2.5. Principal, Maximal and Prime Ideals.
We'll now move more deeper into ring theory. The results prior were already things we've been familiar with, since they were true for groups. It is here that we'll move onto new, deeper concepts regarding the ideal of a ring.
Let \(X \subset R\). Then we can talk about the subring generated by \(X\) as the smallest subring containing \(X\), or equivalently, the intersection of all the subrings containing \(X\). More explicitly, we can define it to be the set of all finite sums of elements of \(X\).
Similarly, we can define the ideal generated by \(X\), which again is the smallest ideal containing \(X\) or equivthalently the intersection of all ideals contianing \(X\). More explicitly, if \(R\) is a ring with identity, then the ideal generated by \(X\) is
while if \(R\) is commutative (and again, has an identity) this becomes
Note: this is not valid for rings without identity.
\textcolor{MidnightBlue}{Why are these formulas correct? We'll show this for the more general case for when \(R\) may not be commutative. Specifically, we'll show that these formulas are not only ideals, but that they are the smallest ideals containing \(X\) as we have claimed. \ \indent If \(r \in R\) then
since \(rr_i \in R\) for each \(i \in \{1, 2, \dots, n\}\). Similarly,
since again, \(s_ir \in R\) for each \(i \in \{1, 2, \dots, n\}\). Thus this is an ideal. Now let \(X'\) be an ideal which contains \(X\). Pick an arbitrary element \(\displaystyle \sum_{i=1}^{n}r_ix_is_i \in \left< X \right>\). Observe that since \(X'\) is an ideal containing \(X\), we know that \(r_ix_is_i \in X'\) for each \(i \in \{1, 2, \dots, n\}\), and hence the sum itself, \(\sum_{i=1}^{n}r_ix_is_i\), must be in \(X'\). \ \indent Thus for any ideal \(X'\) containing \(X\), we see that \(\left< X \right> \subset X'\). Hence, \(\left< X \right>\) is the smallest ideal containing \(X\). The proof is similar, and easier, for the case of \(\left< X \right>\) when \(R\) is commutative. }
Let \(R\) be commutative and \(X = \{a\}\), where \(a \in R\). Then the ideal generated by \(X\) given by
is said to be a principal ideal generated by \(a\). Note that since \(R\) is commutative, we could have also written \(\left< X \right> = \{ar \mid r \in R\}\).
One may also view a principal ideal generated by \(a\) as the set \(Ra\) (or again, equivalently as \(aR\)).
What's an example of this? Consider the ring \(\ZZ\). Then the subring \(2\ZZ\) is a principle ideal generated by the element 2. That is
which is pretty basic fact that we already know. But note that every ideal \(I\) of \(\ZZ\) is of this form.
\textcolor{NavyBlue}{To see this, consider an ideal \(I\) of \(\ZZ\) and suppose \(i\) is the smallest positive element of \(I\). First observe that there is no \(j \in I\) such that \(i < j < 2i\). Suppose there was. Then \(j = i + k\) for some \(0 < k < i\). Since \(I\) is closed, we know that
is a member of \(I\). But this contradicts our assumption that \(i\) was the smallest positive element of \(I\). Therefore, there is no \(j \in I\) such that \(i < j < 2i\), so that \(I\) is of the form
Hence, \(I\) is generated by \(i\), a single element, so that \(I\) is principal. } Thus every ideal of \(\ZZ\) is principal. Rings who exhibit this type of behavior get a special name.
Let \(R\) be an integral domain. Then \(R\) is a principal ideal domain (PID) if every ideal of \(R\) is principal. As we just showed, \(\ZZ\) is a principal ideal domain.
Let \(R\) be a ring. Then an ideal \(M \ne R\) is called maximal if, for any other ideal \(I\) such that \(M \subset I \subset R\) we have that \(M = I\) or \(I = R\).
An example of this is the ring \(\ZZ\) with the ideal \(p\ZZ\) where \(p\) is prime. To show this, suppose \(I\) is an ideal such
and further that there exists an \(i \in I\) such that \(i \not\in p\ZZ\).
Since \(i\) is not divisible by \(p\), we know by Fermat's Little Theorem that
Thus for some \(n \in \ZZ\), \(pn - i^{p-1} = 1.\) Since \(p\ZZ \subset I\) and \(i \in I \implies i^{p-1} \in I\), we see that \(pn - i^{p-1} = 1 \in I\). Since \(1 \in I\), we can repeatedly add and subtract \(1\) to generated all of \(\ZZ\), and since this must all be contained in \(I\), we have that \(I = \ZZ\). Thus \(p\ZZ\) is maximal.
Let \(R\) be a ring and \(I\) a proper ideal (i.e., \(I \ne R\).) Then there is a maximal ideal of \(R\) containing \(I\). Furthermore, if \(R\) is a ring with identity, then there are always maximal ideals.
How do we know when a given ideal is maximal or not? That is, how do we know there aren't "bigger" proper ideals which contain the one we are interested in?
Let \(R\) be a commutative ring with identity. Then an ideal \(M\) of \(R\) is maximal if and only if \(R/M\) is a field. In other words,
\vspace{-0.8cm}
(\(\implies\)) Suppose \(M\) is a maximal ideal. To show that \(R/M\) is a field, we first realize that it is commutative since \(R\) is commutative. Thus we just need to show it is a divison ring.
Let \(a + M \in R/M\) where \(a \not\in M\). Since \(aR\) and \(M\) are both ideals, we have that \(aR + M\) is an ideal. Note that \(M \subset aR + M\), which implies that \(aR + M = R\). Therefore, there exists an element \(r \in R\) and \(m \in M\) such that \(ar + m = 1\).
Since \(ar = 1 - m\), consider \(r + R \in R/M\), and observe that
so that \(r + M\) is the desired inverse of \(a + M\). Since \(a \not\in M\), this shows that \(R/M\) is a division ring. Since it is commutative, it is a field.
(\(\impliedby\)) Now suppose that \(R/M\) is a field. It is a commutative division ring, so that by Lemma \ref{divison_ring_thm} we know that its only ideals are either \(0\) or \(R/M\). But by the Fouth Isomorphism Theorem, we know that these ideals of \(R/M\) correspond to \(M\) and \(R\). Thus there are no other ideals of \(R\) containing \(M\) other than \(M\) and \(R\), proving that \(M\) is maximal.
Let \(R\) be a commutative ring. Then an ideal \(P\) is said to be prime if \(P \ne R\) and if \(ab \in P\), then either \(a \in P\) or \(b \in P\). Furthermore, if there exists a \(p \in R\) such that \(Rp = \left< p \right>\) is a prime ideal, then \(p\) is said to be prime.
\textcolor{NavyBlue}{You may wonder why we would make this definition, since ideals tend to suck in elements of \(R\) (i.e., \(ri \in R\) for all \(i \in I\), \(r \in R\)). However, just because \(ab \in I\) does not mean \(a \in I\) or \(b \in I\). Consider for instance the ring \(4\ZZ\). Obviously, \(4 \in 4\ZZ\), but \(2\cdot2 = 4\) and yet \(2 \not\in 4\ZZ\).}
Prime ideals have a similar theorem that maximal ideals have.
Let \(R\) be a commutative ring with identity. Then an ideal \(P\) of \(R\) is prime if and only if \(R/P\) is an integeral domain. That is,
\vspace{-0.8cm}
(\(\implies\)) Suppose \(P\) is a prime ideal. To show that \(R/P\) is an integral domain, we must show that it has no zero divisor (as we already know it is commutative). Thus let \(r + P \in R/P\) where \(r \not\in P\). Suppose for the sake of contradiction that
for some \(r' + P \in R/P\) where \(r' \not\in P\) (i.e., that there are zero divisors). Then this implies that \(rr' \in P\). Since \(P\) is prime, we have that either \(r \in P\) for \(r' \in P\), which is a contradiction since our hypothesis was that \(r, r' \not\in p\). Therefore \(R/P\) is commutative and has no zero divisors, so it is an integral domain.
(\(\impliedby\)) Now suppose that \(R/P\) is an integral domain. Then
for any \(r\) and \(r' \not\in P\). In other words, if \(r, r' \not\in P\) then \(rr' \not\in P\). Taking the contrapositive of this statement, we have equivalently that if \(rr' \in P\) then \(r\) or \(r' \in P\) (notice how that "and" changed to an "or" upon negation) which proves that \(P\) is prime.
Finally, we can combine all of these theorems into one useful criterion for primeness.
Let \(R\) be a commutative ring with identity. Let \(I\) be an ideal. If \(I\) is maximal then \(I\) is prime. In other words,
\vspace{-0.8cm}
By Theorem 1.\ref{maximal_theorem}, if \(I\) is maximal then \(R/I\) is a field. But since \(R/M\) is a field, it is an integral domain. Hence by Theorem 1.\ref{prime_theorem}, we have that \(I\) is a prime ideal, which proves the theorem.
The corollaries of these theorems are immediate.
Let \(R\) and \(S\) be commutative rings with identity, and suppose \(\phi: R \to S\) be a surjective ring homomorphism. Then
- 1. If \(S\) is a field, then \(\ker(\phi)\) is a maximal ideal of \(R\).
- 2. If \(S\) is an integral domain then \(\ker(\phi)\) is a prime ideal of \(R\).
- 1. If \(\phi\) is surjective, then by the First Isomorphism Theorem \(R/\ker(\phi) = \im(\phi) = S\). Therefore \(R/\ker(\phi)\) is a field, so by Theorem 1.\ref{maximal_theorem} we have that \(\ker(\phi)\) is maximal.
- 2. In this case, we again have that \(R/\ker(\phi) = S\), so that \(R/\ker(\phi)\) is an integral domain. Therefore \(\ker(\phi)\) is a prime ideal of \(R\) by Theorem 1.\ref{prime_theorem}.