Let S and T be subrings of a ring R. In (a) and (b), if the answer is "yes," prove it. If the answer is "no," give a counterexample.

(a) Is S and T a subring of R?

(b) Is S or T a subring of R?

Let's analyze both parts of the question, addressing whether \( S \cap T \) and \( S \cup T \) are subrings of \( R \).

### (a) Is \( S \cap T \) a subring of \( R \)?

**Answer: Yes.**

**Proof:**
To prove that \( S \cap T \) is a subring of \( R \), we need to verify that \( S \cap T \) satisfies the subring criteria, which are closure under addition, closure under multiplication, and containing the additive identity (and also the existence of additive inverses).

Let \( x, y \in S \cap T \). This means that \( x \in S \) and \( x \in T \), and similarly, \( y \in S \) and \( y \in T \).

1. **Closure under addition**: Since \( S \) and \( T \) are subrings of \( R \), both are closed under addition. Hence, \( x + y \in S \) and \( x + y \in T \). Therefore, \( x + y \in S \cap T \).

2. **Closure under multiplication**: Similarly, since \( S \) and \( T \) are closed under multiplication, we have \( xy \in S \) and \( xy \in T \). Hence, \( xy \in S \cap T \).

3. **Additive identity**: Since both \( S \) and \( T \) are subrings, they both contain the additive identity element \( 0 \in R \). Therefore, \( 0 \in S \cap T \).

4. **Additive inverses**: If \( x \in S \cap T \), then \( x \in S \) and \( x \in T \), so \( -x \in S \) and \( -x \in T \) (because subrings are closed under additive inverses). Hence, \( -x \in S \cap T \).

Since all subring properties are satisfied, \( S \cap T \) is indeed a subring of \( R \).

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### (b) Is \( S \cup T \) a subring of \( R \)?

**Answer: No.**

**Counterexample:**
Consider the ring \( \mathbb{Z} \) (the integers) and let \( S = 2\mathbb{Z} \) (the set of all even integers) and \( T = 3\mathbb{Z} \) (the set of all integers divisible by 3). 

- Both \( S \) and \( T \) are subrings of \( \mathbb{Z} \).
- The union \( S \cup T = 2\mathbb{Z} \cup 3\mathbb{Z} \) consists of all numbers that are either divisible by 2 or divisible by 3. 

Consider \( x = 2 \in S \cup T \) and \( y = 3 \in S \cup T \).
- \( x + y = 2 + 3 = 5 \), but 5 is neither divisible by 2 nor by 3, so \( 5 \notin S \cup T \).

Since \( S \cup T \) is not closed under addition, it cannot be a subring of \( R \). Hence, \( S \cup T \) is not a subring.

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### Summary:
- (a) **Yes**, \( S \cap T \) is a subring of \( R \).
- (b) **No**, \( S \cup T \) is not necessarily a subring of \( R \).

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Do you want more details on any part? Here are 5 related questions to explore further:

1. Can \( S \cup T \) be a subring under any conditions?
2. What are some examples of subrings other than intersections?
3. How do you prove that a subset is not a subring in general?
4. What happens if \( S \) and \( T \) are ideals instead of subrings?
5. Are there any stronger conditions where both \( S \cup T \) and \( S \cap T \) could be subrings?

**Tip:** Always check closure properties carefully when verifying if a set is a subring!

Let S and T be subrings of a ring R. In (a) and (b), if the answer is 'yes,' prove it. If the answer is 'no,' give a counterexample.

(a) Is S ∩ T a subring of R?

(b) Is S ∪ T a subring of R?

Explore whether S ∩ T and S ∪ T are subrings of a ring R. This problem focuses on the subring criteria, closure properties, and set operations in abstract algebra. Includes a proof for S ∩ T and a counterexample for S ∪ T. Suitable for university undergraduate students studying abstract algebra.

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