Proving the Ideal Property of a Set in a Commutative Ring with Identity

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SUMMARY

The discussion focuses on proving that the set I = {r1c1 + r2c2 + ... + rncn | r1, r2, ..., rn ∈ R} is an ideal in a commutative ring R with identity. Key steps include demonstrating closure under addition and absorption properties. The proof confirms that if two elements from I are added, their sum remains in I, and for any r in R, the product of r with any element in I also belongs to I. Thus, I satisfies the conditions to be classified as an ideal in R.

PREREQUISITES
  • Understanding of commutative rings with identity
  • Knowledge of ideal properties in ring theory
  • Familiarity with closure properties (addition and multiplication)
  • Basic proficiency in algebraic manipulation of ring elements
NEXT STEPS
  • Study the definition and examples of ideals in commutative rings
  • Learn about the properties of ring homomorphisms
  • Explore the concept of quotient rings and their relationship to ideals
  • Investigate the role of generators in the formation of ideals
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Mathematics students, particularly those studying abstract algebra, and educators looking to deepen their understanding of ring theory and ideal properties.

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Homework Statement


I need to prove this theorem Let R be a commutative ring with identity and c1,c2,...cn E (element of) R Then the set I={r1c1+r2c2+...+rncn|r1,r2,...,rn E R} is an ideal in R




Homework Equations



Well I do know I need to prove closure under subtraction, closure under multiplication and absorption r E R and a E I show ra E I and ar E I I know because R is commutative I do not have to prove both ways. and I know that just proving absorption will prove multiplication.

The Attempt at a Solution



so, I guess I am just confused on what the set I looks like.

to prove closure under subraction I need to elements from I. What would these two elements look like?
 
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The set "looks like" exactly what it says: All things of the form r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n where c_1, c_2, \cdot\cdot\cdot, c_n are given (fixed) members of ring R and r_1, r_2, \cdot\cdot\cdot, r_n can be any members of the ring.

So you want to prove:
(1) Closed under addition: If r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n and s_1c_1+ s_2c_2+ \cdot\cdot\cdot+ s_nc_n are members of this set, it their sum also a member? That is can it be written in the same way: find t_1, t_2, \cdot\cdot\cdot, t_n such that the sum of those two is t_1c_1+ t_2c_2+ \cdot\cdot\cdot+ t_nc_n.

(2) "Absorption" of the entire ring: If r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n is in the set, show that r(r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n), for r any member of the ring, is of that same form.

Those are both just as easy as they look.
 
Ok that is easy.

So under addition, because R is a ring then r1+s1 will be in R so the (r1+s1)c1+(r2+s2)c2... is in I

And because R is a Ring and closed under multiplication r*r1, r*r2,... are in R so r*r1c1+r*r2c2+... is in I

So it is a ideal in R

Thanks
 

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