MA3412S2_Hil2014.pdf

# Thus if m and n are natural numbers satisfying m n

• Notes
• 38

This preview shows pages 4–6. Sign up to view the full content.

Thus if m and n are natural numbers satisfying m < n , then x n divides x m but x m does not divide x n . Let I = { r R : x n | r for some n N } . Then I is an ideal of R . We claim that this ideal is not finitely generated. Let g 1 , g 2 , . . . , g k be a finite list of elements of I . Now there exists some natural number m large enough to ensure that that x m | g j for j = 1 , 2 , . . . , k . If I were generated by these elements g 1 , g 2 , . . . , g k , then x m | r for all r I . In particular x m would divide all x n for all n N , which is impossible. Thus the ideal I cannot be finitely generated. We have shown that if the set S defined above is a proper subset of some integral domain R , then R contains some ideal that is not finitely generated. The result follows. 2.2 Euclidean Domains Definition Let R be an integral domain, and let R * denote the set R \{ 0 R } of non-zero elements of R . An integer-valued function ϕ : R * Z defined on R * is said to be a Euclidean function if it satisfies the following properties:— (i) ϕ ( r ) 0 for all r R * ; 14

This preview has intentionally blurred sections. Sign up to view the full version.

(ii) if x, y R * satisfy x | y then ϕ ( x ) ϕ ( y ); (iii) given x, y R * , there exist q, r R such that x = qy + r , where either r = 0 R or ϕ ( r ) < ϕ ( y ). Definition A Euclidean domain is an integral domain on which is defined a Euclidean function. Example Let Z * denote the set of non-zero integers, and let ϕ : Z * Z be the function defined such that ϕ ( x ) = | x | for all non-zero integers x . Then ϕ is a Euclidean function. It follows that Z is a Euclidean domain. Example Let K be a field, and let K [ x ] be the ring of polynomials in a single indeterminate x with coefficients in the field K . The degree deg p of each non-zero polynomial p is a non-negative integer. If p and q are non-zero polynomials in K [ x ], and if p divides q , then deg p deg q . Also, given any non-zero polynomials m and p in K [ x ] there exist polynomials q, r K [ x ] such that p = qm + r and either r = 0 K or else deg r < deg m . We conclude from this that the function that maps each non-zero polynomial in K [ x ] to its degree is a Euclidean function for K [ x ]. Thus K [ x ] is a Euclidean domain. Example A Gaussian integer is a complex number of the form x + y - 1, where x and y are integers. The set of all Gaussian integers is a subring of the field of complex numbers, and is an integral domain. We denote the ring of Gaussian integers by Z [ - 1]. We define ϕ ( z ) = | z | 2 for all non-zero Gaussian integers z . Then ϕ ( z ) is an non-negative integer for all non-zero Gaussian integers z , for if z = x + y - 1, where x, y Z , then ϕ ( z ) = x 2 + y 2 . If z and w are non-zero Gaussian integers, and if z divides w in the ring Z [ - 1], then there exists a non-zero Gaussian integer t such that w = tz . But then ϕ ( w ) = ϕ ( t ) ϕ ( z ), where ϕ ( t ) 1, and therefore ϕ ( z ) ϕ ( w ). Let z and w be non-zero Gaussian integers. Then the ratio z/w lies in some square in the complex plane, where the sides of the square are of unit length, and the corners of the square are given by Gaussian integers. There is at least one corner of the square whose distance from z/w does not exceed 1 / 2. Thus there exists some Gaussian integer
This is the end of the preview. Sign up to access the rest of the document.
• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern