Notes - Linear Algebra MT23, Rings
Flashcards
Can you define $\mathbb F [x]$, the polynomals over $\mathbb F$?
Define a ring.
A ring is a triple $(R, +, \times)$ where:
- $(R, +)$ is an Abelian group
- $(R\{0\}, \times)$ is a monoid, i.e. associative and has identity
- Distributive laws
- $a(b + c) = ab + ac$
- $(a + b)c = ac + bc$
Can you define the ring $\mathbb F[A]$ where $A \in M _ {n \times n} (\mathbb F)$?
Define a ring homomorphism.
Suppose $R, S$ are rings, $\phi : R \to S$ a homomorphism if
\[\phi(r_1 + r_2) = \phi(r_1)+ \phi(r_2)\]and
\[\phi(r_1 r_2) = \phi(r_1)\phi(r_2)\]Define the $\ker$ and $\text{Im}$ of a ring homomorphism $\phi : R \to S$, and state what is “special” about each
this is a subring.
\[\text{Im } \phi = \\{\phi(r) : r \in R\\}\]this is an ideal.
What are the elements of a ring with multiplicative inverses called?
A unit.
What is an ideal $I$ of a ring $R$, denoted $I \trianglelefteq R$?
A nonempty subset of $R$ with the following properties:
- $I$ is closed under addition
- If $s$ is in $I$ and $r$ is in $R$, then $rs$ and $sr$ is in $I$.
Suppose $R$ is a commutative ring and $I \trianglelefteq R$ is an ideal. Can you define the new commutative ring, $R / I$?
where
\[(r + I) + (r' + I) = (r + r') + I\]and
\[(r + I)(r' + I) = rr' + I\]Suppose $r \in R$ is an element of a ring and $S$ is a subset. What does the notation $r + S$ mean?
Can you state the first isomorphism theorem for rings?
Suppose $\phi : R \to S$ is a ring homomorphism. Then
- $\ker \phi$ is an ideal
- $\text{Im } \phi$ is a subring, and
- $R/\ker \phi \cong \text{Im } f$
What ring isomorphism do you use to show
\[R / \ker \phi \cong \text{Im }\phi\]
?
What does it mean for an ideal $I$ to be a principal ideal?
The ideal is generated by a single element.
Can you state the “Bézout lemma” for polynomial rings?
Let $a(x), b(x) \in \mathbb F[x]$ be non-zero polynomials with $c(x) = \gcd(a(x), b(x))$. Then there exists $s(x), t(x)$ such that
\[a(x) s(x) + b(x) t(x) = c(x)\]Quickly prove the “Bézout lemma” for polynomials over a field, i.e. let $a(x), b(x) \in \mathbb F[x]$ be non-zero polynomials with $c(x) = \gcd(a(x), b(x))$. Then there exists $s(x), t(x)$ such that
\[a(x) s(x) + b(x) t(x) = c(x)\]
Wlog, we can assume $c = \gcd(a, b) = 1$ and that $\deg(a) \ge \deg(b)$ (why? if $\gcd(a, b) \ne 1$, divide both $a$ and $b$ by $c$, and we can assume $\deg(a) \ge \deg(b)$ by symmetry).
Induct on $\deg(a) + \deg(b)$.
By the division algorithm, $\exists q, r \in \mathbb F[x]$ such that
\[a = qb + r\]with $\deg b > \deg r$, then
\[\deg b + \deg r < \deg a + \deg b\]and $\gcd(b, r) = 1$.
If $r = 0$, then $b(x) = \lambda$ is a constant since $\gcd(a, b) = 1$. Hence
\[a(x) \cdot 1 + b(x) \left(\frac 1 \lambda (1-a(x))\right) = 1\]Assume $r \ne 0$. Then by the inductive hypothesis,
\[bs' + rt' = 1\]Hence
\[bs' + (a - qb)t' = 1\]which implies
\[at' + b(s'-qt') = 1\]So we can take $t = t’$ and $s = s’ - qt’$.
Proofs
Suppose $R$ is a commutative ring and $I \trianglelefteq R$ is an ideal. The new commutative ring, $R / I$, is defined
\[R/I = \\{r + I : r \in R\\}\]
where
\[(r + I) + (r' + I) = (r + r') + I\]
and
\[(r + I)(r' + I) = rr' + I\]
Show that this is a well-defined definition.
Todo.
Prove the first isomorphism theorem for rings, i.e. suppose $\phi : R \to S$ is a ring homomorphism. Then
- $\ker \phi$ is an ideal
- $\text{Im } \phi$ is a subring, and
- $R/\ker \phi \cong \text{Im } f$
Todo.
Prove that the division algorithm works in a ring of polynomials, i.e. if $f(x) \in R[x], g(x) \not\equiv 0 \in F[x]$, then $\exists q(x), r(x)$ with
\[f(x) = g(x)q(x) + r(x)\]
where $r \equiv 0$ or $\deg r(x) < \deg g(x)$.
Todo, induct on $\deg f - \deg g$.