Final answer:
Finite fields with specified numbers of elements can be created by factoring polynomial rings using irreducible polynomials. For example, fields with 4, 8, 9, 25, and 361 elements can be obtained by using polynomials over Z2, Z2, Z3, Z5, and Z19 respectively.
Step-by-step explanation:
The question asked to construct fields with exactly 4, 8, 9, 25, and 361 elements respectively. In mathematics, a field is a set equipped with two operations, typically called addition and multiplication, satisfying certain properties. Fields are important in algebra, number theory, and many other areas. A common way to construct finite fields is to consider them as factor rings of polynomial rings.
(a) A field with 4 elements (usually notated as F4 or GF(4)) can be constructed by taking the factor ring F2[x]/(f(x)), where F2 is the field with two elements (0, 1), 'x' is a variable, and f(x) is an irreducible polynomial of degree 2 over F2. A simple example of f(x) is x² + x + 1.
(b) Similarly, a field with 8 elements (F8 or GF(8)) can be constructed by factoring F2[x] with another irreducible polynomial of degree 3, such as x³ + x + 1.
(c) A field with 9 elements (F9 or GF(9)) can be created by factoring F3[x] with an irreducible polynomial of degree 2 over F3, like x² + 1.
(d) For a field with 25 elements (F25 or GF(25)), one could factor F5[x] with an irreducible polynomial of degree 2 over F5, for instance, x² + 2x + 2.
(e) Finally, to make a field with 361 elements (F361 or GF(361)), we would factor F19[x] using an irreducible polynomial of degree 2 over F19, for example, x² + x + 1.