Polynomial and Matrix Rings

The Ring of polynomials

Let R be a commutative unitary ring. A polynomial is a formal sum $\sum_{i = 0}^n a_iX^i$ where each $a_i \in R$ and $a_n \neq 0$. The leading term is $a_nX^n$, the degree is n, the leading coefficient is $a_n$, and it is monic if $a_n = 1$.

$R[x]$ is a ring of polynomials of x with coefficients from R. Addition is defined by $(\sum_{i=0}^n a_ix^i)+(\sum_{i=1}^n b_iX^i) = \sum_{i=1}^n (a_i+b_i)X^i$. Multiplication is defined by: $(\sum_{i=0}^n a_iX^i)(\sum_{j=0}^n b_jX^j) = \sum_{i=0}^n\sum_{j=0}^n a_ib_jX^{i+j}$ (note that it is commutative because R is commutative). It is associative because $(\sum_{i=0}^n a_iX^i)\cdot (\sum_{j=0}^nb_jX^j)(\sum_{k=0}^nc_kX^k) =$ $(\sum_{i=0}^n a_iX^i)(\sum_{j=0}^n \sum_{k=0}^n b_jc_kX^{j+k}) =$ $\sum_{i=0}^n\sum_{j=0}^n\sum_{k=0}^n a_ib_jc_kX^{i+j+k} =$ $(\sum_{i=0}^na_iX^i)(\sum_{j=0}^nb_jX^j)(\sum_{k=0}^nc_kX^k)$ distributivity also holds, 1 is the multiplicative identity. Hence $R[x]$ is a commutative unitary ring.

Example: $\mathbb{Q}[x]$ is the ring of polynomials with rational coefficients. $\mathbb{Z}/3\mathbb{Z}[x]$ is a ring of polynomials with coefficients in $\{0,1,2\}$. Some sample computations in $\mathbb{Z}/3\mathbb{Z}[x]: (x^2+2x+1)+(x^3+x+2) = x^3+x^2+3x+2 = x^3+x^2+2$ (addition) $(x^2+2x+1)(x^3+x+2) =$ $x^5+2x^4+3x^3+4x^2+5x+2 =$ $x^5+2x^4+2x^3+x^2+2x+2$ (multiplication). In $\mathbb{Z}/2\mathbb{Z}[x]$, $x^2+2x+1 =$ $(x+1)^2$ so $x^2+1$ is a square (however it is not a square in $\mathbb{Z}[x]$).

Matrix rings:

Let R be a ring and $n > 0$ an integer. $M_n(R)$ is the ring of all $n \times n$ matrices with entries from R. Addition is defined coordinate wise: $((a_{ij})_{i,j} +(b_{ij}))_{i,j} = (a_{ij}+b_{ij})_{i,j}$. Multiplication: $((a_{ij})_{i,j} \cdot (b_{ij}))_{i,j} = \sum+{k =0}^{n-1} a_{ik}b_{kj}$.

Why is it a ring?

• $(M_n(R), +)$ is an abelian group
• Distributivity: $(a_{ij})\cdot (b_{ij}) + (c_{ij}) =$ $(a_{ij})_{i,j}(b_{ij}+c_{ij})_{i,j} =$ $(\sum_{k=0}^{n-1}a_{ik}(b_{kj}+c_{kj}))_{i,j} =$ $(\sum_{k=0}^{n-1}a_{ik}b_{kj}+\sum_{k=0}^{n-1}a_{ik}c_{kj}) =$ $(a_{ij})(b_{ij})+(a_{ij})(c_{ij})$
• It is associative (the proof is left as an exercise)

Notes:

• R can be identified with $\{ \begin{bmatrix}a & 0\\0 & a\end{bmatrix} : a \in R \}$
• if $n \geq 2, M_n(R)$ is not commutative (unless we’re considering a ring where for all a and b, $a \cdot b = 0$)
• $M_n(R)$ has nonzero zero divisers

If R is unitary, the group of units in $M_n(R)$ is generally called $GL_n(R)$, where GL stands for general linear.