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Homomorphisms and Ideals

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Ring Homomorphism

A ring homomorphism is a map between rings which is compatible with addition, multiplication and maps the unit of to that of . A ring homomorphism which admits an inverse is an isomorphism. Equivalently, it is a bijective homomorphism.

Proposition

For every ring there is a unique homomorphism .

Proof We must have ; and then ( ones) for ; and . It is easy to check that this is multiplicative as well.

Substitution Principle II

Let be a ring homomorphism and . There is a unique ring homomorphism that agrees with on constant polynomials, and that maps to . More generally, given there is a unique homomorphism that agrees with on constants and maps to .

Prop .

Def The kernel and image of a ring homomorphism areThe image is a subring of because , , . The kernel is an additive subgroup of .

Ideals

Ideal

An ideal in a ring is a subset of such that

  • is closed under addition
  • If then for all .

Equivalently, it is a non-empty subset such that for we have for every . We write in such case.

Ideals to Rings are Normal Subgroups to Groups

Ideals are the ring-theoretic analogue of normal subgroups in group theory. Just as normal subgroups allow us to form quotient groups, ideals allow us to form quotient rings.

Generated Ideal

Let . The set is an ideal in called the ideal generated by . The notation is .

Principle Ideal

An ideal is principal if it is generated by one element. That is, there exists such that . It is denoted .

e.g. In every ideal is principal and takes the form . (Proof see theorem).

Product Ideal

For , define the product ideal as

Proposition

If are ideals so are and .

Quotient rings

Quotient Ring

Let be a ring and an ideal. Then, as an additive subgroup, is normal in and we can form the quotient with addition , zero , and inverse . There is a unique way to upgrade this abelian group to a ring. Define

Def Canonical Map There is a unique ring structure on such that the map given by is a ring homomorphism. The kernel of is .

e.g.

Thrm Mapping Property of Quotients

Let be a ring homomorphism with kernel and let be another ideal. Let be the canonical map. Then if ,there exists a unique ring homomorphism such that .

Thrm The First Isomorphism Theorem

Let be a ring homomorphism with kernel . If is surjective then is an isomorphism.

e.g. .

Thrm The Correspondence Theorem

Let be a surjective ring homomorphism with kernel . There is a bijective correspondence:

Misplaced & \{\text{Ideals of } R \text{ containing } K\} &\longleftrightarrow\{\text{Ideals of } R^\prime\} \\ I &\shortmid\!\longrightarrow \varphi(I) \\ I^\prime &\longleftarrow\!\shortmid \varphi^{-1}(I^\prime) \end{aligned} $$ If the ideal $I$ corresponds to the ideal $I^\prime$ then $R/I \cong R^\prime/I^\prime$. **Corollary** Let $R$ be a ring, suppose $I \triangleleft R$. Then the ideals of $R/I$ corresponds to ideals of $R$ containing $I$. ## Chinese Remainder Theorem > [!definition] Product Ring > > Given two rings $R_{1}$ and $R_{2}$, the product ring $R_{1}\times R_{2}$ is the ring with addition and multiplication being elementwise, defined as follows: $$(a,b)+(c,d)=(a+c,b+d),\quad (a,b)(c,d)=(ac,bd)$$ > [!theorem] Chinese Remainder Theorem > > Let $I, J \triangleleft R$ such that $I + J = R$. then $I J =I \cap J$ and $R/I J\cong R/I ×R/J$. *Proof* We first show that $IJ=I\cap J$. It is clear that $IJ\subseteq I\cap J$. Conversely, for all $a\in I\cap J$, note that since $R=I+J$, we have $1=x_{0}+y_{0}$ for some $x_{0}\in I,y_{0}\in J$. Then $a=x_{0}a + y_{0}a \in IJ$. Thus $I\cap J\subseteq IJ$. > [!remark] > > Given $I + J = R$, $IJ=I\cap J$ requires $R$ being a commutative unital ring, while $R/(I\cap J)\cong R/I\times R/J$ holds for any ring $R$. >

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