Prove that the set of all the ordered pairs `(a;b); a,b in RR` for which the addition and the multiplication operations are defined as follows: `(a;b) o+ (c;d)=(a+c;b+d)` ; `(a;b) ox (c;d)=(ac;bd)` cannot be a field.

This structure is not a field because there is no multiplicative inverse for some elements excluding zero.

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There are many rules (axioms) which a field must satisfy. For example, addition must be commutative and associative, as well as multiplication:

`a + b = b + a , ` `a + ( b + c ) = ( a + b ) + c , ` `a b = b a , ` `a ( b c ) = ( a b ) c .`

Also, multiplication must possess the distributive property with respect to addition: `a ( b + c ) = a b + a c . ` There must be also an additive identity element (zero) and a multiplicative identity element (one). Addition has an inverse operation (subtraction), i.e. `- a ` always exists.

All these properties are easily inherited by Cartesian product of two algebraic structures (and here we actually have it: each operation works over one of "dimensions").

But one of field properties is not a pure "operation" because it is not applied to any field element. This action is the multiplicative inverting: `a ^ ( - 1 ) , ` or `1 / a ` must exist for any field element except zero. Here, Cartesian product appears broken: although there is only one zero in the product, the pair `( 0 , 0 ) , ` there are many pairs that have no multiplicative inverses.

They are the pairs of the form `( 0 , c ) ` and `( a , 0 ) . ` Try to invert them:)

Because of this, the structure in question is not a field, although it is still a commutative ring.

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