Midterm 1 Notes:Math 361, Spring 2014

Math 361 Midterm 1 Notes

Rings and Fields

Brief review: a ring is a set \(R\) with two operations, \(+\) and \(*\). \(R\) is an abelian group with respect to \(+\), and \(*\) is associative. Lots of useful things are rings - integers, rational numbers, real numbers, complex numbers, matrices, and polynomials.

Anyway, this is algebra, so we need to be able to compare structures with homomorphisms. Since a ring is an abelian group with an extra operation, it makes sense for the ring homomorphisms to be group homomorphisms with some extra structure. So, if \(R\) and \(S\) are rings, and we have \(\phi:R \rightarrow S\), then we want \(\phi\) to have:

$$ \phi(0_R) = 0_S\\ \phi(a+b) = \phi(a)+\phi(b) $$

These are just the requirements for a group homomorphism. Things on the left hand side are in \(R\), things on the right hand side are in \(S\). To extend \(\phi\) to be a ring homomorphism, we're just going to make it preserve \(\cdot\) as well: $$ \phi(a * b) = \phi(a) * \phi(b) $$

The definitions of epimorphism, monomorphism, and isomorphism extend to the rings as you would expect (surjection, injection, bijection respectively).

When we get homomorphisms, we always want to ask what kinds of things can be kernels. When we did this for groups, the answer was normal subgroups, and we ended up using them to build quotient groups. We're going to do something similar with rings.

Let's say we have a ring homomorphism \(\phi:R \rightarrow S\). Let's say that the kernel of \(\phi\) is the set \(I\). What can we say about \(I\).

Well, \(\phi\) is a group homomorphism, so we know that \(0\in I\), and that \(I\) is a normal subgroup under addition. But \(R\) is commutative - every subgroup is normal, so \(I\) is a subgroup under addition. Namely, \(I\) is closed under addition, i.e. $$ a,b \in I \rightarrow a+b \ in I $$

This makes sense. If \(\phi(a) = 0\) and \(\phi(b) = 0\) then the \(\phi(a+b) = \phi(a) + \phi(b) = 0 + 0 = 0\).

What about multiplication. Well, obviously \(I\) is going to be closed under multiplication - if \(\phi(a) = 0\) and \(\phi(b)=0\), then \(\phi(ab) = \phi(a)\phi(b) = 0*0 = 0\). So now we can say that \(I\) is at least a subring of \(R\).

But we can say more. In rings, anything times 0 is 0. So let's take \(c \in R\) - any element of the entire ring - and \(a \in I\) - an element in the kernel of \(\phi\). Where is \(ca\)? Well: $$ \phi(ca) = \phi(c)\phi(a) = \phi(c) * 0 = 0 $$

So not only is \(I\) a subring, but it absorbs products. If we take something in the kernel, multiply it by anything else, we get something in the kernel.

These are actually the only properties kernels need - contain 0, closure under addition, and absorb multiplication. Any subset of \(R\) that has these properties is a ideal of \(R\).

So ideals are the equivalent of normal subgroups for rings. With normal subgroups, we created quotient groups. We're going to use ideals to create quotient rings, and in basically the same way.

We're going to mod out by an ideal. Ideals are normal subgroups, though, so really we just need to extend our definition of quotient groups.

So we're going to create cosets. I'll work with \(\mathbb{Z}\) as an example ring, and \(\langle 5 \rangle\) (the multiples of 5) as an ideal (I promise this is an ideal.) We already know how to create a quotient group using \(\mathbb{Z}\) and \(\langle 5 \rangle\), so we do that and get things of the form: $$ n + \langle 5 \rangle $$

We know how to add cosets of ideals. We also know how to tell if two ideals are equal: $$ n + \langle 5 \rangle = m + \langle 5 \rangle \Longleftrightarrow n-m \in \langle 5 \rangle $$

So \(4 + \langle 5 \rangle = 14 + \langle 5 \rangle\) because \(14 - 4 = 10 = 2*5 \in \langle 5 \rangle\).
But this is still just a quotient group. To get a quotient ring, we need some way to multiply. Fortunately, multiplying is easy: $$ n + \langle 5 \rangle * m + \langle 5 \rangle = nm + \langle 5 \rangle $$ Just make sure to reduce $mn$ mod \(\langle 5 \rangle\) at the end.
To reiterate: We get a ring \(R\). \(I\) is an ideal of \(R\). That means that \(I\) is a subring of \(R\) that absorbs products, or that \(I\) is the kernel of a ring homomorphism \(\phi:R \rightarrow S\).
Given such an \(R\) and \(I\), we can form the quotient ring \(R/I\), which consists of the cosets \(n + I\), where \(n\) is any element in \(R\). We need to define equality of elements of \(R/I\), addition in \(R/I\), and multiplication in \(R/I\). $$ a + I = b + I \Longleftrightarrow a-b \in I\\ (a+I) + (b+I) = (a+b + I)\\ (a+I)(b+I) = (ab + I) $$

Remember how we had the fundamental theorem on group homomorphisms? The image of a group homomorphism is isomorphic to the domain modulo the kernel? Well turns out the same thing is true for rings.

To give a little more detail: we have a ring homomorphism \(\phi:R \rightarrow S\). The kernel of \(\phi\) is some ideal \(I\). We can create a monomorphism \(\psi:R/I \rightarrow S\). Furthermore, the image of \(\psi\) is equal to the image of \(\phi\).

When we send \(R\) into \(S\), the image of \(R\) is some ring. Maybe it's \(R\), maybe it's smaller, but in some way we compress \(R\) into a subring of \(S\). What this theorem says is that the new ring is equal to \(R/I\), to the original ring modulo the kernel of the homomorphism.

Some Useful Theorems

The image of a ring homomorphism is a ring.

Imagine that we have two ideals, \(I\) and \(J\), and that \(I \subset J\). Then \(\pi[J]\) is an ideal of \(R/I\). So when we mod \(J\) by \(I\), we still get an ideal.

The opposite is also true. If \(K\) is an ideal of \(R/I\), then \(\pi^{-1}[K]\) is an ideal of \(R\), and \(I \subset J\).

To explain those last two - let \(J\) be an ideal of \(R\) that contains \(I\). Then when we compress \(J\) using \(\pi: R \rightarrow R/I\), the result is still an ideal. \(\pi[J]\) will contain \(0+I\), because it contained all of \(I\), and \(I\) is what gets compressed to \(0+I\). Also, since \(\pi\) is a homomorphism, it will preserve addition and multiplication, so \(\pi[J]\) will still be closed under addition and absorb multiplication. I don't feel like explaining why it works the other way, but it should be pretty clear.

If an ideal \(I\) contains a unit, it is all of \(R\). Let's say \(u \in R\) is a unit, and \(u\) is in \(I\). By the definition of \(u\), there is an element \(v \in R\) such that \(uv =1\). But \(I\) absorbs multiplication, so \(uv = 1 \in I\). We can extend this to say that everything is in \(I\). If \(x\) is any element of \(R\), then \(1x\) must be in \(I\), because 1 is in \(I\) and \(I\) absorbs multiplication. But \(1x = x\), so every \(x\) is in \(I\), and \(I\) must be all of \(R\).

All fields have exactly two ideals - the improper ideal and the zero ideal. Every element of a field,except zero, is a unit. If an ideal contains a unit, it is the improper ideal. So the only way to construct an ideal in a field is by taking just zero (the zero ideal), or by taking everything (the whole field/ the improper ideal).

As an extension of this - for any field there are really only two homomorphisms. This is because there are only two possible kernels, because there are only two possible ideals. If we take the kernel to be the zero ideal, then we get a monomorphism, and we inject \(F\) into some other field. If we take the kernel to be the improper ideal, then we compress \(F\) completely, and we get the zero ideal.

Types of Ideals

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?
Imagine that \(R/I\) was an integral domain. What does that imply?
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 \(ab \not \in \).
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 \(b \in I\).
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 prime ideal.
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.
If \(I\) is an ideal, then\(I\) is a maximal ideal if:
-\(I\) is not all of \(R\)
-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.
If \(I\) is a maximal ideal, then \(R/I\) is a field. If \(R/I\) is a field, then \(I\) is maximal.

More Theorems

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

Why This All Matters

We want to be able to solve any polynomial, i.e. we want to be able to find all the roots of a polynomial. If we get a polynomial in \(F[x]\), those roots don't necessarily exist in \(F\). So our goal is to make a new field that does contain the roots.
To be more specific, given a polynomial \(p\ \in F[x]\), we want to find all \(a \in F\) such that \(p(a) = 0\). Alternatively, we want to factor \(p\) as a product of linear polynomials \(x-a\).
This is not necessarily possible in \(F\). We want to find another field \(E\) where it is.
First we need to define what it means to extend a field. Hence, field extensions.
A field extension is a triple \((F,E,\iota)\). \(F\) and \(E\) are fields, and \(\iota:F\rightarrow E\) is a monomorphism. So there is a copy of \(F\) in \(E\).
This also means that \(F[x]\) exists in \(E[x]\). So we can take our original polynomial \(p\in F[x]\) and make a new polynomial in \(q \in E[x]\). We want to make sure these polynomials are equivalent, in the sense that all the roots of \(p\) are roots of \(q\). Turns out they are. Because \(\iota\) is a monomorphism.
Also, \(F\) is called the base field and \(E\) is called the extension field.

Finding the Right Extension Field

The first question is whether or not a useful extension field even exists. The answer is yes. If we have a polynomial \(p \in F[x]\), we can always find a extension field \(E\) such that \(p\) has a root in \(E\). This is Kronecker's Theorem.
You might be thinking that Kronecker's Theorem only guarantees the existence of one root. Can we find a field where all the roots exist?
Again, yes. Remember that if \(p(a) = 0\), then \(x-a\) divides \(p\). So let's say we found a field \(E\), and we found a root \(a \in E\). That means that we can divide \(p\) by \(x-a\). When we do this, we'll get a new polynomial. But by Kronecker's theorem, some field has to exist that contains a root of this new polynomial too. And obviously any root of the new polynomial is a root of \(p\), because the new polynomial is a factor of \(p\). So when we find a root for the new polynomial, we also find a root for \(p\). And of course we can iterate this process. Eventually we will find a field \(K\) where we can express \(p\) as a product of linear polynomials, i.e. \(x-a_1*x-a_2*\ldots*x-a_n\).
The next question is: can we actually find an extension field? Kronecker's Theorem says that a useful extension field exists, so can we find it? Yes. Yes we can.
Specifically, if we have an irreducible polynomial \(p \in F[x]\), then \(\langle p\rangle\) is maximal over \(F[x]\). That means that \(F[x]/\langle p \rangle\) is a field. Turns out that \(p\) has a root over this field. Specifically, the element \(x + \langle p \rangle\) is a root of \(p\) over this field.
There is a very easy way to get confused here. \(F[x]/\langle p \rangle\) is a field. Its elements are cosets of \(\langle p \rangle\). So we take a polynomial \(q \in F[x]\), and we create \(q + \langle p \rangle\).
For example, \(x^2-2\) is irreducible over \(\mathbb{Q}\). So we can create the field \(\mathbb{Q}[x]/\langle x^2-2 \rangle\). One element of this field is \(x-4 + \langle x^2-2 \rangle\).
Now, that element looks an awful lot like a polynomial. But it isn't. It's a coset.
Computation in this field is tedious. You have to do polynomial arithmetic, and then mod out by the ideal \(\langle x^2-2\rangle\). I will give a very brief example.
Two polynomials, \(x-4 +\langle x^2-2 \rangle) and \(x+5 + \langle x^2-2 \rangle\). We're going to multiply them.
The rule for coset multiplication is \((a+I)(b+I) = ab + I\). In our example, \(a = x+4\) and \(b = x+5\) and \(I = \langle x^2-2 \rangle\). So we need to multiply \(x+4\) by \(x+5\). This is just normal polynomial arithmetic: $$ (x+4)(x+5) = x^2 + 9x + 20 $$ Now we have to reduce modulo \(x^2 -2\). This is exactly the same as reducing modulo an integer - divide by the integer and take the remainder. So we divide \(x^2 + 9x + 20\) by \(x^2 -2\). This gives us \(9x + 22\). So our final answer is: $$ (x+4 + \langle x^2-2 \rangle)(x+5 + \langle x^2-2 \rangle) = 9x + 22 + \langle x^2-2 \rangle $$ There's a trick for doing this. What exactly is \(x^2 -2 + \langle x^2-2 \rangle\)? Well, it's \(0 + \langle x^2-2 \rangle\). But \(x^2 - 2 + \langle x^2-2 \rangle\) is really just \((x^2 + \langle x^2-2 \rangle\) - (2 + \langle x^2-2 \rangle)\). So, in this field, \(x^2-2 = 0\), meaning \(x^2 = 2\). So instead of actually doing out the division and finding the remainder, we can just look at \(x^2 + 9x + 20\), and replace every \(x^2\) by a 2. This gives us \(2 + 9x + 20 = 22\). (Note that I'm dropping the \(+\langle x^2-2 \rangle\) for the sake of my laziness.)
Here's where it gets confusing. We can take this field, and we can create polynomials from the elements of this field. So we can take \(\mathbb{Q}[x]/\langle x^2-2\rangle\), and we can create the polynomial ring: $$ \frac{\mathbb{Q}[x]}{\langle x^2-2 \rangle}[x] $$ So now our cosets become coefficients of polynomials. For example: $$ (x-4 + \langle x^2-2\rangle) + (3x-10 + \langle x^2-2 \rangle)x + (x-5 + \langle x^2-2 \rangle)x^2 $$ is an element of this field. To make this horrifying monstrosity slightly more comprehensible, we're going to rename the coset \(x + \langle x^2-2 \rangle\) to be \(\alpha + \langle x^2 -2 \rangle\). So at least now we only have to deal with one set of 'x's.
I am not going to do a computation in this polynomial ring. But I will give a possible problem. Calculate: $$ ((\alpha -1 + \langle x^2-2 \rangle) + (\alpha+5 + \langle x^2-2 \rangle)x^2)) * ((\alpha + 2 + \langle x^2-2 \rangle)x $$ This is just multiplying a quadratic polynomial by a linear polynomial. If you know how to do this, I would suggest you don't bother. If it is not immediately obvious how to calculate this, you should try it.

Theorems

If \(p\) is linear, then \(F[x] / \langle p \rangle \cong F\). Modding out by a linear polynomial does nothing.
Given a field \(F[x]/\langle p \rangle\), where \(p\) is irreducible. the element \(\alpha = x+ \langle p \rangle\) is a root of \(p\). This also means that \(x - \alpha\) is a divisor of \(p\).
To elaborate on that last point - if we have an irreducible polynomial \(p \in F[x]\), we can find a root of \(p\) in the field \(F[x]/\langle p \rangle\).

An Example

Let's say we have a polynomial \(p =x^5-9x^4+20x^3-2x^2+18x-40\), over \(\mathbb{Q}\). We want to find all of its roots. First we find its roots over \(\mathbb{Q}\). These are 5 and 4. So we didn't find all of the roots, meaning we're going to need to extend our field. In order to extend our field, we're going to need an irreducible polynomial. Specifically, we need the irreducible factors of \(p\). \(p\) itself has two linear factors, \(x-4\) and \(x-5\).(Assuming Wolfram Alpha didn't lie to me.) (Assuming Wolfram Alpha didn't lie to me.) When we \(p\) by both of those factors, we get the polynomial \(x^3 - 2\). This new polynomial is irreducible over \(\mathbb{Q}\), meaning we can use it to create an extension field of \(\mathbb{Q}\). So we get, as an extension field, \(\mathbb{Q}/\langle x^3 -2\rangle\).
But we're not done. We know this new field contains at least one root of \(p\). Does it contain all of them?
To find out, we need to do more. We've been given a polynomial over a field, we need to find a root. This is the same problem we started with, except instead of working with \(p\), we're working with a factor of \(p\). We can just do what we did before, only we have a smaller polynomial. We'll get an even smaller polynomial as a result, and polynomials can only get so small, so eventually we'll have found all the roots.
We know that \(x - \alpha\) is a factor of \(x^3 -2\) in \(\mathbb{Q}/\langle x^3 -2 \rangle\). So, like we did before, we divide our starting polynomial by its linear factors. This gives us \(x^2 + \alpha x + \alpha^2\). Does this have any roots in \(\mathbb{Q}[x]/\langle x^3 -2 \rangle\)? No, it doesn't. I can't prove it, but it doesn't. So we have to extend our field again, by modding out by \(\langle x^3 + \alpha x + \alpha^2\). I am not going to demonstrate this - the notation is just too crazy.

Theorems

Not a theorem, a definition. A monic polynomial is a polynomial whose leading coefficient is 1. \(x^2 + 4\) is monic. \(5x^2 + 20\) is not.
Given a polynomial \(p\), we can find a polynomial \(q\) that is monic and generates \(\langle p \rangle\). So we have \(5x^2 + 20\). Well, if we divide by 5 we get \(x^2 + 4\), and \(\langle x^2 + 4\rangle = \langle 5x^2 + 20 \rangle\).

More on Extension Fields

First some definitions.
If \(E\) is an extension field of \(F\), and \(a\in E\), and \(a\) is a root of a (non-zero) polynomial \(p\in F[x]\), then \(a\) is algebraic over \(F\). So \(\sqrt{2}\) is algebraic over \(\mathbb{Q}\), because it is a root of \(x^2 -2\), but \(\pi\) is not algebraic, because there is no rational polynomial that kills \(\pi\). Note that \(\pi\) is algebraic over \(\mathbb{R}\), because it is a root of \(x-\pi\).
Let's say \(a\in E\) is algebraic over \(F\). It must be the root of some non-zero polynomial \(p \in F[x]\). But then it also has to be a root of \(p^2\), and \(p^3\), and \(p+p\). Actually, it has to be a root of every multiple of \(p\). (This should be fairly clear. Let \(p = x^2 - 2\). If we multiply \(p\) by \(q\), we can always factor \(pq\) to be \((x^2 -2)q\), and \(\sqrt{2}\) is obviously a root of this polynomial.) Now \(p\) might not generate all of the polynomials that kill \(a\). For instance, if we took \(p = (x^2-2)^2\)\), we would still get an ideal of polynomials that are zero on \(a\), but not every polynomial.
So let's say we take the set of all polynomials that are zero on \(a\). This is still an ideal. We call this ideal the annihilators of \(a\) over \(F\): \(ann(a,F)\). But this is an ideal of polynomials, i.e. an ideal in \(F[x]\). But \(F[x]\) is a principal ideal domain, meaning \(ann(a,F)\) is principal, meaning it is generated by a single element. Actually, we can go farther than that. We can find a monic polynomial that generates \(ann(a,F)\). We call this monic generator of \(ann(a,F)\) the minimal polynomial of \(a\) over \(F\). It is written \(irr(a,F)\). So \(irr(\sqrt{2},\mathbb{Q})\) is \(x^2-2\).
Another definition:

Theorems

If \(p\) is any annihilator of \(a\) over \(F\), then \(irr(a,F)\) must be a factor of \(p\). So \((x^2-2)^2\) is an annihilator of \(\sqrt{2}\), and the minimal polynomial of \(\sqrt{2}\) over \(\mathbb{Q}\) is \(x^2-2\), which is obviously a factor of \((x^2-2)^2\).