Stuff I'll look at later

[!Fact]
Let $k$ be a field. Then, $S=k[x_1,\dots ,x_n]$ is a UFD. This follows from the fact that any regular local ring is a UFD. A Local ring $R$ is one which has a unique maximal ideal $\mathfrak{m}$. If a ring is regular local, then $\mathfrak{m}/{\mathfrak{m}}^2$ is a vector space over $R/\mathfrak{m}$.

[!Exercise]
$\dim R[x]\le 1+\dim R$.

It follows that $\dim \mathbb{Z}[x]=2$.

$\mathbb{C}[x]$ is a Euclidean domain. Define the degree of a power series to be the degree of the smallest non-zero term.

[!Definition]
Let $R$ be a ring and $S$ be a subset of $R$. An element $d\in R$ is a gcd of $S$ if

1. $d|a$ for all $a\in S$, and
2. If $c|a$ for all $a\in S$, then $c|d$.

[!Theorem]
Let $a_1,\dots ,a_n\in R$. Let $S=\{a_1,\dots ,a_n\}$. Then,

$$d\in R\text{ is gcd of }S\text{ such that }d=\sum _{i=1}^n{r}_i{a}_i\text{ for }{r}_i\in R⟺(d)=(a_1)+(a_1)+\cdots +(a_n)=(a_1,\dots ,a_n)$$

[!Proof]-

Let $d=\gcd S$ such that $d={\sum }_{i=1}^n{r}_i{a}_i$. Since $d\in (a_1)+\cdots +(a_n)$, we have $(d)\subseteq (a_1)+\cdots +(a_n)$. Also, since $d|a_i$ for all $i$, $(a_i)\subseteq (d)$ for all $i$, and the reverse inclusion holds.

Conversely, suppose $(d)=(a_1)+\cdots +(a_n)$. $d={\sum }_{i=1}^n{r}_i{a}_i$ and $d|a_i$ immediately follow. show property $(2)$. Show that if there exists another gcd $d^{\prime }$, $d$ and $d^{\prime }$ are associates.

[!Example]
Let $R$ be a ring. Suppose $d$ is the gcd of $S=\{a_1,\dots ,a_n\}$. It is possible that $d$ may not be expressible as a $R$ linear combination of the elements of $S$. For example, $1$ is the gcd of $\{x,y\}$ in $k[x,y]$, but $1\notin ⟨x,y⟩$.

$\mathbb{Z}[x]/(2,x+1,x^2+5)$.

Show that $(x^2+5,3)$ is not a prime ideal.

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[!Definition]
Let $\alpha \in \mathbb{C}$ not be an integer. Let $\alpha$ satisfy a monic irreducible polynomial $f(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0$, where $a_i\in \mathbb{Z}$. We define the norm and trace of $\alpha$ as $\text{Norm}(\alpha )=(-1{)}^n{a}_0$, $\text{Tr}(\alpha )=-a_{n-1}$.

For example, every element in $\mathbb{Z}[i]$ is of the form $\alpha =a+bi$; $\text{Tr}(\alpha )=2a$, $\text{Norm}(\alpha )=a^2+b^2$.

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[!Remark]
Every ideal in a ring is a module over $R$. Look up nakayama’s lemma Singh (2011, p. 16).

[!Remark]
Let $f:R\to S$ be a homomorphism and $M$ be a $S$-module $⟹$ $M$ is an $R$-module. If $M$ is an $R$-module, then for any ideal $J\subseteq ann(M)$, $M$ is $R/J$ module.

$ann(M)=\{r\in R|rM=0\}$.

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[!Theorem]
Every PID is a UFD.

[!Proof]-
factorization exists: Suppose $R$ is a PID. We prove by contradiction. Suppose there exists nonzero nonunit $x$ which cannot be factors as a finite product of irreducible elements. Let $S$ be the set of all such elements. Let $\alpha \in S$. Then $(\alpha )\subseteq M$ where $M$ is a maximal ideal and $M=(\beta )$ where $\beta$ is irreducible. So $(\alpha )\subseteq (\beta )$, $\alpha =f\beta$. $f$ can’t be a unit, since that would mean $\alpha \notin S$. For the same reason, $f$ can’t be written as the product of a unit and finitely many irreducibles. Thus, $f$ is forced to be in $S$. So, $(\alpha )\subset (\beta )$, and the inclusion is proper. Continue this process to get an infinite chain; this is a contradiction.

Uniqueness of factorization:jjj

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[!Proposition]
Let $R$ be a UFD. Then a non-zero element if prime iff it is irreducible.

[!Proof]-
Prime implies irreducible in any ID. Have to show irreducible implies prime. Let $p$ be an irreducible element and let $p|ab$. We need to show that $p|a$ or $p|b$.

$ab=rp$ for some $r\in R$. By hypothesis, we can write $a=u{p}_1\dots p_s$, $b=w{q}_1\dots q_t$, where $u,w$ are units and $p_i$ and $q_i$ are irreducibles. We now have

$$(uw)p_1\dots p_s{q}_1\dots q_t=rp.$$

Thus, $p$ is an associate of at least one of $p_1,\dots ,p_s,q_1,\dots ,q_t$. If $p$ is an associate of some $p_i$, then $p|a$. If $p$ is an associate of some $q_i$, $p|b$.

[!Proposition]
If $R$ is a UFD and $a,b\ne 0$, then gcd of $a$ and $b$ exists.

[!Proof]-
Let $a=u{p}_1^{a_1}\dots p_s^{a_s}$ and $b=w{q}_1^{b_1}\dots q_t^{b_t}$. Consider $\{p_{i_1},\dots ,p_{i_k}\}$ where $p_{i_j}$ is an associate of some …

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In $R=k[x_1,\dots ,x_n]$, If $I=(m_1,\dots ,m_t)$ and $d$ is the gcd of the generators of $I$, then $I=(d)(I:(d))$.

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References

Singh, B. (2011). Basic Commutative Algebra. World Scientific.