Thursday, June 22, 2017

The Quotient Ring


Since moving my website and establishing a blog, I've posted things on science fiction, physics, and debunking conspiracy theories. What I haven't posted about yet is my own field of interest, namely commutative algebra. And what better place to start than with The Quotient Ring.

The Quotient Ring

While more generic rings (i.e. noncommutative rings) exist, I'm not really interested in them. So, when I talk about rings I always mean (unless otherwise stated) a commutative ring with unity.

Commutative rings are nice for a lot of reasons. An immediate benefit for us is that we can speak of ideals rather than left and right ideals. And, of course, having $1$ makes life a lot nicer.

Remember that for a commutative ring with unity $R$, a subset $I\subset R$ is an ideal of $R$ if, for all $x,y\in R$ the following properties hold.

  • $x+y\in I$ ($I$ is closed under addition)
  • $x\cdot y\in I$ ($I$ is closed under multiplication)
  • $-x\in I$ (every element in $I$ has an additive inverse)

It should be obvious that $(I,+)$ is a subgroup of $R$. If it isn't obvious, it would be a good exercise to prove that $(I,+)$ is indeed a subgroup of $R$.

We can now define the quotient group $R/I=\{I+x\mid x\in R\}$ which is simply the group of cosets of $I$ under addition, but we don't just want a group--we want a ring. So, we can define multiplication for any $x,y\in R/I$ as follows. $$(I+x)\cdot(I+y)=I+(xy)$$ Now we have the Quotient Ring $R/I$.

Units and Nilpotent Elements

There are a lot of results we could prove about the quotient ring, but the one I'm interested right now concerns the preservation of units under the map $\phi:R\to R/I$ when $I\subset nil(R)$. Remember that $nil(R)=\{x\in R\mid x^n=0, n\in \mathbb{Z}, n>0\}$ is the nilradical of $R$, and the elements $x\in nil(R)$ are called nilpotent elements. In particular, I want to look at what a unit in $R/I$ can tell us about elements in $R$. This is formalized in the proposition below.

Proposition. For a commutative ring with unity $R$ and $I\subset nil(R)$ consider the natural map $\phi:R\to R/I$. If $\phi(a)$ is a unit in $R/I$ then $a$ is a unit in $R$.

Proof. Since $a$ is a unit in $R/I$ that means we have some $b\in R$ such that $ab=1-x\in R/I$ with $x\in I$. Technically, we should write $\phi(a)\phi(b)$, but the notation becomes cumbersome and I think the meaning is clear enough we can simplify the notation.

Now, I know that for some $n$ we have $x^n=0$, and since I want to show that $a$ times something is $1$, I'm going to be careful about how I choose that something. So, I take $\gamma=b(1+x+\cdots+x^{n-1})$. Then


But, $ab=(1-x)$ so we have

$$(1-x)(1+x+\cdots+x^{n-1})\\ 1+x+\cdots+x^{n-1}-x-x^2-\cdots-x^{n-1}-x^n$$

We immediately notice that everything cancels out except $1-x^n$. Since $x^n=0$ we are left with

$$a\gamma=1\in R$$

Thus $a$ is a unit in $R$.    q.e.d.

There isn't anything Earth shattering here, just a very basic introduction to the quotient ring and (what I think is) a nice little result.

I hope you found this interesting and informative. I'm going to be doing a lot more math and physics on my blog, so if you have any requests or suggests let me know in the comments below.