Rings, Ideals & Polynomial Rings
Rings, Ideals & Polynomial Rings
A ring $(R,+,\cdot)$ combines an abelian group under addition with an associative multiplication that distributes over addition. Ideals are the ring analogue of normal subgroups: a subset $I\subseteq R$ such that $R\cdot I\subseteq I$. Quotient rings $R/I$ and ring homomorphisms obey analogous isomorphism theorems. Polynomial rings $R[x]$ are the fundamental construction for algebraic geometry and Galois theory.
Rings & Ideals
A ring $R$ satisfies: $(R,+)$ is abelian group; multiplication is associative and distributes over addition. An ideal $I\trianglelefteq R$: $(I,+)$ is a subgroup; $r\in R,a\in I\Rightarrow ra,ar\in I$. The quotient ring $R/I$ with $(a+I)(b+I)=(ab+I)$ is a ring iff $I$ is an ideal. Examples: $n\mathbb{Z}\trianglelefteq\mathbb{Z}$; $\mathbb{Z}/n\mathbb{Z}$ is a ring; $\langle f\rangle\trianglelefteq k[x]$ for any polynomial $f$; $k[x]/\langle f\rangle$ is a ring.
Principal Ideal Domains & UFDs
An integral domain $R$ (no zero divisors) is a PID (principal ideal domain) if every ideal is principal ($I=\langle a\rangle$ for some $a\in R$). Every PID is a UFD (unique factorization domain): every non-zero non-unit factors uniquely into irreducibles. $\mathbb{Z}$ and $k[x]$ (polynomials over a field) are PIDs. In $k[x]$: irreducibles are the irreducible polynomials; GCDs exist; Bezout's identity holds. The ideal $\langle f,g\rangle=\langle\gcd(f,g)\rangle$.
Example 1
Factor $x^4-1$ over $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$.
Solution: Over $\mathbb{Q}$: $x^4-1=(x-1)(x+1)(x^2+1)$; $x^2+1$ is irreducible over $\mathbb{Q}$ (no rational root). Over $\mathbb{R}$: same ($x^2+1$ is irreducible over $\mathbb{R}$). Over $\mathbb{C}$: $x^4-1=(x-1)(x+1)(x-i)(x+i)$ (splits completely). The factorization reflects the root structure: 4th roots of unity $\{1,-1,i,-i\}$.
Example 2
Show $\mathbb{Z}[i]/(1+i)\cong\mathbb{Z}/2\mathbb{Z}$ (ring of Gaussian integers mod $1+i$).
Solution: $(1+i)$ is a prime ideal in $\mathbb{Z}[i]$: norm $N(1+i)=2$, and $\mathbb{Z}[i]/(1+i)$ has order $N(1+i)=2$. Homomorphism $\phi:\mathbb{Z}[i]\to\mathbb{F}_2$: $\phi(a+bi)=(a+b)\bmod 2$. This is a ring homomorphism (check: $\phi((a+bi)(c+di))=\phi(ac-bd+i(ad+bc))=(ac-bd+ad+bc)\bmod 2=(a+b)(c+d)\bmod 2=\phi(a+bi)\phi(c+di)$ ✓). $\ker\phi=\{a+bi:a\equiv b\pmod 2\}=\langle 1+i\rangle$. By FIT: $\mathbb{Z}[i]/(1+i)\cong\mathbb{F}_2$.
Practice
- Prove every field is a PID.
- Show $\mathbb{Z}[\sqrt{-5}]$ is not a UFD by exhibiting two distinct factorizations of 6.
- Prove: $k[x]/\langle f\rangle$ is a field iff $f$ is irreducible over $k$.
- Find all ideals of $\mathbb{Z}/12\mathbb{Z}$ and identify which are prime/maximal.
Show Answer Key
1. In a field $F$, every nonzero element is a unit. For ideal $I\neq(0)$, pick $a\in I$, $a\neq0$. Then $a^{-1}\in F$, so $1=a^{-1}a\in I$, giving $I=F$. So the only ideals are $(0)$ and $F$, making $F$ a PID (every ideal is principal).
2. $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$. All four factors are irreducible in $\mathbb{Z}[\sqrt{-5}]$ (check norms: $N(2)=4$, $N(3)=9$, $N(1\pm\sqrt{-5})=6$; none has norm-2 or norm-3 divisors). So $\mathbb{Z}[\sqrt{-5}]$ is not a UFD.
3. ($\Rightarrow$) If $f$ is irreducible and $g\notin\langle f\rangle$, then $\gcd(f,g)=1$, so $\exists a,b$: $af+bg=1$ in $k[x]$, meaning $\bar{g}$ is invertible in $k[x]/\langle f\rangle$. ($\Leftarrow$) If $f=gh$ with $\deg g,\deg h\ge1$, then $\bar{g}\cdot\bar{h}=0$ but neither is zero, contradicting $k[x]/\langle f\rangle$ being a field.
4. Ideals of $\mathbb{Z}/12\mathbb{Z}$ correspond to divisors of 12: $\langle1\rangle$, $\langle2\rangle$, $\langle3\rangle$, $\langle4\rangle$, $\langle6\rangle$, $\langle0\rangle$. Prime ideals: $\langle2\rangle$ and $\langle3\rangle$ (quotients $\mathbb{Z}_2$ and $\mathbb{Z}_4$... actually $\mathbb{Z}/12/\langle2\rangle\cong\mathbb{Z}_2$, $\mathbb{Z}/12/\langle3\rangle\cong\mathbb{Z}_3$). Maximal: $\langle2\rangle$ and $\langle3\rangle$ (since the quotients are fields).