Definition 1 An ideal \( I \) of \( R \) is prime if the quotient \( R / I \) is an integral domain. It is maximal if \( R / I \) is a field.
Lemma 2 An ideal \( I \) is prime if, and only if, for every pair of elements \( r , s \) in \( R \) such that \( rs \) is in \( I \), either \( r \) is in \( I \) or s is in \( I \).
Proof.
- An ideal \( I \) of \( R \) is prime \( \implies \) \( R / I \) is an integral
domain \( \implies \forall a,b \in R/I, ab=0+I \implies a=0+I \lor b=0+I \)
Since \( \forall a,b \in R/I, \exists x,y \in R : a=x+I,b=y+I \). So, \( ab=xy+I=0+I \implies x+I=0+I \lor y+I=0+I \). Thus, \( \forall xy-0 \in I \implies x-0=x \in I \lor y \in I \).
- \( xy \in I \implies x \in I \lor y \in I \)
\( \forall x+I,y+I \in R/I, (x+I)(y+I)=xy+I=0+I \implies xy \in I \implies x+I=0+I \lor y+I =0+I \). Thus, \( R / I \) is an integral domain and \( I \) is prime.
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Lemma 3 The only ideals of a field are the zero ideal and the unit ideal.
Proof. Assume \( I \) is an ideal of the field \( F \) that has a non-zero element \( r \). By definition, \( \exists r^{-1} \in F : rr^{-1}=1\in I \). Since \( 1 \in I \), unit ideal must be a subset of \( I \). But every ideal is a subset of the unit ideal, so \( I \) must be equivalent to the unit ideal.
Clearly, zero ideal is also an ideal of \( F \).
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Lemma 4 An ideal \( I \) is maximal if, and only if, the only ideals of \( R \) containing \( I \) are \( I \) and the unit ideal.
Proof.
- \( I \) is maximal \( \implies \) \( R / I \) is a field
Assume there exist an ideal \( J \) of \( R \) such that \( I \subseteq J \subseteq R \). Note \( J/I \subseteq R/I \). Since \( J \) is an ideal of R, \( J/I \) is an ideal of \( R/I \). \( R / I \) is a field. Based on the previous lemma, \( J/I \) is zero ideal or unit ideal which means \( J=I \) or \( J=R. \)
- For any \( r \in R \setminus I \), \( I \) is an ideal of \( I+\langle r \rangle \). \( r \not\in I \), \( I+\langle r \rangle \) must be the unit ideal. Thus, \( 1 \in (I+\langle r \rangle) \). There exists \( t \in R, i \in I : i+rt=1 \). Thus, \( rt \equiv 1 \bmod I \) and \( r \) has an inverse in \( R/I \). Note, \( \forall i \in I, i\equiv 0 \bmod I \). Therefore, \( R / I \) is a field.
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