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Chapter 14: Problem 18

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 \(I=R\).

Short Answer

Expert verified
Both parts of the proofs essentially show that if a commutative ring with unity \(R\) has an ideal \(I\) that either contains the unity or a unit of \(R\), then \(I\) must contain every element of \(R\), hence \(I=R\).

Step by step solution

01

Proof of part (a)

If the unity \(u \in I\) which is an ideal of \(R\), then it follows that every element \(r \in R\) can be represented as \(r = r \cdot u\). Since both \(r\) and \(u\) are elements of \(R\) and \(I\) is an ideal which means it is closed under ring multiplication, \(r\) must also be in \(I\). Thus, \(I=R\).
02

Proof of part (b)

Assumptions are made that \(I\) contains a unit of \(R\), meaning there exists an element \(v \in I\) such that \(v \cdot r = r \cdot v = r\) for all \(r \in R\). Since \(v \in I\), and \(I\) is an ideal, it is closed under right and left multiplication by any element from \(R\). Hence, for any \(r\), \(r \cdot v\) and \(v \cdot r\) both will be in \(I\). However, \(r \cdot v = r\) and \(v \cdot r = r\), so any \(r \in R\) is also in \(I\). Consequently, it can now be concluded that \(I = R\).

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Most popular questions from this chapter

a) Determine the multiplicative inverse of the matrix \(\left[\begin{array}{ll}1 & 2 \\ 3 & 7\end{array}\right]\) in the ring \(M_{2}(\mathbf{Z})\) - that is, find \(a, b, c, d\) so that $$ \left[\begin{array}{ll} 1 & 2 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$ b) Show that \(\left[\begin{array}{ll}1 & 2 \\ 3 & 8\end{array}\right]\) is a unit in the ring \(M_{2}(Q)\) but not a unit in \(M_{2}(\mathbf{Z})\).

If \(n \in \mathbf{Z}^{*}\) and \(n>2\) prove that $$ \sum_{i=1}^{n-1} i=\left\\{\begin{array}{lll} 0 & (\bmod n), & n \text { odd } \\ \frac{n}{2} & (\bmod n), & n \text { even. } \end{array}\right. $$

For \(p\) a prime determine all elements \(a \in \mathbf{Z}_{p}\) where \(a^{2}=a\).

In the solution of Example \(14.17\), part of the problem required the calculation of $$ \left(\left[\begin{array}{rr} 1.3 & 3.7 \\ -3.7 & 1.3 \end{array}\right]\right)^{-1} $$ Algebraically this equals $$ \left(\left[\begin{array}{rr} 1.3 & 3.7 \\ -3.7 & 1.3 \end{array}\right]^{-1}\right)^{4} $$ Is there any advantage, with respect to computer computation, in using one form instead of the other?

Let \((R,+, \cdot)\) be a commutative ring, and let \(z\) denote the zero element of \(R\). For a fixed element \(a \in R\), define \(N(a)=\\{r \in R \mid r a=z\\}\). Prove that \(N(a)\) is an ideal of \(R\).
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