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# System prepended metadata

title: MTH 350 Homework 10
tags: [' homework', mth350]

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tags: mth350, homework
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# MTH 350 Homework 10

Instructions for this Homework set are on Blackboard where this is posted. The due date is on the class calendar, which is linked on Blackboard. 



## Problem 1

:::warning
This problem is an **independent** problem. As described in [the Syllabus](https://docs.google.com/document/d/1Rest_DodWnDy7Y8EVhp2lEOVnWcTEjLAGfWV2khuQQY/edit?usp=sharing) (top of page 14) this means that **on this problem, the only help you can get is from the professor.** 
:::

Prove the following **modified version of Theorem 2.3.1**.  This does not appear in the main problem list, but we've been using it a lot and so we should prove it. 

>Let $R$ be a commutative ring with identity. Then $R$ is an integral domain if and only if for all $a,b,c \in R$ with $c \neq 0$ and $ac = bc$, we have $a = b$. 

**Notes and hints:**
- This is an "if and only if" theorem. Remember that in these situations (a "biconditional statement") there are two directions of proof. Begin by writing down what you will prove in each direction; then for each direction, identify what you are going to assume and what you are going to prove. 
- Note that in each direction, you are making the global assumption that $R$ is a commutative ring with identity. 
- This theorem is modified from your textbook because the version in the text is false! The statement of Theorem 2.3.1 in the text just assumes that $R$ is a ring --- it does not assume that $R$ is commutative or that it has an identity. Therefore according to the text, any ring with the "cancellation" property (that $ac = bc$ implies $a=b$)  should be an integral domain. However, $2\mathbb{Z}$ is a ring that has this property, but $2 \mathbb{Z}$ is not an integral domain because it doesn't have an identity. So one direction of the theorem has a counterexample. The authorities have been notified. 
- I'm not entirely sure the "cancellation" property implies that $R$ is commutative, either, but I couldn't find a counterexample, so maybe that's correct. See Problem 2!


## Problem 2
:::success
This problem is a **collaborative** problem. This means that **you can work on it in a small group, as long as you stay within the bounds of academic honesty** laid out in [the Syllabus](https://docs.google.com/document/d/1Rest_DodWnDy7Y8EVhp2lEOVnWcTEjLAGfWV2khuQQY/edit?usp=sharing) (starting on page 13). 
:::

Work **exactly one** of the following: 

1. Prove or disprove: If $R$ is a ring with identity that has the "cancellation property" --- that is, for all $a,b,c \in R$ with $ac = bc$, then $a=b$ --- then $R$ is commutative. This is the issue I mentioned in the notes on Problem 1. 
2. Solve **Problem 2.29**: Let $R$ be a ring and $a \in R$. Then $\langle a \rangle = \langle ua \rangle$ where $u$ is any unit in $R$. 

:::info
More problems may be added as options here following class on April 7. But no more will be added after Friday April 8.
:::


## Submission instructions

Turn in a single PDF with: 

- Your solution for Problem 1
- Your solution for Problem 2

Remember to type up your work using $\LaTeX$, and to put a page break in between the solution for Problem 1 and the solution for Problem 2. 

Then upload your PDF to the "Homework 10" assignment area, and remember to hit **Submit**.

:::warning
**Reminder**: You will be given feedback on your work **only if your solution is a complete, good-faith efforts at correct work.** Work that is partial or insubstantial will result in the entire submission being returned to you with no comment, and you'll need to spend a revision to resubmit before getting any feedback on it.

However it is OK to submit only one of the two problems; just realize that the missing problem will be treated as a revision.
:::

 

**Handy $\LaTeX$ commands:** 

The symbols for the angle-brackets on principal ideals such as $\langle a \rangle$, are `\langle` and `\rangle`, short for "left angle" and "right angle". For example to typeset $\langle x^2 + 1 \rangle$ you would type 

```latex 
\langle x^2 + 1 \rangle
```

Make sure not to get lazy and just use the less-than and greater-than symbols from the keyboard, because those look janky. For example `< x^2 + 1 >` produces $< x^2 + 1 >$ which doesn't look as nice. 