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Chapter 14: Problem 11

Let \(S\) and \(T\) be subrings of a ring \(R\). Prove that \(S \cap T\) is a subring of \(R\).

Short Answer

Expert verified
The intersection of \(S\) and \(T\), \(S \cap T\), is indeed a subring of \(R\) as it satisfies all subring test properties, i.e., it is non-empty and is closed under subtraction and multiplication.

Step by step solution

01

Non-emptiness

First check for non-emptiness. In any ring \(R\), \(0\) is an element in the ring. Since \(S\) and \(T\) are subrings, \(0\) is also in \(S\) and \(T\). Thus, by definition of intersection, \(0\) is in \(S \cap T\). So, \(S \cap T\) is not empty.
02

Closed under subtraction

Take two arbitrary elements \(a\) and \(b\) in \(S \cap T\). Since \(S\) is a subring, \(a - b\) is in \(S\). Similarly, since \(T\) is a subring, \(a - b\) is in \(T\). Therefore, by definition of intersection \(a - b\) is in \(S \cap T\). So, \(S \cap T\) is closed under subtraction.
03

Closed under multiplication

Similarly for multiplication, if we take the same arbitrary elements \(a\) and \(b\) in \(S \cap T\), since \(S\) is a subring, \(a \cdot b\) is in \(S\). And since \(T\) is a subring, \(a \cdot b\) is in \(T\). Therefore, by definition of intersection \(a \cdot b\) is in \(S \cap T\). So, \(S \cap T\) is closed under multiplication.

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Most popular questions from this chapter

Let \(R\) be a commutative ring with unity \(u\), and let \(I\) be an ideal of \(R\). (a) If \(u \in I\), prove that \(I=R\). (b) If \(I\) contains a unit of \(R\), prove that \(I=R\).

Prove that in any list of \(n\) consecutive integers, one of the integers is divisible by \(n .\)

If \(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \in M_{2}(\mathbf{R})\), prove that \(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is a unit of this ring if and only if \(a d-b c \neq 0\).

Let \((S,+, \cdot)\) and \(\left(T,+^{\prime}, *^{\prime}\right)\) be two rings. For \(R=S \times T\), define addition " \((\leftrightarrow)\) " and multiplication " \(\odot\) " by $$ \begin{aligned} &\left(s_{1}, t_{1}\right) \oplus\left(s_{2}, t_{2}\right)=\left(s_{1}+s_{2}, t_{1}+^{\prime} t_{2}\right), \\ &\left(s_{1}, t_{1}\right) \odot\left(s_{2}, t_{2}\right)=\left(s_{1} \cdot s_{2}, t_{1} \cdot t_{2}\right) . \end{aligned} $$ a) Prove that under these closed binary operations, \(R\) is a ring. b) If both \(S\) and \(T\) are commutative, prove that \(R\) is commutative, c) If \(S\) has unity \(u_{5}\) and \(T\) has unity \(u_{T}\), what is the unity of \(R\) ? d) If \(S\) and \(T\) are fields, is \(R\) also a field?

Let \(k, m\) be fixed integers. Find all values for \(k, m\) for which \((\mathbf{Z}, \oplus, \odot)\) is a ring under the binary operations \(x \oplus y=x+y-k, x \odot y=x+y-m x y\), where \(x, y \in \mathbf{Z}\).
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