**Step 1: Let \( I \) be an ideal of a field \( F \). If \( I \neq \{0\} \), then \( I \) contains some nonzero element \( a \).** **Explanation:** By the definition of an ideal, \( I \) is a nonempty subset of \( F \) closed under addition and under multiplication by any element of \( F \). If \( I = \{0\} \), we are done. If not, there exists \( a \in I \) with \( a \neq 0 \). --- **Step 2: Show that \( I = F \) if \( I \neq \{0\} \).** **Explanation:** Since \( a \neq 0 \) and \( F \) is a field, \( a \) has a multiplicative inverse \( a^{-1} \in F \). Because \( I \) is an ideal and \( a \in I \), for any \( r \in F \), \( r = r a^{-1} a \in I \) (since \( a \in I \) and ideals absorb multiplication from field elements). Therefore, \( I \) contains every element of \( F \), so \( I = F \). --- **Final Answer:** The only ideals of a field \( F \) are \( \{0\} \) and \( F \) itself.

Certainly! Here's the problem broken down into detailed parts with step-by-step reasoning: --- **Part 1: State the goal** Prove that the only ideals in a field \( F \) are the zero ideal \( \{0\} \) and the entire field \( F \). --- **Part 2: Assume an arbitrary ideal \( I \) of \( F \)** Let \( I \) be an ideal of \( F \). **Part 3: Consider the case when \( I = \{0\} \)** If \( I = \{0\} \), then the statement is true for this case. **Part 4: Suppose \( I \neq \{0\} \)** If \( I \neq \{0\} \), then \( I \) contains some element \( a \neq 0 \). **Part 5: Show that \( I \) contains an invertible element** Since \( a \neq 0 \) and \( F \) is a field, \( a \) has a multiplicative inverse \( a^{-1} \in F \). **Part 6: Demonstrate that \( I = F \)** Because \( a \in I \), and \( I \) is closed under multiplication by any element of \( F \), for any \( r \in F \), we have: \[ r = r \cdot 1 = r \cdot (a^{-1} a) = (r a^{-1}) a \] Since \( r a^{-1} \in F \), and \( a \in I \), the product \( (r a^{-1}) a \in I \) (because ideals absorb multiplication from the field). Thus, \( r \in I \) for all \( r \in F \). **Part 7: Conclude \( I = F \)** Since every element \( r \in F \) is in \( I \), \( I = F \). --- **Final conclusion:** Because any ideal is either \( \{0\} \) or \( F \), the only ideals of a field \( F \) are exactly these two. --- **Summary:** - If an ideal contains a nonzero element, it must be the whole field. - If it contains only zero, it's the zero ideal. - No other ideals exist in a field. **Final answer:** The only ideals of a field \( F \) are \( \{0\} \) and \( F \).