Question

Let \(R\) be a ring (commutative, with 1 ) with a valuation \(v\), with the special property that \(v(a) \leq 1\) for all \(a \in R\). Show that if \(a \in R\) is a unit, then \(v(a)=1\).

Short answer

If \( a \) is a unit in a ring \( R \) with valuation \( v \leq 1 \), then \( v(a) = 1 \).

Step-by-step solution

Understanding the Problem

We are given a ring \( R \) with a valuation \( v \) where for all elements \( a \in R \), \( v(a) \leq 1 \). We need to show that if \( a \) is a unit in \( R \), then \( v(a) = 1 \).

Definition of a Unit

Recall that \( a \) is a unit in a ring \( R \) if there exists an element \( b \in R \) such that \( ab = 1 \), where 1 is the multiplicative identity in \( R \). This is crucial for our proof as it allows us to utilize the properties of units.

Using the Valuation Property

Since \( v(a) \leq 1 \) for all \( a \in R \), this property naturally applies to the unit \( a \). Thus, \( v(a) \leq 1 \) and \( v(a^{-1}) \leq 1 \) since \( a^{-1} \) is also in \( R \) as a unit.

Applying the Valuation of the Product

For the valuation, we have under the given setup, \( v(ab) = v(1) = 1 \). In a valuation, the product of two elements' valuations is the valuation of the product, i.e., \( v(a) v(a^{-1}) = v(1) = 1 \).

Solving for \( v(a) \)

From \( v(a) v(a^{-1}) = 1 \), since both \( v(a) \) and \( v(a^{-1}) \) are less than or equal to 1, the only possibility for their product to be 1 is if both are precisely equal to 1, hence \( v(a) = 1 \).

Conclusion

Thus, since \( v(a) v(a^{-1}) = 1 \) can only be true if \( v(a) = 1 \), we have shown that if \( a \) is a unit in \( R \), then \( v(a) = 1 \).

Key concepts

Units in a Ring

In ring theory, a unit is an especially important type of element. **But what exactly does it mean for an element to be a unit?** Simply put, an element \( a \) in a ring \( R \) is called a unit if there exists another element \( b \) in the ring such that their product is the multiplicative identity, often simply referred to as "1" in ring notation. Thus, \( ab = 1 \) if \( a \) is a unit.
Understanding units is critical because they help us grasp invertibility in a ring context. If you think about numbers, units act like multiplicative inverses. Just like how multiplying a number by its reciprocal equals 1, in ring terms, units reflect similar behavior. This concept is fundamental when you explore how elements in a ring can interact and contribute to the ring's structure. Without a unit, ensuring the possibility of inverting an element, would limit many properties and operations possible within the ring.

Valuation in Algebra

A valuation is a fascinating concept found in algebra. It helps us measure or 'evaluate' elements within a ring, to understand them better. Think about it as assigning a value to ring elements, symbolized as \( v - \) so each element \( a \) in ring \( R \) gets a value \( v(a) \).
For the ring \( R \) in our exercise, we have the special condition that the valuation \( v(a) \) for any element \( a \) is always less than or equal to 1. This tells us about the 'size' or 'strength' of each element in contexts where valuations are involved. It's like having a rule where no element in the ring exceeds a specific 'boundary' set by 1.
Understanding how a valuation interacts with elements, particularly units, showcases how algebraic structures maintain balance and consistency. When you find that a unit's valuation must equal 1, it underlines the robustness of units in achieving this maximal 'bound' value compared to other potential ring elements.

Multiplicative Identity

The multiplicative identity is a simple yet powerful concept. In any ring, there exists a neutral element for multiplication, denoted as "1." This '1' ensures that when any element of the ring is multiplied by it, the result remains the same element.
So when we say \( ab = 1 \), \( 1 \) is playing its role as the multiplicative identity, maintaining the element's 'neutrality' and integrity during multiplication. This trait of achieving an unaltered result is what makes '1' the multiplicative identity in rings.
The presence of a multiplicative identity is vital for understanding units in a ring. It also ties directly to the unit's definition, where multiplying a unit with its inverse returns the multiplicative identity. Recognizing and appreciating this role of '1' helps us understand how rings are structured to enable operations like inverses, stability, and consistency within the algebraic framework.