Intro to Rings and Fields

Mike
Math, Rings, Fields
Intro to Rings and Fields
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It’s about time we introduced rings and fields.

Definition of Ring

Let \(R\) be a set with two binary operators, usually written \(+\) and \(\cdot\), where the following conditions hold

  1. \(R\) under the action of \(+\) is an Abelian group with identity element \(0\).
    1. For all \(x, y \in R\), we have \( x + y = y + x\)
    2. For all \(x \in R\), we have \( x + 0 = 0 + x = x\)
    3. For all \(x, y, z \in R\), we have \( (x + y) + z = x + (y + z)\)
    4. For all \(x \in R\), there is a \(y \in R\), such that \(x + y = 0\). \(y\) is written as “\(-x\)” (recall that the inverse of an element of a group is unique, so \(-x\) is well-defined for any \(x\))
  2. \(R\) under the action of \(\cdot\) is associative. This means that for all \(x,y,z \in R\), we have \( (x \cdot y) \cdot z = x \cdot (y \cdot z) \)
  3. There are distributive laws that hold between the two operators. This means that for all \(x, y, z \in R\) we have
    1. \( x \cdot (y + z) = (x \cdot y) + (x \cdot z) \)
    2. \( (y + z ) \cdot x = (y \cdot x) + (z \cdot x) \)
  4. Some rings (but not all!) have an element, usually written “\(1\)”, such that for all \(x\in R\), \(x \cdot 1 = 1 \cdot x = x\). Those are sometimes called “rings with identity”, others call them “rings with units”. They are very common. We will often fail to mention that a ring has \(1\), and instead point out when a ring doesn’t have \(1\).
  5. Some rings (but not all!) commute with respect to \(\cdot\). This means that for all \(x,y \in R\), we have \( x \cdot y = y \cdot x\). Rings where this property holds are called commutative rings.

Definition of a Field

A field, \(F\) is a commutative ring with identity such that for every \(x\) in \(R\) which is not equal to \(0\), there exists a \(y\) in \(R\), such that \(x \cdot y = 1\).

This means that \(F \setminus \{0\}\), sometimes written \(F^{*}\), is a group with respect to the action of \(\cdot\). This means that the \(y\) mentioned in the definition is unique for each \(x\). It is often written \(x^{-1}\).

A note on notation

We’ll often omit the “\( \cdot \)” operator. It should come as no surprise that “\(\cdot\)” is often identified with multiplication, so we’ll omit it like we do with multiplication when the meaning is clear (which is most of the time).

We’ll write \( x - y\) in place of \(x + (-y)\), and sometimes we’ll write \( \frac{x}{y}\) instead of \( xy^{-1}\).

We’ll write \(x^n\) in place of \(xxx … x\), where \(x\) occurs exactly \(n\) times.

Let’s establish some basic facts about rings.

Mini-Lemma 1: Times 0

Let \(x\) be an arbitrary element of a ring, \(R\). Then we have \(x0 = 0x = 0\).

Proof

We’ll prove that \(x0 = 0\). The proof that \(0x = 0\) is nearly identical.

Notice that since \(0 = 0 + 0\), we have \(x0 = x(0 + 0)\). By the first distribution law above, this is equal to \(x0 + x0\), which means that \(x0 = x0 + x0\). We can then add \(-x0\) to both sides of the equation to obtain

\[ x0 - x0 = (x0 + x0) - x0 \tag{1}\label{1}\]

By part (1) subpart (3) of the definition of a ring, we have

\[ (x0 + x0) - x0 = x0 + (x0 - x0) \tag{2}\label{2} \]

combining equations (1) and (2) gives us

\[ 0 = x0 - x0 = x0 + 0 \tag{3}\label{3}\]

and by the definition of \(0\), the right-hand side of (3) is equal to \(x0\).

QED

Mini-lemma 2: Times negative

Let \(x\) and \(y\) be arbitrary elements of a ring \(R\). Then we have

  1. \(x(-y) = -xy\)
  2. \((-x)y = -xy\)

Proof

We’re only going to prove part 1. The proof of part 2 is very similar.

By mini-lemma 1, we have \(0 = x0 = x(y + (-y))\). Applying the distributive law gives us \(xy + x(-y)\).

Therefore, since \(0 = xy + x(-y)\), we have \(-xy = x(-y)\).

QED

Mini-corollary: neg 1 squared

Let \(R\) be a ring with unity, then \((-1)^2 = 1\).

Proof

We’ll show that \( (-1) + (-1)^2 = 0\), which will give us our result. Notice that \( (-1) + (-1)^2\) is equal to \( (-1) \cdot 1 + (-1)(-1) \). By the distributive law (1) above, this is equal to \( (-1)(1 - 1) \), which is equal to \((-1) \cdot 0 = 0\).

QED

We’ve already used the binomial coefficient in our proof of Sylow’s first theorem. We’ll define it here, anyway.

Definitions: Factorial and Binomial coefficient

Let \(n\) be a non-negative integer. Then \(n\) factorial, written \(n!\), is defined so that

\[ n! := \prod\limits_{m = 1}^{n} m \tag{4}\label{4} \]

whenever \(n > 0\). However, \(0! = 1\) by definition.

Let \(n\) be a non-negative integer and \(k\) a non-negative integer such that \(k \leq n\), then

\[ \binom{n}{k} := \frac{n!}{(n-k)!k!} \tag{5}\label{5} \]

Notice that by equation (5) we can conclude that

\[ \binom{n}{k} = \binom{n}{n-k} \tag{6}\label{6} \]

We’re going to call the next lemma Pascal’s Triangle because you can use it to prove that Pascal’s Triangle is composed of binomial coefficients.

Lemma: Pascal’s Triangle

Let \(n\) and \(k\) be integers, such that \( 1 \leq k < n\), then

\[ \binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k} \tag{7}\label{7} \]

Proof

Applying equation (5) to the left side of equation (7) gives us

$$\begin{align} &\frac{(n-1)!}{((n - 1) -k)!k!} + \tag{8}\label{8} \\ &\frac{(n-1)!}{((n -1) - (k-1))!(k-1)!} \tag{9}\label{9} \\ \end{align}$$

Notice that \( (n-1) - (k-1) = n - k\), so (9) is equal to

\[ \frac{(n-1)!}{(n - k)!(k-1)!} \tag{10}\label{10} \]

and (8) is equal to

\[ \frac{(n-1)!}{(n -k - 1)!k!} \tag{11}\label{11} \]

Now we have to find a common denominator for (10) and (11) so that we can add them. We can do that by multiplying \( (k-1)! \) in (10) by \(k\) to get \(k!\). Similarly, we multiply \( (n-k - 1)!\) in (11) by \( n-k\) to get \( (n-k)!\).

When we do that (10) becomes

\[ \frac{k(n-1)!}{(n-k)!k!} \tag{12}\label{12} \]

and (11) becomes

\[ \frac{(n-k)(n-1)!}{(n-k)!k!} \tag{13}\label{13} \]

Adding (12) and (13) we get

\[ \frac{n(n-1)!}{(n-k)!k!} = \frac{n!}{(n-k)!k!} \tag{14}\label{14} \]

and that is equal to \( \binom{n}{k} \).

QED

And now, the reason why they’re called binomial coefficients.

Lemma Binomial Expansion

Let \(x\) and \(y\) be arbitrary elements of a commutative ring with unity, \(R\), then we have

\[ (x + y)^n = \sum\limits_{k = 0}^{n} \binom{n}{k}x^{n-k}y^k \tag{15}\label{15} \]

Note: When we have a ring with unity, then we identify \(x^0 = 1\).

Proof

We’ll do this one by induction on \(n\). Notice that if \(n = 1\), then both sides are equal to \(x + y\).

Now, suppose that this lemma holds for all \(m < n\). We have \( (x + y)^{n} = (x + y)^{n-1}(x + y)\). By induction,

\[ (x + y)^{n-1} = \sum\limits_{k = 0}^{n-1} \binom{n-1}{k} x^{n - 1 - k}y^k \tag{16}\label{16} \]

multiplying the right side of (16) by \(x+y\) gives us

$$\begin{align} & \sum\limits_{k=0}^{n-1} \binom{n-1}{k} x^{n-k}y^k + \tag{17.1}\label{17.1} \\ & \sum\limits_{k=0}^{n - 1} \binom{n-1}{k} x^{n - (k+1)}y^{k+1} \tag{17.2}\label{17.2} \\ \end{align}$$

(17.1) is where we multiplied by \(x\) and (17.2) is where we multiplied by \(y\). In (17.2) we also re-wrote the exponent of \(x\). It works because \(n-1-k\) is equal to \(n-k-1\), which is equal to \(n - (k+1)\).

We can substitute \(l = k+1\) in (17.2) to obtain

\[ \sum\limits_{l=1}^{n} \binom{n-1}{l-1}x^{n - l}y^l \tag{18}\label{18} \]

but since \(l\) is a bound variable – because it is the variable we’re summing over, we can replace it with anything we like. Replacing it in (18) with \(k\) gives us

\[ \sum\limits_{k=1}^{n} \binom{n-1}{k-1}x^{n-k}y^k \tag{19}\label{19} \]

So, what we have shown so far is that \( (x+y)^n \) is equal to (17.1) + (19). That is

$$\begin{align} & \sum\limits_{k=0}^{n-1} \binom{n-1}{k} x^{n-k}y^k + \tag{20.1}\label{20.1} \\ & \sum\limits_{k=1}^{n} \binom{n-1}{k-1} x^{n - k}y^{k} \tag{20.2}\label{20.2} \\ \end{align}$$

Let’s evaluate the sum of (20.1) and (20.2) in three cases.

  1. \(k = 0\). This case only occurs in (20.1), and in that case it evaluates to \(x^n = \binom{n}{0}x^ny^0\).
  2. \(k = n\). This case only occurs in (20.2), and in that case it evalueates to \(y^n = \binom{n}{n}x^0y^n\).
  3. \(0 < k < n\). In this case we have to add terms from both (20.1) and (20.2), where we get

\[ \left [ \binom{n-1}{k} + \binom{n-1}{k-1} \right ] x^{n-k}y^k \tag{21}\label{21} \]

By equation (7) from Pascal’s triangle lemma above, the sum of the two binomial coefficients inside the brackets is equal to \( \binom{n}{k} \).

Therefore, when we combine all three cases, we see that (20.1) plus (20.2) is equal to

\[ \sum\limits_{k=0}^{n} \binom{n}{k}x^{n-k}y^k \]

QED

We’ll conclude this entry by investigating some of the properties of homomorphisms between two rings.

Definition: Ring homomorphism

Let \(R\) and \(S\) be rings. Let \(\phi: R \rightarrow S\) be a function from \(R\) to \(S\), such that for any two \(x,y \in R\), we have

  1. \( \phi(x \cdot y) = \phi(x) \cdot \phi(y)\)
  2. \( \phi(x + y) = \phi(x) + \phi(y) \)

Then \(\phi\) is said to be a ring homomorphism from \(R\) to \(S\).

The definitions of

  • isomorphism
  • automorphism
  • endomorphism
  • monomorphism
  • epimorphism

are analogous to those we discussed here in the context of groups. In fact, analogous morphisms can be found among all sorts of mathematical objects. We’ll dive deeper into this in a later entry when we discuss sets, models, (mathematical) languages, and (mathematical) theories.

Definition: Ideal

Let \(\phi : R \rightarrow S\) be a ring homomorphism with kernel \(I \subseteq R\), then \(I\) is said to be an ideal of \(R\).

Definition: subring

Let \(Q \subseteq R\) be a subset of a ring \(R\), such that for any \(x, y \in Q\), we have

  1. \( x + y \in Q\)
  2. \(x \cdot y \in Q\)
  3. \(-x, -y \in Q\)

Then \(Q\) is said to be a subring of \(R\).

Notice that combining conditions 1 and 3 means that \( 0 = x + (-x) \in Q\).

The following few lemmas are very similar to their group counterparts. You can probably get away with skipping the proofs.

Mini-lemma: subring test

Let \(Q \subseteq R\) be such that for any \(x, y \in Q\), we have

  1. \( x - y \in Q\)
  2. \( x \cdot y \in Q\)

Then \(Q\) is a subring of \(R\).

Proof

Notice that if \(x = y\) in condition 1, then we are guaranteed to have \(0 \in Q\). Then for each \(y \in Q\), \(0 - y = -y \in Q\), which takes care of 3. Condition 2 is the same in both.

QED

Mini-lemma: Ideals are subrings

Let \(\phi: R \rightarrow S\) be a ring homomorphism from \(R\) to \(S\) with kernel equal to the ideal \(I\). \(I\) is a subring of \(R\).

Proof

This is a straight-forward computation. Notice that since

$$ \begin{align} I &= \text{Ker}(\phi) \\ &= \{ x \in R | \phi(x) \\ &= 0 \}\\ \end{align} $$

then for all \(x, y \in \text{Ker}(\phi) \), we have

$$ \begin{align} &\phi(x - y) \\ = &\phi(x) - \phi(y) \\ = &0-0 \\ = &0\\ \end{align} $$

Thus \( x-y \in \text{Ker}(\phi) \) and condition 1 of the subring test is satisfied.

We also have

$$ \begin{align} &\phi(x \cdot y) \\ = &\phi(x) \cdot \phi(y) \\ = &0 \cdot 0 \\ = &0\\ \end{align} $$

Thus \( x \cdot y \in \text{Ker}(\phi) \) and condition 2 is satisfied.

QED

Mini-lemma: Image of phi is a subring

Let \(\phi : R \rightarrow S\) be a ring homomorphism. Then

$$ \begin{align} &\text{Im}(\phi) = \\ &\{ y \in S | \exists x \in R, \phi(x) = y \} \\ \end{align} $$

is a subring of \(S\).

Proof

Let \(x, y \in \text{Im}(\phi)\). Then there exists \(u, v \in R\), such that \( \phi(u) = x\) and \(\phi(v) = y\). Then

$$ \begin{align} &\phi(u - v) \\ = &\phi(u) - \phi(v) \\ = &x - y \in \text{Im}(\phi)\\ \end{align} $$

Therefore \(x - y \in \text{Im}(\phi) \) and condition 1 of the subring test is satisfied.

Similarly, \( \phi(u \cdot v) = \phi(u) \cdot \phi(v)\) which is equal to \( x \cdot y \in \text{Im}(\phi)\), so condition 2 is satisfied.

QED

Lemma: Factor rings are rings

Let \(\phi : R \rightarrow S\) be a ring homomorphism with kernel \(I\). Then we can define an equivalence relation where \( x \equiv y (\mod I)\) if and only if \(x - y \in I\).

This is well-defined and an equivalence relation. Furthermore, the set of equivalence classes is a ring. It is denoted \( R/I\).

Proof

It’s well-defined and an equivalence relation because it is equal to \(R / I\), where \(I\) is the normal subgroup of \(R\) when viewed as an Abelian group with the operator \( +\). We showed that mod a normal subgroup is an equivalence relation and that the set of equivalence classes form a group. (We showed that here and here). This works, because the definition of the equivalence relation only makes reference to \(+\) (in the form of \(-\)).

Now we only have to show that the set of equivalence classes form a ring. We already mentioned that \(R/I\) is a group with the operator \(+\), so we only have to verify the ring axioms that involve \( \cdot\). These are

  1. \(R\) under the action of \(\cdot\) is associative. This means that for all \(x,y,z \in R\), we have \( (x \cdot y) \cdot z = x \cdot (y \cdot z) \)
  2. There are distributive laws that hold between the two operators. This means that for all \(x, y, z \in R\) we have
    1. \( x \cdot (y + z) = (x \cdot y) + (x \cdot z) \)
    2. \( (y + z ) \cdot x = (y \cdot x) + (z \cdot x) \)

In order to prove 1, we must show that the equivalence classes of \(R/I\) are well-defined with respect to \(\cdot\). I.e. \( [x_1] \cdot [y_1] = [x_2] \cdot [y_2]\) whenever \(x_1 \equiv x_2\) and \(y_1 \equiv y_2\). This happens if we can show that \(x_1y_1 - x_2 y_2 \in I\).

\(x_1 \equiv x_2\) because \( x_1 - x_2 \in I \), which means that \( \phi(x_1 - x_2) = 0 \), which means that \( \phi(x_1) = \phi(x_2)\). Similarly, \( \phi(y_1) = \phi(y_2)\). Therefore, \(\phi(x_1y_1) = \phi(x_1) \cdot \phi(y_1)\) \( = \phi(x_2) \cdot \phi(y_2) = \phi(x_2y_2) \). That means that \( \phi(x_1y_1) - \phi(x_2y_2) = 0\) and that means that \( \phi(x_1y_1 - x_2y_2) = 0 \), which implies that \( x_1y_1 - x_2y_2 \in I \).

So we have finished proving 1. Now we will prove the distributive laws.

The same reasoning that we applied in showing that condition 1 holds works in this case as well. Since distributivity holds in \(R\), we are guaranteed that

$$ \begin{align} &x \cdot (y + z) \\ = (&x \cdot y) + (x \cdot z) \\ \end{align} $$

Since \(R/I\) is well-defined with respect to \(\cdot\), we are guaranteed to have

$$ \begin{align} &[x] \cdot ([y] + [z]) \\ = (&[x] \cdot [y]) + ([x] \cdot [z]) \\ \end{align} $$

QED

That’s all that we have time for with this entry. We’ll continue our discussion of rings and fields with the next entry. Thanks for reading!