prove that the product of any two ideals of ring R is also an ideal of R
Answers
Answer:
Hello mate..
Step-by-step explanation:
ring theory, a branch of abstract algebra, an ideal is a special subset of a ring. Ideals generalize certain subsets of the integers, such as the even numbers or the multiples of 3. Addition and subtraction of even numbers preserves evenness, and multiplying an even number by any other integer results in another even number; these closure and absorption properties are the defining properties of an ideal. An ideal can be used to construct a quotient ring similarly to the way that, in group theory, a normal subgroup can be used to construct a quotient group.
Among the integers, the ideals correspond one-for-one with the non-negative integers: in this ring, every ideal is a principal ideal consisting of the multiples of a single non-negative number. However, in other rings, the ideals may be distinct from the ring elements, and certain properties of integers, when generalized to rings, attach more naturally to the ideals than to the elements of the ring. For instance, the prime ideals of a ring are analogous to prime numbers, and the Chinese remainder theorem can be generalized to ideals. There is a version of unique prime factorization for the ideals of a Dedekind domain (a type of ring important in number theory).
The concept of an order ideal in order theory is derived from the notion of ideal in ring theory. A fractional ideal is a generalization of an ideal, and the usual ideals are sometimes called integral ideals for clarity.
HOPE IT WILL HELP YOU ❣️❣️..
PLZ MARK AS BRAINLIST ANSWER ❣️❣️..
Answer:
Let K be an ideal of the direct product R×S.
Define
I={a∈R∣(a,b)∈K for some b∈S}
and
J={b∈S∣(a,b)∈K for some a∈R}.
We claim that I and J are ideals of R and S, respectively.
Let a,a′∈I. Then there exist b,b′∈S such that (a,b),(a′,b′)∈K.
Since K is an ideal we have
(a,b)+(a′,b′)=(a+a′,b+b)∈k.
It follows that a+a′∈I.
Also, for any r∈R we have
(r,0)(a,b)=(ra,0)∈K
because K is an ideal.
Thus, ra∈I, and hence I is an ideal of R.
Similarly, J is an ideal of S.
Next, we prove that K=I×J.
Let (a,b)∈K. Then by definitions of I and J we have a∈I and b∈J.
Thus (a,b)∈I×J. So we have K⊂I×J.
On the other hand, consider (a,b)∈I×J.
Since a∈I, there exists b′∈S such that (a,b′)∈K.
Also since b∈J, there exists a′∈R such that (a′,b)∈K.
As K is an ideal of R×S, we have
(1,0)(a,b′)=(a,0)∈K and (0,1)(a′,b)=(0,b)∈K.
It yields that
(a,b)=(a,0)+(0,b)∈K.
Hence I×J⊂K.
Putting these inclusions together gives k=I×J as required.
Step-by-step explanation:
Remark.
The ideals I and J defined in the proof can be alternatively defined as follows.
Consider the natural projections
π1:R×S→R and π2:R×S→S.
Define
I=π1(K) and J=π2(K).
Since the natural projections are surjective ring homomorphisms, the images I and J are ideals in R and S, respectively.