Topics in Algebra, Chapter 3.4, Part 3
There are no exercises from 3.3, so this section covers both 3.3 ("Homomorphisms") and 3.4 ("Ideals and Quotient Rings"). This page is split into parts so as to not get excessively long. **See also**: [part 1](/journal/topics-in-algebra-chapter-3-section-4a) (3.4.1 through 3.4.10) and [part 2](/journal/topics-in-algebra-chapter-3-section-4b) (3.4.11 through 3.4.16). Throughout, $R$ is a ring and $p$ is a prime integer. The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book. ### Herstein 3.4.17: With $a\in R$, let $r(a)=\{x\in R\mid ax=0\}$. Show that $r(a)$ is a right ideal of $R$. $r(a)$ is the set of elements which $a$ annihilates from the left. If $x,y\in r(a)$, then $a(x+y)=ax+ay=0$ so that $x+y\in r(a)$ and $r(a)$ is an additive subgroup of $R$. With $r\in R$, we also have $a(xr)=(ax)r=0$ so that $xr\in r(a)$. Thus $r(a)$ is a right ideal of $R$. ### Herstein 3.4.18: Let $L$ be a left ideal of $R$ and let $\lambda(L)=\{x\in R\mid xa=0$ for each $a\in L\}$. Show that $\lambda(L)$ is an ideal (two-sided) of $R$. $\lambda(L)$ is the set of elements which annihilate the entire left ideal $L$. If $x,y\in\lambda(L)$ and $a\in L$, then $(x+y)a=xa+ya=0$ so that $\lambda(L)$ is an additive subgroup of $R$. Let $r\in R$ and consider $(rx)a=r(xa)=0$, which shows $\lambda(L)$ to be a left ideal, and $(xr)a=x(ra)=xa'=0$, which shows $\lambda(L)$ to be a right ideal, with $a'=ra\in L$ because $L$ is a left ideal. Therefore $\lambda(L)$ is a two-sided ideal. ### Herstein 3.4.19*: Suppose $x^3=x$ for every $x\in R$. Prove that $R$ is commutative. This problem is pretty tricky, and I must give credit to [this page](http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/herstein) for filling in some gaps in my argument that I couldn't fill myself even after considerable effort. First observe that $2x=(2x)^3=8x^3=8x$ so that $6x=0$ for every $x\in R$. This is not a statement about characteristic, because we do not assume $R$ to be an integral domain. We do not even assume that $R$ is unital, so recall that $6x$ is simply shorthand for $x+x+x+x+x+x$. Consider $$x+x^2=(x+x^2)^3=x^3+3x^4+3x^5+x^6=4x+4x^2,$$ so that we have $3(x+x^2)=0$ for any $x\in R$. In particular, plugging in $x+y$ gives \begin{multline*} 0=3(x+y+(x+y)^2)=3(x+y+x^2+xy+yx+y^2)\\ =3(x+x^2)+3(y+y^2)+3(xy+yx)=3(xy+yx). \end{multline*} By the above comment, we can subtract $6yx=0$ from the equation for free, and this leaves us with $$3(xy-yx)=0.$$ The simplest example of a ring with $x^3=x$ for all $x\in R$ is $R=\mathbb{Z}/3\mathbb{Z}$ which has characteristic $3$, and we immediately see that this result is worthless in that case. Clearly we need an additional result to prove that $R$ is commutative even when we can't "cancel" the three. A natural thing to compute is $$x\pm y=(x\pm y)^3=x\pm y\pm x^2y\pm yx^2+xy^2+y^2x\pm xyx+yxy.$$ Subtracting the second ($-$) equation from the first ($+$) gives $2x^2y+2yx^2+xyx=0$. If we multiply that equation on the left by $x$ we find $$2xy+2xyx^2+2x^2yx=0,$$ whereas if we multiply by $x$ on the right instead, $$2x^2yx+2yx+2xyx^2=0.$$ Subtracting one equation from the other again, we see that $$2(xy-yx)=0.$$ Finally, subtracting that from the previous result, we arrive at the desired statement that $xy=yx$ for all $x,y\in R$, i.e. that $R$ is commutative. With a solution in hand, one still has to wonder why this problem is in section 3.4, ideals and quotient rings. It is worth noting that $\phi:R\to R$ given by $\phi(x)=3x$ is a ring homomorphism, in a not-entirely-trivial way. This may lead somewhere useful. ### Herstein 3.4.20: Let $R$ and $S$ be rings, with $R$ unital, and let $\phi:R\to S$ be a surjective ring homomorphism. Prove that $\phi(1)$ is the unit element of $S$. For any $s\in S$, there exists $r\in R$ such that $s=\phi(r)$, because $\phi$ is surjective. Then we also must have that $$s=\phi(1r)=\phi(1)\phi(r)=\phi(1)s$$ and that $$s=\phi(r1)=\phi(r)\phi(1)=s\phi(1)$$ so that $\phi(1)$ is the unit element of $S$. Note that $\phi(1)$ will be the unit element of the sub-ring $\phi(R)\subset S$ regardless, but if $\phi$ is not surjective then there may be elements of $S$ for which $\phi(1)$ does not act as a unit. ### Herstein 3.4.21: Let $R$ and $S$ be rings with $R$ unital and $S$ an integral domain. Let $\phi:R\to S$ be homomorphism such that $\ker\phi\ne R$. Prove that $\phi(1)$ is the unit element of $S$. Clearly there is a problem if $1\in\ker\phi$, so first let $r\in R$ be outside the kernel of $\phi$ and consider $$0\ne\phi(r)=\phi(1r)=\phi(1)\phi(r).$$ This shows that $1\not\in\ker\phi$. Now we also see from the above computation that $\phi(1)$ acts as a unit element for $\phi(R)$ (which is commutative by the definition of integral domain), and the question is whether it also acts as a unit element for the other elements of $S$. Again letting $r\not\in\ker\phi$ and letting $s\in S$ be arbitrary, we have that $\phi(r)s=\phi(r)\phi(1)s$ so that $$\phi(r)\Big(s-\phi(1)s\Big)=0.$$ By assumption, $\phi(r)\ne 0$, so, because $S$ is an integral domain, we conclude that $s=\phi(1)s$ for all $s\in S$.
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