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 Algebra Notes
04-03-2015, 04:40 PM
Post: #5
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
1.5 Polynomial rings over fields

Fix a field $K$ in this section
1.5.1 Proposition (Division Algorithm for polynomials over a field).
$\qquad\forall f,g\in F[t]\;((f\ne 0)\implies \exists !\,q\in F[t]\;(g = qf + r\;(\deg(r ) < \deg(f))))$

1.5.2 Definition Let $f,g\in F[t]$. We say that $f$ divides $g$ (write $f\mid g$) if $\exists h\in F[t]\;(g = fh)$.
$\qquad$We write $f\not\mid g$ if $f$ does not divide $g\underset{\,}{.}$
$\qquad$A member of $\underset{\,}{\,}D(f,g) = \{ h\in R[t]: (h\mid f)\wedge(h\mid g)\}$ is called a common divisor of $f$ and $g$.
$\qquad d(\in D(f,g))$ is a highest common divisor (hcd) of $f,\;g$ if $h\in D(f,g)\implies h\mid d\underset{\;}{.}$
$\qquad$we say $f,\;g$ are relatively prime.If $1$ is a hcd of $f$ and $g$.

1.5.3 Proposition hcd of $f,\,g$ exists hence $\forall f,g\in F[t]\;(D(f,g)\ne \varnothing)\underset{\,}{.}$
Proof WLOG, assume $\{f,g\}\ne \{0\}.$ then $C = \{h\mid h = af + bg\ne 0, \; a,b\in F[t] \}\ne \varnothing\underset{\,}{.}$
$\qquad \therefore\;\exists d\in C\,\forall h\in C\;(\deg(d)\le \deg(h)).\;\;\{\lambda d\mid \lambda\in K-\{0\}\}$ is clearly the set of all hcd's.

1.5.4 Definition $f\in R[t] - R$ is irreducible over a ring $R$ if $\forall g,h\in R[t]\;(f = gh\implies \{g,h\}\cap R \ne\varnothing)$
1.5.5 Examples All polynomials of degree $1$ are irreducible. $t^2 - 2$ is irreducible over $\mathbb{Q}$
$\qquad$but reducible over $\mathbb{R}$.

1.5.6 Remark Clearly each $f\in R[t] - R$ is expressible as the product of irreducible polynomials.
$\qquad$If $f$ is monic, then it is a product of monic irreducibles.

We return our attention to the case where $R = F$ is a field.

1.5.7 Lemma If $f,g,h\in F[t]$ and $f$ is irreducible, then $(f\mid gh)\implies (f\mid g)\vee (f\mid h)$
Proof. A hcd of $f,\;g$ divides irreducible $f$ hence must be either $\lambda$ or $\lambda f\;(\lambda\in F^{\times}).$ The latter case
$\qquad$implies $f\mid g$. Otherwise $f,\;g$ are relatively prime so Proposition 1.5.3 shows that $af + bg = 1$
$\qquad$for some $a,b\in F[t]$ hence $h = afh + bgh$ and so $f\mid h.\quad\square$

1.5.8 Theorem Every $f\in F[t] - F$ is a product of irreducible polynomials, unique up to order
$\qquad$and multiplication by constants. If $f$ is monic, it is a product of monic irreducible polynomials,
$\qquad$unique up to order.

1.5.9 Definition If $R$ is a ring, $\alpha\in R,\;f\in R[t],\;\; f(\alpha) = 0$ then $\alpha$ is a zero or root of $f$.

1.5.10 Lemma $\alpha \in F$ is a zero of $f\in F[t]$ if and only if $(t -\alpha) \mid f$.

1.5.11 Example If $\underset{\,}{\,}f\in F[t],\;\deg(f).$ Then $f$ is irreducible if and only if $\forall \alpha\in F\; (f(\alpha) \ne 0)$.
$\qquad$For instance $f = t^3 + t^2 + 2 \in \mathbb{Z}_3[t]$ is irreducible since $f(0)=2 = f(2),\,f(1) = 1.$
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 Messages In This Thread Algebra Notes - elim - 04-02-2015, 11:29 AM 1.2 Prime subfields - elim - 04-02-2015, 02:02 PM 1.3 Fields of fractions - elim - 04-02-2015, 04:40 PM 1.4 Polynomial rings - elim - 04-02-2015, 05:11 PM 1.5 Polynomial rings over fields - elim - 04-03-2015 04:40 PM RE: Algebra Notes - elim - 04-04-2015, 08:22 PM Notes - elim - 04-05-2015, 07:51 AM Notes - elim - 04-05-2015, 08:02 AM Notes - elim - 04-06-2015, 07:05 AM 2 Field Extensions - elim - 04-06-2015, 07:24 AM Notes - elim - 04-08-2015, 02:58 PM Notes - elim - 04-08-2015, 02:58 PM

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