# Abstract Nonsense

## Definition and Basics of Ideals

Point of Post: In this post we define left, right, and two-sided ideals.

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Motivation

As was alluded to when we discussed subrings, subrings themselves aren’t the important thing when discussing rings, in the same way that it’s not subgroups but normal subgroups that play a pivotal role in group theory. In particular, what we’d eventually like to do is to define the quotient objects for rings–the ‘quotient rings’. But, as normal we can’t quotient out by any old subset, or even any old subring, of a ring. It turns out, much the same as the case for groups that what defines these special subrings is that they are the kernels of ring homomorphisms. But, this is all in the future. What we shall show in this post is that kernels of ring homomorphisms satisfy a certain ‘absorption’ property and then define these analogs of normal subgroups as being subrings that satisfy this property

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Ideals

We begin by noting that kernels of ring homomorphisms satisfy a certain absorption principle which is much stronger than being multiplicatively closed. Namely:

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Theorem: Let $R$ and $S$ be rings and $f:R\to S$ a ring homomorphism. Then, for any $r\in R$ and any $s\in\ker f$ it’s true that $rs,sr\in\ker f$.

Proof: We merely note that $f(rs)=f(r)f(s)=f(r)0=0$ and $f(sr)=f(s)f(r)=0f(r)=0$ and so $rs,sr\in\ker f$ as claimed. $\blacksquare$

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We abstactify this property and call a subring $I$ of a ring $R$ an ideal if $rs,sr\in I$ whenever $r\in R$ and $s\in I$. We generally denote ideals by lower case Fraktur letters such as $\mathfrak{a,b,c,d,e,f},\cdots$. More generally we call a subring $J$ of a ring $R$ a left ideal if $rs\in J$ whenever $r\in R$ and $s\in J$, right ideals are defined similarly. When convenient we shall call an ideal a two-sided ideal.

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Let $A\subseteq R$. We define $\displaystyle RA=\left\{r_1a_1+\cdots+a_nr_n:a_k\in A,\; r_k\in R\text{ and }n\in\mathbb{N}\right\}$ (i.e. the set of all finite combinations of elements of $A$ with coefficients in $R$). We define $AR$ similarly as the set $\left\{a_1r_1+\cdots+a_nr_n:a_k\in A,\; r_k\in R\text{ and }n\in\mathbb{N}\right\}$ as well as $RAR=\left\{r_1a_1s_1+\cdots+r_na_ns_n:a_k\in A,\; r_k,s_k\in R\text{ and }n\in\mathbb{N}\right\}$ (this last one evidently being equal to $R(AR)=(AR)R$). If $A=\{a\}$ (i.e. if $A$ is a singleton) we denote $RA,AR$, and $RAR$ as $Ra,aR$, and $RaR$ respectively.

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For any ring $R$ there are always two ideals, namely $\{0\}$ and $R$. We call these the trivial ideals of $R$. An ideal $\mathfrak{a}$ is proper if $\mathfrak{a}\subsetneq R$, a proper left and right ideal are defined similarly.

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There are some fundamental facts about ideals, in particular:

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Theorem: Let $R$ and $S$ be rings and $f\in\text{Hom}(R,S)$, then $\ker f$ is an ideal of $R$.

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Theorem: Let $R$ be a ring and $a\in R$, then $Ra$, $aR$, and $RaR$ is a left, right, and two-sided ideal respectively.

Proof: We merely note that $R(Ra)=\left\{r(sa):r\in R,\text{ and }s\in R\right\}\subseteq\left\{ra:r\in R\right\}=Ra$ and thus $Ra$ is a left ideal as claimed. The other two cases are done similarly.

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Theorem: Let $R$ be a unital ring and $J$ a left or right ideal. Then, $J=R$ if and only if $J$ contains a unit.

Proof: Evidently if $J=R$ then $1_R\in J$. Conversely, if $u\in J\cap R^\times$ then for any $r\in R$ one has that $r=ru^{-1}u\in J$ or $r=uu^{-1}r\in J$ depending on whether $J$ is a left or right ideal, either way $r\in J$ and so the conclusion follows. $\blacksquare$

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Corollary: Let $R$ be a ring and $\mathfrak{a}$ an ideal. Then, $\mathfrak{a}=R$ if and only if $\mathfrak{a}$ contains a unit.

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Theorem: Let $R$ be a unital ring. Then, $R$ is a division alebra if and only if the only left ideals of $R$ are the trivial ones.

Proof:

Lemma: Let $(M,\cdot,e)$ be a monoid where every element $m\in M$ has a left inverse $m^\ast$. Then, every element of $M$ has a two-sided inverse $m^{-1}$ and $m^{-1}=m^\ast$–in other words $(M,\cdot,e)$ is a group.

Proof: Let $m\in M$ be arbitrary. Since $m^\ast m=e$ it suffices to prove that $mm^\ast=e$. To do this merely note that

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$mm^\ast=emm^\ast=\left(m^\ast\right)^\ast m^\ast mm^\ast=\left(m^\ast\right)^\ast em^\ast=\left(m^\ast\right)^\ast m^\ast=e$

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So $m^\ast$ is a two-sided inverse of $m$. Since $m$ was arbitrary the conclusion follows. $\blacksquare$

Suppose first that $R$ has no non-trivial left ideals. Since $R^\times$ is a multiplicative monoid we see from the lemma it suffices to show that every element of $R-\{0\}$ is left invertible. So, let $r\in R-\{0\}$ be arbitrary. We know from the above theorems that $Rr$ is a left ideal which is non-zero (since $1_Rr=r\in rR$) and thus by assumption $rR=R$. But, in particular this means that $1_R\in R=Rr$ and so there exists some $s\in R$ for which $sr=1_R$, and thus $r$ is left invertible. Since $r$ was arbitrary the conclusion follows.

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Conversely, suppose that $R$ is a division ring and let $J$ be a non-zero left ideal. Then, by assumption there exists some $x\in J\cap(R-\{0\})$ but by definition $R-\{0\}=R^\times$ and thus $J$ contains a unit, so by the above theorem we have that $J=R$. $\blacksquare$

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Corollary: Let $R$ be a commutative unital ring. Then, $R$ is a field if and only if all ideals are trivial.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

June 21, 2011 -

1. R^\times is a division ring this term is not true

I assume you meant the “Conversely, suppose that $R^\times$ is a division ring..” in the last proof? If so, it was indeed a typo, it clearly should have been “Conversely, suppose that $R$ is a division ring…”. It’s fixed now! Thank you very much for pointing that out!

Best,
Alex

Comment by Alex Youcis | June 21, 2011 | Reply

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