View source for Midterm 1 Notes:Math 361, Spring 2014

==Types of Ideals==
	<p>
	Once we know how to create rings, we want to know how to create specific types of rings.  Specifically, how do we create integral domains and fields?
	<br/>
	Imagine that \(R/I\) <em>was</em> an integral domain. What does that imply?
	<br/>
	Well, we can't have any zero divisors. So given \(a + I \neq 0 + I\) and \(b + I \neq 0 + I), then \(ab + I \neq 0 + I\). But what it really means, if \(a +I \neq 0+ I\), is that \(a-0 \not \in I\) (by the definition of equality). So \(a \not \in I\). And \(b \not \in I\) and \(ab \not \in \).
	<br/>
	So what we've shown, really, is that if we want to get an integral domain from \(R/I\), we need \(ab\in I\) to imply either \(a\in I\) or \(b \in I\).
	<br/>
	I'm not going to prove this next part, but this is an if and only if relationship. So if \(I\) is any ideal with this property, then \(R/I\) is an integral domain, and if \(R/I\) is an integral domain, then \(I\) has this property. We call \(I\) a <em>prime ideal</em>.
	<br/>
	Fields are more complicated, and honestly hard to explain so I'm just going to say how to get them and not try to explain why it works.
	<br/>
	If \(I\) is an ideal, then\(I\) is a maximal ideal if:
	<br/>
	-\(I\) is not all of \(R\)
	<br/>
	-and \(I\) isn't contained by any other ideal except \(R\). Basically there's no way to make \(I\) bigger without either making it all of \(R\) or not an ideal.
	<br/>
	If \(I\) is a maximal ideal, then \(R/I\) is a field. If \(R/I\) is a field, then \(I\) is maximal.
	</p>

===More Theorems===
		<p>
		In \(F[x]\), the prime ideals and the maximal ideals are the same.
		<br/>
		In \(F[x]\), the maximal ideals/ prime ideals are the ideals generated by irreducible polynomials.
		<br/>
		So if we have \(\mathbb{Q}[x]\) and \(x^2-2\), we get a field from \(\mathbb{Q}[x]/\langle x^2-2\rangle\).
		</p>