LEC ALG3 2

Quotients

Definition 107.1(Definition).

Let $I\subseteq R$ and $(I,+)$ be an abelian group. $I$ is a left/right ideal if it is closed under left/right multiplication by elements of $R$.

Let $I$ be a subgroup of the abelian group $(R,+)$ of a ring $R$. Subgroups of abelian groups are automatically normal, so we have a quotient group $R/I$, whose elements are cosets $r+I$ of $I$. Further, we have a surjective group homomorphism

$$\pi :R\to R/I,r\mapsto r+I.$$

==What requirements should $I$ meet, in order to have a ring structure on $R/I$, such that $\pi$ becomes a ring homomorphism?== If $\pi$ is a ring homomorphism, there is only one way to define a ring structure on $R/I$:

$$(a+I)(b+I)=\pi (a)\pi (b)=\pi (ab)=ab+I.$$

Thus, there is only one sensible ring structure on $R/I$, given by $(a+I)(b+I)\equiv ab+I$ (note that if this operation is well defined, $R/I$ is, in fact, a ring). When is this operation well defined?

Claim 107.2(Claim).

The operation $(a+I)(b+I)\equiv ab+I$ on $R/I$ is well defined iff $I$ is an ideal of $R$.

Proof.

Assume the operation is well defined, making$R/I$ into a ring, and $\pi$ into a ring homomorphism. $\ker \pi =I$. The absorption properties are easily verified: for all $a\in I$ and $r\in R$,

$$\begin{array}{r}\pi (ra)=\pi (r)\pi (a)=\pi (r)\cdot 0=0,\\ \pi (ar)=\pi (a)\pi (r)=0\cdot \pi (r)=0.\end{array}$$

Thus, $I$ is an ideal.

Conversely, assume $I$ is an ideal. Let $\alpha ,\beta \in I$.

$$\begin{array}{rl}& (a+\alpha +I)(b+\beta +I)\\ & =(a+\alpha )(b+\beta )+I\\ & =ab+\alpha b+a\beta +\alpha \beta +I\\ & =ab+I,\end{array}$$

which proves that the operation is well defined.□

Thus, $R/I$ is a ring, in such a way that the canonical projection $\pi :R\to R/I$ is a ring homomorphism, iff $I$ is an ideal of $R$.

The mapping property of quotient groups provides the scaffolding for its analogue in $\text{Ring}$: the needed (group) homomorphism exists and is unique by the group theoretic theorem; verifying it is a ring homomorphism is immediate.

Theorem 107.3(Aluffi (2009) III.3.8).

Let $I$ be a two-sided ideal of a ring $R$. Then for every ring homomorphism $\phi :R\to S$ such that $I\subseteq \ker \phi$ there exists a unique ring homomorphism $\tilde{\phi }:R/I\to S$ so that the diagram

commutes.

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This allows for the canonical decomposition and the first isomorphism theorem for rings.

Correspondence and third isomorphism

Observe that the ideals of $R$ containing $I$ are in bijective correspondence with the ideals of the quotient $R/I$: we already know that the subgroups $J$ of $R$ containing $I$ are in bijective correspondence with the subgroups of $R/I$, viz., $J\leftrightarrow J/I$. It is easily verified that $J/I$ is an ideal of $R/I$ iff $J$ is an ideal of $R$.

Theorem 107.4(Correspondence theorem, Artin (2011) 11.4.3).

Let $\phi :R\to \mathcal{R}$ be a surjective homomorphism with kernel $K$. Let $I$ be an ideal of $R$ containing $K$ and $\mathcal{I}$ be an ideal of $\mathcal{R}$. Then,

1. $\phi (I)$ is an ideal of $\mathcal{R}$.
2. ${\phi }^{-1}(\mathcal{I})$ is an ideal of $R$, and it contains $K$.
3. $\phi ({\phi }^{-1}(\mathcal{I}))=\mathcal{I}$, and ${\phi }^{-1}(\phi (I))=I$.
4. If $\phi (I)=\mathcal{I}$, then $R/I\cong \mathcal{R}/\mathcal{I}$.

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As in the case of groups, quotients by corresponding ideals are isomorphic:

Theorem 107.5(Aluffi (2009) III.3.11).

Let $I$ be an ideal of a ring $R$, and let $J$ be an ideal of $R$ containing $I$. Then $J/I$ is an ideal of $R/I$, and

$$\frac{R/I}{J/I}\cong \frac{R}{J}.$$

Proof.

Since$I\subseteq J=\ker (R\to R/J)$, Thm 3 gives us an induced ring homomorphism $\phi :R/I\to R/J$. Explicitly, $\phi (r+I)=r+J$. Clearly, $\phi$ is surjective.

$$\begin{array}{rl}\ker \phi & =\{r+I|r+J=J\}\\ & =\{r+I|r\in J\}\\ & =J/I.\end{array}$$

Thus, $J/I$ is an ideal, and the stated isomorphism follows from the first isomorphism theorem.□

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Warning

$J/I$ is not a ring!

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Ideals

Algebra of ideals

Definition 107.6(Definition).

Let $R$ be a commutative ring with $1$. Let $I,J$ be ideals of $R$.

1. $I+J:=\{i+j|i\in I,j\in J\}$. This is a subring of $R$.
2. $IJ:=\{i_1{j}_1+\cdots +i_n{j}_n|i_k\in I,j_k\in J,n\ge 1\}$.

1. Clearly, $IJ\subseteq I\cap J$.
2. If $n=p_1^{a_1}\dots p_n^{a_n}$, then $n\mathbb{Z}=(p_1^{a_1}\mathbb{Z})\cap \cdots \cap (p_n^{a_n}\mathbb{Z})=(p_1^{a_1}\mathbb{Z})\cdot \cdots \cdot (p_n^{a_n}\mathbb{Z})$.

Definition 107.7(Definition).

Let $I$ be an ideal of a ring $R$. $X=\{a_i|a_i\in I\}\subseteq I$ is said to generate $I$ if every element of $I$ can be written as a finite sum $r_{i_1}{a}_{i_1}+\cdots +r_{i_k}{a}_{i_k}$, $r_i\in R$.

Proposition 107.8(Proposition).

1. If $I=(a_1,\dots ,a_n)$ and $J=(b_1,\dots ,b_k)$, then $I+J=(a_1,\dots ,a_n,b_1,\dots ,b_k)$ and $IJ=(a_i{b}_j|1\le i\le n,1\le j\le k)$.
2. $I(J_1+\cdots +J_k)=I{J}_1+\cdots +I{J}_k$.
3. $(IJ)K=I(JK)$.

[!Example]
Let $R=\mathbb{C}[x,y,z]$ or $\mathbb{R}[x,y,z]$ or $\mathbb{Q}[z,y,z]$. $I=(x^3-yz,y^2-xz,z^2-x^{2}y)$, $J=(x,y)$. Show that $I\cap J=(x^3-yz,y^2-xz)$.

Properties of ideals under homomorphisms

Proposition 107.9(Proposition).

1. The inverse image of an ideal under a homomorphism is an ideal.

2. The image of an ideal under a surjective homomorphism is an ideal.

Proof.

Let$\phi :R\to S$ be a homomorphism.
$(1)$ Let $I\subseteq S$ be an ideal. Let $a,b\in {\phi }^{-1}(I)$. Then, $\phi (a+b)=\phi (a)+\phi (b)\in I$, so $a+b\in {\phi }^{-1}(I)$. Next, $0=\phi (0)=\phi (a)+\phi (-a)$, so $\phi (-a)=-\phi (a)\in I$. Thus, $({\phi }^{-1}(I),+)$ is an abelian group. Let $r\in R$. Then, $\phi (ar)=\phi (a)\phi (r)\in I$, and $\phi (ra)=\phi (r)\phi (a)\in I$, so ${\phi }^{-1}(I)$ is closed under left and right multiplication by $R$.

$(2)$ Let $I\subseteq R$ be an ideal. $\phi (I)$ is clearly an abelian group. Let $(i+\ker \phi )\in \phi (I)$. Any element of $S$ may be expressed as $(s+\ker \phi )$. Since

$$(i+\ker \phi )(s+\ker \phi )=(is+\ker \phi )\in \phi (I),$$

$\phi (I)$ is closed under left and right multiplication by $S$.□

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Analogous to this proposition from group theory.

For any ring $R$ with ideal $I$, we have a natural surjective homomorphism $\pi :R\to R/I$: $r\mapsto r+I$. If $J$ is an ideal in $R$, then $\pi (J)(R/I)$ (the ideal generated by $\pi (J)$ in $R/I$) is equal to $\pi (J)$, and is the ideal $(J+I)/I$.

Proposition 107.10(Proposition).

Let $\phi :R\to S$ be a homomorphism. Let $I\subseteq R$ and $J\subseteq S$ be ideals such that $\phi (I)\subseteq J$. Then, $\overline{\phi }:R/I\to S/J$ is a homomorphism.

Proposition 107.11(Proposition).

Let $\phi :R\to S$. Let $I$ be an ideal in $R$. Let $\phi (I)\cdot S$ be the ideal in $S$ generated by $\phi (I)$. Then, ${\phi }^{-1}(\phi (I)\cdot S)$ is an ideal in $R$ containing $I$.

If $J$ is an ideal of $S$, $\phi ({\phi }^{-1}(J))\cdot S$ is an ideal in $S$ contained in $J$.

It follows that every ideal $J\subseteq \mathbb{Z}[x]$ contains $n\mathbb{Z}$ for some $n$ on considering the inclusion homomorphism $\mathbb{Z}\to \mathbb{Z}[x]$.

Principal ideals

Let $a\in R$ be any element of a ring. Then the subset $I=Ra$ of $R$ is a left-ideal of $R$. Indeed, for all $r\in R$ we have $rI=rRa\subseteq Ra$. Similarly, $aR$ is a right ideal.

Definition 107.12(Definition).

A principal ideal is an ideal that is generated by a single element.

1. A left principal ideal of $R$ is a subset of $R$ given by $Ra$ for some element $a$.
2. A right principal ideal of $R$ is a subset of $R$ given by $aR$.
3. A two-sided principal ideal of $R$ is a subset of $R$ given by $RaR=\left\{{\sum }_{i=1}^n{r}_{i}a{s}_i|r_1,s_1,\dots ,r_n,s_n\in R\right\}$ for some element $a$, namely, the set of all finite sums of elements of the form $ras$.

When $R$ is commutative, the three notions coincide and are denoted $(a)$. This is the principal ideal generated by $a$.

The zero-ideal $(0)=\{0\}$ and the whole ring $(1)=R$ are both principal ideals.

Lemma 107.13(Lemma).

If $\{I_{\alpha }{\}}_{\alpha \in A}$ is a family of ideals of a ring $R$, then the sum ${\sum }_{\alpha }{I}_{\alpha }$ is an ideal of $R$.

Definition 107.14(Definition).

If $a_{\alpha }$ is any collection of elements of a commutative ring $R$, then

$$(a_{\alpha }{)}_{\alpha \in A}:=\sum _{\alpha \in A}(a_{\alpha })$$

is the ideal generated by the elements $a_{\alpha }$.

In particular,

$$(a_1,\dots ,a_n)=(a_1)+\cdots +(a_n)$$

is the smallest ideal of $R$ containing $a_1,\dots ,a_n$; the elements of this ideal are of the form

$$r_1{a}_1+\cdots +r_n{a}_n$$

for $r_1,\dots ,r_n\in R$.

Definition 107.15(Definition).

An ideal $I$ of $R$ is finitely generated if $I=(a_1,\dots ,a_n)$ for some $a_1,\dots ,a_n\in R$.

Example 107.16(Example).

Let $R$ be a commutative ring, and let $a,b\in R$. Let $\overline{b}$ denote the class of $b$ in $R/(a)$. Then,

$$\frac{R/(a)}{(\overline{b})}\cong R/(a,b).$$

This follows from Thm 5, since

$$\begin{array}{rl}(\overline{b})& =(b+(a))(R/(a))\\ & =\{br+(a)|r\in R\}\\ & =\frac{(a,b)}{(a)}.\end{array}$$

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Definition 107.17(Definition).

1. A commutative ring $R$ is Noetherian if every ideal of $R$ is finitely generated.
2. An integral domain $R$ is a PID (Principal Ideal Domain) if every ideal of $R$ is principal.

$\mathbb{Z}$ is clearly a PID.

Theorem 107.18(Theorem).

If $k$ is a field, then $k[x]$ is a PID.

Proof.

Let$I\subseteq k[x]$ be any ideal. If $I=(0)$, then $I$ is principal. Otherwise, let $f(x)$ be the minimal degree monic polynomial in $I$ ¹. Let $g(x)$ be any polynomial in $I$. We can express $g(x)$ as

$$g(x)=f(x)p(x)+r(x),$$

where $p(x)\in k[x]$ and $r(x)$ has degree less than $f(x)$. But, $f(x)$ has minimal degree, so $r(x)=0$ ². It follows that $I=(f(x))$.□

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$\mathbb{Z}[x]$, however, is not a PID: the ideal $(2,x)$ cannot be generated by a single element.

Footnotes

1. Minimal degree monic polynomials in an ideal, if they exist, are unique in polynomial rings. They must exist in this case since $k$ is a field. See van Leeuwen (2013) for more. ↩

2. With the convention that the degree of the polynomial $0$ is $-\infty$, $\text{deg}r(x)<\text{deg}f(x)$ is satisfied by $r(x)=0$ for all nonzero $f(x)$. ↩

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

Artin, M. (2011). Algebra (2. ed). Pearson Education, Prentice Hall.

van Leeuwen, M. (2013). Answer to “a Principal Ideal Contains a Monic Polynomial of Least Degree n.” In Mathematics Stack Exchange. https://math.stackexchange.com/a/538125/1677240