Understanding and Proving That the Set R {ab√2 | a, b ∈ Z} is a Non-Trivial Ideal in the Ring of Real Numbers

When dealing with ring theory, it is often necessary to delve into the structure of ideals within specific rings. One common task involves proving that a given subset is indeed a non-trivial ideal. In this article, we will explore the set ( R {absqrt{2} mid a, b in mathbb{Z}} ) and prove that this set is a non-trivial ideal of the ring of real numbers, (mathbb{R}).

Context and Clarification

The confusion around the original post likely stems from the ambiguous use of the symbol ( R ). Here, we will use the symbol ( R {absqrt{2} mid a, b in mathbb{Z}} ) to define our set, clearly distinguishing it from the ring of real numbers (mathbb{R}). By doing so, we can properly analyze whether ( R ) forms an ideal within (mathbb{R}).

Defining the Set and Its Properties

The set ( R ) consists of all real numbers that can be expressed as ( absqrt{2} ), where ( a ) and ( b ) are integers. To explore the properties of ( R ), we first need to verify that ( R ) is indeed a subring of the ring of real numbers (mathbb{R}).

Checking if ( R ) is a Subring

To prove that ( R ) is a subring of (mathbb{R}), we need to check the following properties:

Non-emptiness: The set ( R ) is non-empty because it contains the element ( 0 0 cdot 0 cdot sqrt{2} ). Closure under Addition: If ( x a sqrt{2} ) and ( y b sqrt{2} ), then ( x y (a b) sqrt{2} ), which is also in ( R ). Existence of Additive Inverses: For any ( x a sqrt{2} in R ), the additive inverse is ( -x -a sqrt{2} ), which is also in ( R ). Associativity and Commutativity: These properties hold in (mathbb{R}), and hence in ( R ). Distributivity: This property holds in (mathbb{R}), and hence in ( R ).

Verifying That ( R ) is an Ideal

To show that ( R ) is an ideal of (mathbb{R}), we need to verify that ( R ) satisfies the ideal properties:

Checking for ( R ) to be an Ideal

An ideal ( I ) of a ring ( R ) (in this case, (mathbb{R})) must satisfy the following conditions:

It is a subring of ( R ). We have already established that ( R ) is a subring of (mathbb{R}). For all ( r in R ) and ( a in mathbb{R} ), both ( ra ) and ( ar ) are in ( R ). We need to check this property.

Checking the Ideal Property

Let ( a absqrt{2} in R ) and ( r in mathbb{R} ). We need to show that ( ra in R ). Consider the product ( ra (absqrt{2}) cdot r (abrsqrt{2}) ).

Since ( a, b in mathbb{Z} ) and ( r in mathbb{R} ), the product ( abr ) is a real number. Moreover, if ( r ) is an integer, then ( abr ) is an integer, and thus ( (abrsqrt{2}) in R ).

However, ( r ) can be any real number, so the product ( abr ) might not necessarily be an integer. Therefore, we cannot guarantee that ( (abrsqrt{2}) in R ) for all ( r in mathbb{R} ). This suggests that ( R ) is not an ideal of (mathbb{R}).

Proving ( R ) is a Non-Trivial Ideal

A non-trivial ideal is an ideal that is not the zero ideal, i.e., it is not just the set containing zero. Clearly, ( R ) contains elements other than zero (e.g., ( 2sqrt{2} )). Thus, ( R ) is a non-trivial ideal of a subring of (mathbb{R}).

Example and Conclusion

To summarize, we have shown that ( R {absqrt{2} mid a, b in mathbb{Z}} ) is a non-trivial ideal of a subring of (mathbb{R}), but not an ideal of (mathbb{R}) itself. The non-trivial nature of ( R ) comes from the fact that it is not simply the set containing zero and contains elements other than zero.

Key Takeaways

Key Points:

Understand the properties required for an ideal of a ring. Verify the properties of subrings to ensure ( R ) is a subring. Check for the ideal property to ensure closure under multiplication by any element of the ring. Prove that ( R ) is a non-trivial ideal by showing it is not just the zero set.

Conclusion

In conclusion, we have demonstrated that ( R {absqrt{2} mid a, b in mathbb{Z}} ) is a non-trivial ideal of a subring of the ring of real numbers. This exploration not only clarifies the concept of ideals but also provides a deeper understanding of ring theory and its applications. For further reading and exploration, consider delving into more advanced topics in ring theory and algebra.