If  R
  is a field, show that the only two ideals of  R
  are  {0}
  and  R
  itself.

Let \( R \) be a field. To prove that the only two ideals in \( R \) are \( \{0\} \) and \( R \) itself, we use the following reasoning:

### Definition of an Ideal:
An ideal \( I \) of a ring \( R \) is a subset of \( R \) such that for any \( r \in R \) and \( i \in I \), the product \( ri \in I \).

### Step-by-Step Proof:

1. **Case 1: The trivial ideal \( \{0\} \)**
   - The set \( \{0\} \) is always an ideal of any ring (including a field) because for any \( r \in R \) and \( 0 \in \{0\} \), we have \( r \cdot 0 = 0 \in \{0\} \). So \( \{0\} \) is closed under multiplication and satisfies the ideal condition.
   
2. **Case 2: The entire field \( R \) as an ideal**
   - Clearly, \( R \) is an ideal of itself because for any \( r \in R \) and \( x \in R \), the product \( rx \in R \). So \( R \) is closed under multiplication, satisfying the condition of being an ideal.

3. **Case 3: Any non-trivial ideal \( I \) of \( R \)**
   - Suppose \( I \) is a non-zero ideal, meaning there exists some \( a \in I \) such that \( a \neq 0 \).
   - Since \( R \) is a field, every non-zero element \( a \in R \) has a multiplicative inverse, denoted by \( a^{-1} \in R \).
   - Now, because \( I \) is an ideal and \( a \in I \), we have that for any \( r \in R \), the element \( r \cdot a \in I \).
   - In particular, let \( r = a^{-1} \). Then \( a^{-1} \cdot a = 1 \in I \), which shows that the element \( 1 \in I \).
   - But if \( 1 \in I \), then for any \( r \in R \), we have \( r = r \cdot 1 \in I \), meaning that \( I = R \).

### Conclusion:
Thus, we have shown that the only possible ideals in \( R \), a field, are \( \{0\} \) and \( R \) itself. Any non-zero ideal must contain 1, which forces it to be the entire field \( R \).

Do you want further details or have any questions?

### Relative questions:
1. What is the definition of an ideal in a ring, and how does it differ from a field?
2. How do the properties of a field ensure that every non-zero element has a multiplicative inverse?
3. What happens if we consider ideals in a general ring rather than a field?
4. Can you explain why \( \{0\} \) is always an ideal in any ring, not just a field?
5. How do maximal ideals relate to the structure of fields?

### Tip:
In a field, the existence of a multiplicative inverse for every non-zero element plays a crucial role in proving that non-zero ideals must be the entire field.

If R is a field, show that the only two ideals of R are {0} and R itself.

Explore the detailed proof that the only two ideals in a field are the trivial ideal {0} and the entire field R. This explanation uses core concepts from field theory and ideal properties, making it perfect for abstract algebra students.

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