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For a nonzero ring \(R\), we define:
- \(\{0\}\) to be the trivial ideal.
- \(R\) to be the improper ideal.
- Any other ideal is a proper non-trivial ideal.
Let \(R\) be a ring with unity. If an ideal \(I\) contains an unit, then \(I=R\).
Proof
Let \(u\) be a unit contained in \(I\). Then \(1=u^{-1}u\in I\). It follows that \(r=r1\in I\) for all \(r\in R\). ◼
A field contains no proper nontrival ideals.
Proof
Sine any nontrivial ideal of a field contains a unit, it must be the whole field.
Let \(R\) be a commutative ring with unity. A principal ideal \(Ra\) is equal to \(R\) iff \(a\) is a unit.
看看這！
Proof
Use this theorem.
A proper ideal of a ring \(R\) is a maximal ideal if no proper ideal \(N\) of \(R\) properly contains \(M\). That is, if \(N\) is an ideal such that \(M\subset N\subseteq R\), then \(N=R\).
Let \(p\) be a prime. Then \(p\Bbb Z\) is a maximal ideal of \(\Bbb Z\).
Proof
Let \(I\) be an ideal of \(\Bbb Z\) properly containing \(p\Bbb Z\).
We have
\[p = [\Bbb Z: p\Bbb Z] = [\Bbb Z:I][I:p\Bbb Z].\]Since \(p\Bbb Z\subset I\), \([I:p\Bbb Z]>1\). This implies that \([I:p\Bbb Z]=p\) for \(p\) is prime. Therefore, \(I=\Bbb Z\). ◼
Let \(R\) be a commutative ring with unity. Then \(M\) is a maximal ideal iff \(R/M\) is a field.
Check this.
Proof
\((\Rightarrow)\):
Suppose \(M\) is a maximal ideal. Since \(R/M\) is a commutative ring with the unity \(1+M\). It remains to show that every \(a+M\not=M\) is a unit.
Since \(a+M\not=M, a\not \in M\). Consider the canonical homomorphism \(\gamma: R\to R/M\) defined by \(\gamma(r) = r+M\). The preimage of \(\langle a+M \rangle\) is \(Ra\), and by this theorem, \(Ra+M\) is also an ideal. Moreover, since \(Ra+M\) contains \(M\) and \(M\) is maximal, \(Ra+M = R\). Therefore, there exist \(m\in M\) and \(r\in R\) such that \(ra+m=1\). It follows that \(ra=1-m \in 1+M\). Thus, \(a+M\) has a multiplicative inverse \(r+M\) in \(R/M\). ◼
\((\Leftarrow)\):
Suppose \(R/M\) is a field, and let \(N\) be an ideal of \(R\) properly containing \(M\). We want to show that \(N=R\).
By our assumption, there exists an element \(a\in N\backslash M\). Then \(a+M\not=M\) in \(R/M\), and thus \(a+M\) has a multiplicative inverse \(b+M\) in \(R/M\). This means that \(1-ab\in M\subset N\). Since \(N\) is an ideal, \(ab \in N\), and thus \(1=(1-ab)+ab\in N\). This implies \(N=R\). Therefore \(M\) is maximal. ◼
Alternative proof for \((\Leftarrow)\):
Since \(N\) is an ideal of \(R\) and \(N\) properly contains \(M\), with the canonical homomorphism \(\gamma\) defined above, \(N/M = \gamma(N)\) is a non-trivial ideal of \(\phi(R) = R/M\). To see why \(N/M\) is a non-trivial ideal of \(\phi(R)\), recall from this theorem, and thus we know the trivial ideal of \(\phi(R)\) is \(M\) itself. (Note that by that theorem, \(M\) is the kernel of \(\gamma\).) Since \(R/M\) is a field, the non-trivial ideal \(N/M\) must be the whole field \(R/M\), by this corollary. We conclude that \(N=R\). ◼
A proper ideal \(P\) in a commutative ring \(R\) is a prime ideal if \(ab\in P\) implies \(a\in P\) or \(b\in P\).
Why this is called prime? See the example below.
Let \(R=\Bbb Z\). If \(n=p\) is a prime and \(ab\in p\Bbb Z\), then \(p\mid ab\), which implies that \(p\mid a\) or \(p\mid b\). Thus, \(ab\in p\Bbb Z\) does imply \(a\in p\Bbb Z\) or \(b\in p\Bbb Z\), i.e., \(p\Bbb Z\) is a prime ideal of \(\Bbb Z\).
質因數！
Let \(R\) be a commutative ring with unity. Then \(P\) is a prime ideal of \(R\) iff \(R/P\) is an integral domain.
Check this.
Proof
Since \(R\) is a commutative ring with unity, its quotient ring \(R/P\) is also a commutative ring with unity. We can see that
\[\begin{align*} &R/P \text{ is an integral domain} \\ \iff &(a+P)(b+P)=0+P \implies (a+P=0+P) \lor (b+P=0+P) \\ \iff &ab\in P \implies (a\in P) \lor (b\in P) \\ \iff &P\text{ is a prime ideal.} \tag*{$\blacksquare$} \end{align*}\]Every maximal ideal in a commutative ring with unity is a prime ideal.
反之未必！The trivial ideal \(\{0\}\) of \(\Bbb Z\) is a prime ideal but not a maximal one.
由此可知，maximal 是比 prime 更強的性質。
Let \(R\) be a ring with unity. For all \(a\in R\), if \(\langle a\rangle\) is a prime ideal, then \(a\) is a prime in \(R\).
Proof
Since \(\langle a\rangle\) is a prime ideal, whenever \(xy\in \langle a\rangle\), either \(x\) or \(y\) is in \(\langle a\rangle\). And since all elements in \(\langle a\rangle\) can be represented as \(ra\) for some \(r\) in \(R\), we can conclude that \(a\) is a prime in \(R\). ◼
Let \(\Bbb F\) be a field. Then every ideal \(I\) in \(\Bbb F[x]\) is principal.
Proof
If \(I=\{0\}\), \(I=\langle 0\rangle\) is principal. Otherwise, let \(g(x)\) be a nonzero element of \(I\) of minimal degree. We claim that \(I=\langle g(x)\rangle\).
For any \(f(x)\in I\), using the division algoritm can yield \(f(x) = q(x)g(x) + r(x)\), where \(r(x)=0\) must hold, by the minimality of \(\deg g(x)\). Thus, \(f(x)=q(x)g(x)\in \langle g(x) \rangle\), and thus \(I\subseteq \langle g(x)\rangle\).
Conversely, since \(g(x)\in I\), we have \(\langle g(x)\rangle \subseteq I\), and hence \(I=\langle g(x)\rangle\). Therefore, \(I\) is a principal ideal. ◼
Every \(I\) on \(\Bbb Z\) is also principal!
An ideal \(\langle p(x)\rangle\) in \(\Bbb F[x]\) is maximal iff \(p(x)\) is irredcible over \(\Bbb F\).
Proof
\((\Rightarrow)\):
Let \(\langle p(x)\rangle\) be a maximal ideal. We need to prove that
If \(p(x)\) is a nonzero constant polynomial, then \(p(x)\) is a unit in \(\Bbb F[x]\) and \(\langle p(x) \rangle = \Bbb F[x]\), contradictiing to the assumption that \(\langle p(x)\rangle\) is maxiaml. Thus, \(p(x)\) is not a constant polynomial.
Now suppose \(p(x) = f(x)g(x)\). Then by this corollary, \(\langle p(x) \rangle \subset \langle f(x)\rangle\). Since \(\langle p(x)\rangle\) is maximal, either \(\langle f(x)\rangle=\Bbb F[x]\) or \(\langle f(x)\rangle = \langle p(x) \rangle\). For the former case, \(f(x)\) is a unit; for the latter case, \(f(x)\in \langle p(x)\rangle\) and \(f(x) = p(x)h(x)\) for some \(h(x) \in \Bbb F[x]\). Then, \(p(x) = f(x)g(x) = p(x)[h(x)g(x)]\), and thus \(h(x)g(x)=1\), implying \(g(x)\) is a unit.
\((\Leftarrow)\):
Use a similar argument as above.
The factor ring \(\Bbb F[x]/\langle p(x)\rangle\) is a field iff \(p(x)\) is irreducible over \(\Bbb F\).
By this theorem!
Check here.
Proof (use ideals)
Since \(p(x)\) is irreducible, by the previous theorem, \(\langle p(x)\rangle\) is maximal, which is also a prime ideal. If \(p(x)\mid r(x)s(x)\), then \(r(x)s(x)\in \langle p(x)\rangle\). Thus, \(r(x)\in \langle p(x)\rangle\) or \(s(x)\in \langle p(x)\rangle\), which implies the result. ◼