are below n. But

Transposing we see that f(£) = 0.

We thus have constructed a set E j and an addition and multipli­cation in E j that already satisfies most of the field axioms. E j contains p as subfield and £ satisfies the equation f (£) = 0. We next have to show: If g ( 4 0 and h ( £) are given elements ofE p there is an element

in E j such that

To prove it we consider the coefficients Xj of X (<f) as unknowns and compute nevertheless the product on the left side, always reducing higher powers of £ to lower ones. The result is an expression

where each Li is a linear combination of of the Xj with coefficients in F. This expression is to be equal to h(£); this leads to the n equations with n unknowns:

have only the trivial solution.

where the bj are the coefficients of h (<f ). This system will be soluble if the corresponding homogeneous equations

The homogeneous problem would occur if we should ask for the set of elements X(Q) satisfying g (£) . X ( = 0. Going back for a moment to the old multiplication this would mean that the ordinary product g( £)X (<f) has the remainder 0, and istherefore divisible by f(t). According to the lemma, page 24, this is only possible for X ( = 0.

Therefore E[ is a field.

Assume now that we have also our old extension E with a root a of f(x), leading to the set E,. We see that E0 has in a certain sense the same structure as E if we map the element g (£) of E j onto the element g(a) of E . This mapping will have the property that the image of a sum of elements is the sum of the images, and the image of a product is the product of the images.

Let us therefore define: A mapping o of one field onto another which is one to one in both directions such that is called an isomorphism. If the fields in question are not distinct — i.e., are both the same field — the isomorphism is called an automorphism. Two fields for which there exists an isomorphism mapping one on another are called isomorphic. If not every element of the image field is the image under a of an element in the first field, then a is called an isomorphism of the first field into the second. Under each isomorphism it is clear and that

We see that ED is also a field and that it is isomorphic to E,. We now mention a few theorems that follow from our discussion: THEOREM 7. (Kronecker). If f (x ) is a polynomial in a field F, there exists an extension E of F in which f(x) has a root.

Proof: Construct an extension field in which an irreducible factor of f ( x ) has a root.

THE0.REM 8. Let Q be an isomorphism mapping a field F on a field F' Let f (x ) be an irreducible polynomial in F and f 1 (x ) the cor­responding polynomial inF 1 . If E = F (/3) and E 1 = F ' (fV) are exten­sions of F and F1 , respectively, where f(/3) = 0 in E and f 1 ( /3 ') = 0 in E' , then o can be extended to an isomorphism between E and E 1 .

Proof: E and E' are both isomorphic to Eq.