Polynomial Rings

I took Abstract Algebra I/II in UW-Madison(MATH541) with professor Chenxi Wu and Tonghai Yang. This article is inspired by prof. Wu's lecture note, and many formats, content, and structure of this blog are influenced by and similar to it.

RECAP

Please read this introduction with a complete understanding of ring and concepts including Euclidean / Principal Ideal / Unique Factorization Domains.

Definitions and Basic Concepts

For a given indeterminate $x$ with coefficients from a given ring $R$, the polynomial ring $R[x]$ is the polynomials of the form $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, with $n \geq 0$ and $a_i \in R, \; \; \forall \, 0 \leq i \leq n$. For this polynomial, $n$ is its degree, $a_nx^n$ is its leading term, $a_n$ is its leading coefficient. The polynomial is monic if $a_n = 1$.

The addition in $R[x]$ is

$$ \sum_{i=0}^n a_i x^i + \sum_{i=0}^n b_i x^i = \sum_{i=0}^n (a_i + b_i) x^i $$

and the multiplication in $R[x]$ is

$$ (\sum_{i=0}^n a_i x^i) \times (\sum_{i=0}^m b_i x^i) = \sum_{k=0}^{n+m} (\sum_{i=0}^k a_i b_{k-i}) x^k $$

In this way $R[x]$ is a well-defined communicative ring with identity $1$. Clearly $R$ is a subring of $R[x]$.

Proposition. Let $R$ be an integral domain, then

1. degree $p(x)q(x)$ = degree $p(x)$ + degree $q(x)$ if $p(x)$ and $q(x)$ are both nonzero;
2. the units of $R[x]$ are units of $R$;
3. $R[x]$ is an integral domain.

Proposition. Let $I$ be an ideal of the ring $R$ and let $(I) = I[x]$. Then

$$ R[x] / (I) \cong (R / I) [x] $$

Definition. [Polynomial ring in the variables] $R[x_1, x_2, \ldots, x_n]$ is defined inductively by $$ R[x_1, x_2, \ldots, x_n] = R[x_1, x_2, \ldots, x_{n-1}][x_n] $$

Polynomial Rings Over Fields $I$

Theorem. Let $F$ be a field, the polynomial ring $F[x]$ is a Euclidean Domain.

Colloary. If $F$ is a field, then $F[x]$ is a PID and UDF.

Polynomial Rings that are UFD

Proposition. (Gauss' Lemma) Let $R$ be a UFD with field of fractions $F$, then if $p(x) \in F[x]$ then $p(x)$ is reducible in $R[x]$. Precisely, if $p(x) = A(x)B(x)$ for some nonconstant polynomials $A(x), B(x) \in F[x]$, then there are nonzero elements $r, s \in F$ such that $rA(x) = a(x)$ and $sB(x) = b(x)$ both lie in $R[x]$ and $p(x) = a(x)b(x)$ is a factorization in $R[x]$.

Corollary. Let $R$ be a UFD. Let $R$ be its field of fractions and let $p(x) \in R[x]$. Suppose the gcd of all its coefficients are $1$, then $p(x)$ is irreducible in $R[x]$ if and only if it's irreducible in $F[x]$.

Theorem. $R$ is a UDF if and only if $R[x]$ is a UFD.

Corollary. If $R$ is a UFD, then a polynomial ring in an arbitrary number of variables with coefficients in $R$ is also a UFD.

Irreducibility Criteria

Proposition. Let $F$ be a field and let $p(x) \in F[x]$. Then $p(x)$ has a factor of degree one if and only if $p(x)$ has a root in $F$. That is, there is an $\alpha \in F$ with $p(\alpha) = 0$.

Proposition. A polynomial of degree two or three over a field $F$ is reducible if and only if it has a root in $F$.

Proposition. Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ be a polynomial of degree $n$ with integer coefficients. If $r$ and $s$ are relatively prime and $r/s$ is a root of $p(x)$ (that is, $p(r/s) = 0$), we have $r | a_0$ and $s | a_n$.

Proposition. Let $I$ be a proper ideal in the integral domain $R$. If the image of $p(x)$ in $(R/I)[x]$ cannot be factored in $(R/I)[x]$ into two polynomials of smaller degree, then $p(x)$ is irreducible in $R[x]$.

Proposition. [Eisenstein's Criterion] Let $P$ be a prime ideal of the integral domain $R$ and let $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ be a polynomial in $R[x]$. If $a_{n-1}, \ldots, a_0 \in P$ are all elements of $P$, and $a_0 \notin P^2$, then $f(x)$ is irreducible in $R[x]$.