The goal of these problems is to practice with the definition of an ideal, investigate more ideal properties, and practice computing in a quotient ring.
In \(\R[x]\text{,}\) show that the set \(I\) of polynomials with \(f(0)=0\) is an ideal. Find a polynomial \(g\in\R[x]\) such that \(I=\ideal{g}\text{.}\)
So \(I\) is an ideal. Now we have \(\ideal{x}\subseteq I\) since \(x\) evaluated at 0 is 0. Conversely, if \(f\in I\) then by the root theorem for polynomials we have \(f=(x-0)g\) for some \(g\in \R[x]\text{,}\) so \(f\in \ideal{x}\text{.}\) Thus \(I=\ideal{x}\text{.}\)
Let \(R=\R[x]\) and \(I=\langle x^2+1\rangle\text{.}\) What familiar structure do the distinct cosets of \(R/I\) have? Try repeating the simplification procedure from the Daily Prep example.
We will show \(R/I\) has the structure of \(\C\text{.}\) First, we see that if \(a+bx\) and \(c+dx\) are different polynomials, then \((a-c)+(b-d)x \not\in I \) because a linear polynomial is not a multiple of a quadratic polynomial. Thus the cosets of \(R/I\) with linear representatives are distinct. Now, let \(f\in R\text{.}\) By the division algorithm, we can write \(f=(x^2+1)q+r\) where \(r\) is either 0 or has degree less than 2. If \(r=0\text{,}\) then \(f\in I\) and \(f+I=I\text{.}\) If \(r\neq 0\text{,}\) then \(r=ax+b\) for some \(a,b\in \R\text{.}\) Thus every coset of \(R/I\) has a linear representative, and we have shown that the cosets of \(R/I\) with linear representatives are distinct.
We use the Ideal Test. Let \(a,b\in I+J\) and \(r\in R\text{.}\) Then \(a=i_1+j_1\) and \(b=i_2+j_2\) for some \(i_1,i_2\in I\) and \(j_1,j_2\in J\text{.}\) Now we have
We use the Ideal Test. Let \(a,b\in I\cap J\) and \(r\in R\text{.}\) Then \(a,b\in I\) and \(a,b\in J\text{.}\) Since both \(I\) and \(J\) are ideals, we have \(a-b\in I\) and \(a-b\in J\text{,}\) so \(a-b\in I\cap J\text{.}\) Also, \(ra\in I\) and \(ra\in J\text{,}\) so \(ra\in I\cap J\text{.}\) Similarly, \(ar\in I\cap J\text{.}\) Thus \(I\cap J\) is an ideal.
We use the Ideal Test. Let \(a,b\in IJ\) and \(r\in R\text{.}\) Then \(a=\sum_{i=1}^n a_i b_i\) and \(b=\sum_{j=1}^m c_j d_j\) for some \(a_i,c_j\in I\) and \(b_i,d_j\in J\text{.}\) Now we have
\begin{equation*}
a-b=\sum_{i=1}^n a_i b_i - \sum_{j=1}^m c_j d_j\in IJ
\end{equation*}
since \(I\) and \(J\) are ideals. Also, we have
\begin{equation*}
ra=r\sum_{i=1}^n a_i b_i=\sum_{i=1}^n (ra_i) b_i\in IJ
\end{equation*}
since \(I,J\) are ideals. Similarly, \(ar\in IJ\text{.}\) Thus \(IJ\) is an ideal.
First, \(I\cap J\subset I\) and \(I \subseteq I+J\) (by taking the elements \(i+0\in I+J\)), so \(I\cap J \subseteq I+J\text{.}\) Now, let \(a_1b_1+\dots+a_nb_n\in IJ\text{.}\) Since \(a_i\in I \) and \(I\) is an ideal, each term \(a_ib_i\) is in \(I\text{.}\) So the sum is in \(I\text{.}\) Similarly, each term is in \(J\text{,}\) so the sum is in \(J\text{.}\) Thus \(a_1b_1+\dots+a_nb_n\in I\cap J\text{,}\) and we have shown \(IJ \subseteq I \cap J\text{.}\)
The goal of these problems is to practice working with the definitions of prime and maximal ideals and the theorems relating these ideals to properties of their quotient rings.
Because all ideals are subgroups, there are only a few possibilities for maximal ideals in \(\Z_8\) and \(\Z_{10}\text{.}\) In \(\Z_8\text{,}\) the only proper ideals are \(\ideal{0}=\{0\}\text{,}\)\(\ideal{2}=\{0,2,4,6\}\text{,}\) and \(\ideal{4}=\{0,4\}\text{.}\) The only ideal that is not contained in any other proper ideal is \(\ideal{2}\text{,}\) so this is the only maximal ideal.
In \(\Z_{10}\text{,}\) the proper ideals are \(\ideal{0}=\{0\}\text{,}\)\(\ideal{2}=\{0,2,4,6,8\}\text{,}\)\(\ideal{5}=\{0,5\}\text{,}\) and \(\ideal{10}=\Z_{10}\text{.}\) The only ideals that are not contained in any other proper ideal are \(\ideal{2}\) and \(\ideal{5}\text{,}\) so these are the maximal ideals.
Since \(|R|=30\text{,}\) any ideal of \(R\) has order dividing 30. If an ideal has order 10, then the quotient ring has order 30/10 = 3. Since 3 is prime, the quotient ring is a field, which means the ideal is maximal.
The ideal \(I\) is prime because the quotient ring \((\Z\oplus \Z)/I\) is isomorphic to \(\Z\text{,}\) which is an integral domain. However, \(I\) is not maximal because \(\Z\) is not a field.
We have \(I=\{\sum_i 6a_i8b_i\mid a_i,b_i\in\Z\}\text{.}\) So elements of \(I\) are multiples of 48, and so \(I\subseteq \ideal{48}\text{.}\) Conversely, \(48=6\cdot 8\in I\text{,}\) so \(\ideal{48}\subseteq I\text{.}\) Thus \(I=\ideal{48}\text{.}\)
Elements of \(I\) are both multiples of 6 and multiples of 8, so they are multiples of \(\lcm(6,8)=24\text{.}\) Thus \(I\subseteq \ideal{24}\text{.}\) Conversely, \(24\in I\text{,}\) so \(\ideal{24}\subseteq I\text{.}\) Thus \(I=\ideal{24}\text{.}\)
