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Thus c h divides a, and the result follows. If f is a nonconstant polynomial in D[X], then f is irreducible over D if and only if f is primitive and irreducible over F.

If f is irreducible over D, then f is irreducible over F by 2. This contradicts the irreducibility of f over D.

Recall that F [X] is a Euclidean domain, hence a unique factorization domain. Again 1 1 by 2. Furthermore, the content is either a unit or a product of irreducible elements of the UFD R, and these elements are irreducible in R[X] as well.

This establishes the existence of a factorization into irreducibles. Therefore ai and bi are associates, which makes ci a unit in R, which in turn makes gi and hi associates in R[X], proving uniqueness of factorization.

Thus by 2. Thus we may assume without loss of generality that f is primitive, and prove that f is irreducible over R.

Assume that p fails to divide h0 , so that p divides g0 ; the argument is symmetrical in the other case. But p divides the entire sum ai , so p must divide the last term gi h0.

Consequently, either p divides gi , which contradicts the choice of i, or p divides h0 , which contradicts our earlier assumption.

Thus there can be no factorization of f as a product of polynomials of lower degree; in other words, f is irreducible over R.

The rational root test, which can be useful in factoring a polynomial over Q. Show that for every positive integer n, there is at least one irreducible polynomial of degree n over the integers.

If f is factorable over Z, then fp is factorable over Zp. Therefore if fp is irreducible over Zp , then f is irreducible over Z. Note that Eisenstein does not apply.

Thus if f is irreducible over Z, so is g. Chapter 3 Field Fundamentals 3. Thus any nonzero polynomial f in F [X] can be factored uniquely as a product of irreducible polynomials.

Any root of f must be a root of one of the irreducible factors, but at this point we have no concrete information about the existence of roots and how they might be found.

The dimension of this vector space is called the degree of the extension, written [E : F ]. We do this after a preliminary result.

Then f is a monomorphism. Since f can be factored into irreducibles, we may assume without loss of generality that f itself is irreducible.

Here is a further connection between roots and extensions. Then f and g are relatively prime if and only if f and g have no common root in any extension of F.

If h is a nonconstant divisor of the irreducible polynomials f and g, then up to multiplication by constants, h coincides with both f and g, so that f is a constant multiple of g.

This is impossible because f and g are monic and distinct. Thus f and g are relatively prime, and the result follows from 3.

If every element of E is algebraic over F , then E is said to be an algebraic extension of F. Any two such generators must be multiples of each other, so if we require that m X be monic, then m X is unique.

Property 2 follows from 1. Thus m X is irreducible, and uniqueness of m X follows from 3. Let E be an extension of F , and let S be a subset of E.

As in Problem 1, describe the composite explicitly. Find this minimal polynomial. Theorem 3. In the proof of 3. Show that conversely, if I is a maximal ideal, then f is irreducible.

If we continue the process, eventually we reach a complete factorization of f. In this section we examine this idea in detail.

There is a subtle point that should be mentioned. Continue inductively and use 3. Roots of the same irreducible factor of f yield essentially the same extension, as the next result shows.

We simply use i to transfer f. F -homomorphisms, F -monomorphisms, etc. But the comment at the end of 3. Problems For Section 3.

Let C be a family of polynomials over F , and let K be an extension of F. There are many ways to express this idea. Proceed inductively to show that any nonconstant polynomial is a product of linear factors.

By 2 , f is a product of linear factors. But f is irreducible, so there can be only one such factor. Note that C is minimal among algebraically closed extensions of F.

To prove this, we will need the following result. But by 3. The bi belong to K and are therefore algebraic over F.

Then C is an algebraic closure of F if and only if every nonconstant polynomial in F [X] splits over C. Thus assume that every nonconstant polynomial in F [X] splits over C.

If f is a nonconstant polynomial in C[X], we will show that f has at least one root in C, and it will follow from 1 of 3. Thus f has at least one root in C.

For complete proofs, see the appendix to Chapter 3. Informal argument. Carry out the recursive procedure described in 3.

But an algebraic closure is a minimal algebraically closed extension by 3. Within C, carry out the recursive procedure of 3. Show that the converse of 3.

A transcendental number is a complex number that is not algebraic over Q. Show that there only countably many algebraic numbers, and consequently there are uncountably many transcendental numbers.

Show that the set A of algebraic numbers is an algebraic closure of Q. In this section we investigate the multiplicity of the roots. If f is an arbitrary polynomial, not necessarily irreducible, then we call f separable if each of its irreducible factors is separable.

Here is a method for testing for multiple roots. Thus f is a nonzero polynomial whose degree is less than that of f. Since f is irreducible, the gcd of f and f is either 1 or f , and the latter is excluded because f cannot possibly divide f.

Then f is an automorphism. Thus we have transitivity of separable extensions. In view of 3. Inseparable extensions are less common, but here is one way to construct them.

The product of two nonconstant polynomials in t cannot possibly be t. Give an example of a separable polynomial f whose derivative is zero.

Equivalently, every algebraic extension of F is separable. Show that if F has prime characteristic p, then F is perfect if and only if every element of F is the pth power of some element of F.

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