## The Quotient Ring

### Introduction

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

$$a\gamma=ab(1+x+\cdots+x^{n-1})$$

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.