How about the algebraic closure of Z_3? Or is that too nonconstructuve?ehrenfest said:Homework Statement
I want to find a number that is algebraic with degree 3 over Z_3. To do this, I need to find an extension field of Z_3. Q,R, and Z_p (p greater than 3) definitely will not work because they have different algebra. Anyone?
An algebraic number with degree 3 over Z3 is a number that can be expressed as a root of a polynomial with coefficients from the field Z3 (the integers modulo 3). In other words, the number is a solution to an equation of the form ax3 + bx2 + cx + d = 0, where a, b, c, and d are all integers modulo 3.
To find a number that is algebraic with degree 3 over Z3, you can start by choosing any three integers from 0 to 2 (since Z3 contains only three elements). Then, you can plug these values into the polynomial equation ax3 + bx2 + cx + d = 0 and solve for x. The resulting value of x will be an algebraic number with degree 3 over Z3.
Yes, a number can be algebraic with degree 3 over Z3 and not be an integer. This is because the field Z3 contains only three elements (0, 1, and 2), so any cubic polynomial with coefficients from this field will have solutions that are not integers.
No, not all real numbers are algebraic with degree 3 over Z3. In fact, the majority of real numbers are transcendental, meaning they cannot be expressed as the root of any polynomial with integer coefficients. Only a small fraction of real numbers are algebraic, and an even smaller fraction are algebraic with degree 3 over Z3.
Finding numbers that are algebraic with degree 3 over Z3 is important in many areas of mathematics, such as number theory and algebraic geometry. These numbers help us understand the properties of polynomial equations and can provide insights into more complex mathematical structures. Additionally, they have applications in fields such as coding theory and cryptography.