“It is a useless life that is not consecrated to a great ideal. It is like a stone wasted on the field without becoming a part of any edifice.”
―Jose Rizal
Definition9.1.Ideal.
For a ring \(R\text{,}\) an ideal (or a two sided ideal) of \(R\) is a non empty subset \(I\) such that
\((I,+)\) is a subgroup of \((R,+)\) and
for all \(r \in R\) and \(a \in I\text{,}\) we have \(ra \in I\) and \(ar \in I\text{.}\) This is often called absorption 1
Personally I think it would be splendid if we could rename ideals sponges to match this imagery.
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For a ring \(R\text{,}\) a proper ideal is an ideal \(I\) such that \(I\neq R\text{.}\)
Example9.2.Ideals.
In any ring \(R\text{,}\)\(\{0\}\) and \(R\) itself are ideals.
The ideals of \(\Z\) are \(n\Z\text{.}\)
The sets \(R_i=\left\{\begin{bmatrix} 0 & 0 & \cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\ a_{i1} & a_{i2} & \cdots &a_{in} \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots &0\end{bmatrix}\right\}\) and \(L_j=\left\{\begin{bmatrix} 0 & \cdots & a_{j1} &\cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\0 & \cdots & a_{ji} &\cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\ 0 & \cdots & a_{jn} &\cdots & 0\\\end{bmatrix}\right\}\) are a right ideal and a left ideal of \(\M_n(R)\) respectively. Neither are two-sided ideals.
The set \(\cR\) of all nilpotent elements in a ring \(A\) is an ideal.
Let \(R\) be a commutative ring, and set \(I = \{r \in R | rn = 0 \;\text{for some integer}\; n\}\text{.}\) Then \(I\) is an ideal in \(R\text{.}\)
A fun fact about ideals is that they are subrings.
Proposition9.3.Ideals are Subrings.
Any ideal \(I\) of a ring \(R\) is a subring of \(R\)
The converse need not be true, however.
Exercise9.4.Subrings Need not be Ideals.
Find, with justification, a subring which is not an ideal.
Solution.
In \(\R[x]\text{,}\) the set \(S\) of polynomials for which every term has even degree is a subring (it’s closed under subtraction and multiplication), but it is not an ideal because it is not closed under multiplication by arbitrary polynomials. Indeed, \(p(x)=x^2\in S\text{,}\) but \(xp(x)=x^3\not\in S\text{.}\)
Another fun fact is that we can combine ideals in all sorts of ways to get new ones!
Theorem9.5.Combinations of Ideals.
Let \(R\) be a ring and let \(I,J\) be ideals of \(R\text{.}\) Then
\(I + J := \{a + b \mid a \in I, b \in J\}\) is an ideal
\(I\cap J\) is an ideal
\(IJ=\{\sum_{i=1}^n a_ib_j \mid n\geq 0, a_i\in I, b_j\in J\}\) is an ideal and \(IJ\subseteq I\cap J\text{.}\)
The intersection \(\bigcap_{\a\in J} I_\a\) of any collection of ideals \(I_\a\) of \(R\) is an ideal.
The set of all ideals of a ring \(R\) forms a lattice with respect to the partial order given by containment. In this lattice, the supremum of a pair of ideals \(I,J\) is \(I+J\) and the infimum is \(I\cap J\text{.}\)
Exercise9.6.Union of Ideals Need not be an Ideal.
Give an example
Exercise9.7.Modular Law.
Let \(I,J,K\) be ideals in \(R\) such that \(J\sse I\) or \(K\sse I\text{.}\) Then
\begin{equation*}
I\cap(J+K)=I\cap J+I\cap K.
\end{equation*}
And here are some important properties to wrap up on.
Proposition9.8.Proper Ideals, Fields, and Units.
An ideal \(I\) of a unital ring \(R\) is proper if and only if \(I\) contains no units. Moreover, if \(F\) is a field it has only two ideals \(\{0\}\) and \(F\text{.}\)
Theorem9.9.Homomorphisms and Ideals.
If \(\varphi: R \to S\) is a ring homomorphism, then
the image of \(\varphi\) is a subring of \(S\text{,}\)
the kernel of \(\varphi\) is an ideal of \(R\text{,}\)
\(\varphi\) is injective if and only if \(\ker(\varphi)=\{0\}\text{,}\)
if \(I\) is an ideal of \(R\) then \(\varphi(I)\) is an ideal of \(\varphi(R)\text{,}\) and
if \(J\) is an ideal of \(S\) then \(\varphi^{-1}(J)\) is an ideal of \(R\text{.}\)
Proof.
Since \(\) is a ring homomorphism, it is in particular a group homomorphism \((R,+)\to (S,+)\text{.}\) We know the kernel of a group homomorphism is a subgroup, so \(\ker(f)\leq (S,+)\text{.}\) All that remains to be shown is that for any \(r\in R\)\(r\ker(f)\subseteq \ker(f)\) and \(\ker(f)r\subseteq \ker(f)\text{.}\) Let \(x\in \ker(f)\text{;}\) then \(f(x)=0\) and \(f(rx)=f(r)f(x)=0\text{,}\)\(f(xr)=f(x)f(r)=0\) show \(rx,xr\in \ker(f)\text{.}\)
SubsectionGenerated Ideals
“If you make yourself more than just a man, if you devote yourself to an ideal, you become something else entirely.”
―Liam Neeson, Batman Begins
Definition9.10.Generated Ideals.
If \(A\) is any subset of a ring \(R\text{,}\) the ideal generated by\(A\text{,}\) denoted \(\igen A\text{,}\) is the intersection of all ideals of \(R\) that contain \(A\text{:}\)
\begin{equation*}
\igen A=\bigcap_{I\;\text{ideal of}\;R, A\subseteq I} I.
\end{equation*}
An ideal \(I\) is finitely generated if \(I = \igen A\) for some finite subset \(A\) of \(R\text{.}\)
Remark9.11.
By Theorem 9.5, \(\igen A\) is an ideal. It is also the smallest ideal of \(R\) that contains \(A\text{.}\)
Lemma9.12.Elements in Generated Ideals.
For a subset \(A\) of a ring \(R\) with \(1\text{,}\) the ideal generated by \(A\) is given by
In the commutative ring \(\Z\text{,}\) we have \(\igen{2,3}=\igen 1=\Z\text{.}\) Indeed any element \(n\in \Z\) can be written as \(n=(-n)\cdot 2+n\cdot 3=n\cdot 1\text{.}\) Note that \(1=\gcd(2,3)\text{.}\)
In the commutative ring \(\Z\text{,}\) we have \(\igen{2,4}=\igen 2=2\Z\text{,}\) the set of all even integers. Notice this shows that different sets can generate the same ideal. Also note that \(2=\gcd(2,4)\text{.}\)
Exercise9.14.Finitely Generated Ideals and Nilpotent Elements.
Let \(I\) be a finitely generated ideal of \(R\text{.}\) Suppose every element of \(I\) is nilpotent. Prove that there exists an integer \(n\geq 1\) such that \(b^n=0\) for all \(b\in I\text{.}\)
Exercise9.15.Infinitely Generated Ideal.
Let \(I=\{f\in \cC[0,1]|f(x)=0\text{ on } [0,b]\;\text{for some}\; b>0\}\)
Prove that \(I\) is an ideal of \(\cC[0,1]\text{.}\)
