VeeEight said:Use the division algorithm.
kimberu said:You mean, say that n = qd + r, or for m?
Sorry, I'm totally lost on this problem.
TMM said:Suppose you have an ideal of the form mR+nR. How can you express this as a principal ideal?
So from here...can I say that s must equal 0 and x = aq for all x, since otherwise it's a contradiction because a is the smallest element?VeeEight said:Let a be the smallest positive element in your ideal I. If you have some element x in I, then x = aq + s for some s less then a.
VeeEight said:Yes.
Showing that every ideal of R has the form mR means proving that every ideal in the ring R can be written as a multiple of some fixed element m in R. This is known as the principal ideal generated by m, denoted as mR.
It is important to show that every ideal of R has the form mR because it helps us understand the structure of the ring R. It also allows us to simplify computations and proofs related to ideals in R.
To prove that every ideal of R has the form mR, we can use the ideal membership test. This test states that an element x belongs to an ideal I if and only if x can be written as a linear combination of elements in I. Using this test, we can show that every element in an ideal can be expressed as a multiple of some fixed element m in R, thus proving that the ideal has the form mR.
Yes, there are some exceptions to this statement. If the ring R is not a principal ideal domain, there may exist ideals that cannot be written in the form mR. Additionally, if R has zero divisors (elements that multiply to zero), then not all ideals will have the form mR.
Showing that every ideal of R has the form mR is equivalent to proving that R is a principal ideal domain. This is because a principal ideal domain is defined as a ring where every ideal is a principal ideal. Therefore, by showing that every ideal has the form mR, we can conclude that R is a principal ideal domain.