Hi, today we will discuss some lemmas I felt are often used in Ring theory till now. I have also uploaded my notes made till now here. This is a revision post. Please refer to my notes for detailed explanations.

Contents

Integral domain and fields

  • We know that every field is an integral domain.
Every finite integral domain is a field
Since it is an ID, it is a commutative ring and unital. To show it is a field, we need to show every element has a multiplicative inverse. We use the same "FLT" trick. Let $a_1,\dots,a_n$ be the element of the ID. Say we want to show $a$ has an inverse. Consider $\{aa_1,\dots, aa_n\}$. Note that each element in the new set is distinct if not then $aa_i=aa_j\implies a(a_i-a_j)=0$ but that is not possible as ID. So done.

  • Note that we can embed any integral domain and make it a field. We can do it by making a structure called fractions fields. We denote it by $Q(D)$, where $D$ is the integral domain. It is defined by $Q(D)={a/b\text{ such that} (a,b)\in D\times D{0}}$. It is the smallest field containing $D$ injectively.

  • Note that if $D$ is an Euclidean domain, then $D$ is a PID. ( Euclidean algorithm essentially)

If we consider evaluation map, we get that $F[x]/ker\phi_{\alpha}\cong F[\alpha]$

  • So if $\alpha$ is a transcedental number, then $ker \phi{\alpha}$ is empty. So $F[x]\cong F[\alpha]$
  • Note that if $\alpha$ is algebraic, then $ker\phi_{\alpha}$ is ideal and is generated by one element called the minimal polynomial of $\alpha$. Note that the minimal polynomial is irreducible as $F[x]$ is an ID.

Irreducibble and maximal ideal

  • Note that in $\Bbb{Z}$, $\Bbb{Z}/\langle p \rangle$ is a field.
Suppose $D$ is a ID which is not a field. Then $a$ is irreducible iff $\langle a \rangle$ is a maximal ideal among principal ideals.
Since $a$ is irreducible in $D$, it is not a unit. Consider $\langle a \rangle$. It is a proper ideal as it is not a unit. Say $$\langle a \rangle\subset \langle b \rangle\implies b|a\implies a=bc\implies c\text{ is a unit or } \langle b \rangle= D.$$
  • So if $D$ is PID, then $a$ is is irreducible iff $\langle a \rangle$ is a maximal ideal.

Another lemma which is used often is the following. I have also seen it used in Lie algebras!

Suppose $A$ is a unital commutative ring and $I\unlhd A$. Then $\overline{J}$ is ideal of $A/I$ iff $\overline{J}=J/I$ where $J$ is ideal of $A$ containing $I$.
It is sort of like the correspondence theorem in group theory
Suppose $A$ is a unital commutative ring and $I\unlhd A$. Then $I$ is a maximal ideal iff $A/I$ is a field.
We know a field has only two ideals. So simply show $A/I$ has two ideals.
  • So we get that, if $A$ is a PID, then \(a \text{ is irreducible} \iff \langle a\rangle \text{ is a maximal ideal } \iff A/I \text{ is a field}.\)
So if $\alpha$ is algebraic over $F$ then $$F[x]/ker\phi_{\alpha}\cong F[\alpha]$$ is a field.

Noetherian rings

A ring is called Noetherian if there is no ascending infinite sequence of ideals.
Suppose $A$ is a unital commutative ring. Then $A$ is noetherian iff all the ideals are finitely generated.
Hence a $PID$ is noetherian
Also, note that if any ring is Noetherian, then any element can be written as products of irreducibles.

Primes and UFD

Let $D$ be an ID. Suppose every element of $D$ can be written as a product of irreducibles. Then $D$ is a UFD iff every irreducible is a prime.
- Say $p$ is irreducible. Then say $p|ab$. Then $ab=pd$. Express $a,b,d$ in irreducible form. Compare the factors, by uniqueness, there must be a $p$ in factors of $ab$. So $p|a$ or $p|b$. - If $p_1\dots p_m=q_1\dots q_n$ where $p_i,q_j$s are primes, then $p_1|q_{i_1}$ by prime, but they are irreducible too, so $p_1=q_{i_1}$. Cancel it and use induction.
  • So note that $p$ is prime if $\langle p\rangle$ is a prime ideal

So the following result is immediate.

Suppose $A$ is a unital commutative ring and $I$ is an ideal. Then $I$ is a prime ideal iff $A/I$ is Integral domain.
Note $$ ab\in I\iff ab+I=0+I\iff(a+I)(b+I)=(0+I).$$ So if $I$ is prime ideal, then $a+I$ or $b+I$ is $0+I$. So $A/I$ is ID. Similarly for other direction.
So if $I$ is a maximal ideal, then $A/I$ is a field. So $I$ is prime ideal.

Showing PID $\implies$ UFD

Suppose $D$ is a PID. Then every irreducible element is prime.
If $D$ is a PID, then for irreducible element $p$, $\langle p\rangle$ is maximal ideal, so it is a $\langle p\rangle$ prime ideal, so $p$ is prime.
Suppose $D$ is a ID. Then every prime is irreducible.
If $D$ is a ID, then say $p=ab$. So $p|ab\implies p|a$ or $p|b$. Then say $p|a$. Then $a=pa'$. So $p=pa'b\implies b$ is a unit. Done.
So if $D$ is a PID, then irreducible $=$ primes. But we know that $D$ is UFD if every element of $D$ can be written as a product of irreducibles. But we also show that PID are Noetherian.

So we finally proved \(\text{ ED}\implies \text{PID}\implies \text{ UFD}.\)

And done!