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+# Math4302 Modern Algebra (Lecture 29)
+
+## Rings
+
+### Polynomial Rings
+
+$$
+R[x]=\{a_0+a_1x+\cdots+a_nx^n:a_0,a_1,\cdots,a_n\in R,n>1\}
+$$
+
+Then $(R[x],+,\cdot )$ is a ring.
+
+If $R$ has a unity $1$, then $R[x]$ has a unity $1$.
+
+If $R$ is commutative, then $(R[x],+,\cdot )$ is commutative.
+
+#### Definition of evaluation map
+
+Let $F$ be a field, and $F[x]$. Fix $\alpha\in F$. $\phi_\alpha:F[x]\to F$ defined by $f(x)\mapsto f(\alpha)$ (the evaluation map).
+
+Then $\phi_\alpha$ is a ring homomorphism. $\forall f,g\in F[x]$,
+
+- $(f+g)(\alpha)=f(\alpha)+g(\alpha)$
+- $(fg)(\alpha)=f(\alpha)g(\alpha)$ (use commutativity of $\cdot$ of $F$, $f(\alpha)g(\alpha)=\sum_{k=0}^{n+m}c_k x^k$, where $c_k=\sum_{i=0}^k a_ib_{k-i}$)
+
+#### Definition of roots
+
+Let $\alpha\in F$ is zero (or root) of $f\in F[x]$, if $f(\alpha)=0$.
+
+<details>
+<summary>Example</summary>
+
+$f(x)=x^3-x, F=\mathbb{Z}_3$
+
+$f(0)=f(1)=0$, $f(2)=8-2=2-2=0$
+
+but note that $f(x)$ is not zero polynomial $f(x)=0$, but all the evaluations are zero.
+
+</details>
+
+#### Factorization of polynomials
+
+Division algorithm. Let $F$ be a field, $f(x),g(x)\in F[x]$ with $g(x)$ non-zero. Then there are unique polynomials $q(x),r(x)\in F[x]$ such that
+
+$f(x)=q(x)g(x)+r(x)$
+
+$r(x)$ is the zero polynomial or $\deg r(x)<\deg g(x)$.
+
+<details>
+<summary>Proof</summary>
+
+Uniqueness: exercise
+
+---
+
+Existence:
+
+Let $S=\{f(x)-h(x)g(x):h(x)\in F[x]\}$.
+
+</details>