# Abstract Nonsense

## PIDs (Pt. II)

Point of Post: This is a continuation of this post.

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We now would like to discuss the notion of greatest common divisors in PIDs. Recall that the notation $a\mid b$, $a$ divides $b$, for $a,b$ in some ring $R$ is that there exists some $c\in R$ such that $b=ac$. Note that this is equivalent to saying that $(a)\supseteq (b)$. Now, note that if $R$ is an integral domain and $a\mid b$ and $b\mid a$, say $a=ub$ and $b=u'a$ we see then that $a=ub=uu'a$ and so by cancellation $uu'=1$ so that $u,u'\in R^\times$.  We call two elements related in this way, $a\mid b$ and $b\mid a$, associated and we call them associates. It’s evident that being associate is an equivalence relation on $R$. We then call an element $d\in R$ a greatest common divisor of $a,b\in R$ if $d\mid a$, $d\mid b$ and whenever $c\mid a$ and $c\mid b$ then $c\mid d$. An emphasis was placed on the article ‘a’ because greatest common divisors are not unique. For example, in $\mathbb{Z}$ a greatest common divisor of $2$ and $3$ is $1$, but so is $-1$. In general, greatest common divisors are unique up to associates:

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Theorem: Let $R$ be an integral domain and let $a,b\in R$. Then, if $d$ is a greatest common divisor of $a,b$ then so is $ud$ for any $u\in R^\times$. Moreover, every greatest common divisor is of this form.

Proof: Suppose first that $ud$ is an associate of $d$, then evidently $du\mid a,b$ and if $c\mid a,b$ then $c\mid d$ so that $c\mid ud$ and so $ud$ is a greatest common divisor of $a,b$. Conversely, suppose that $x$ is a greatest common divisor of $a,b$. Since $x\mid a,b$ and $d$ is a greatest common divisor we have that $x\mid d$. But, since $x$ is also a greatest common divisor of $a,b$ applying the same argument shows that $d\mid x$. From previous observation we have that $d$ and $x$ are associates. $\blacksquare$

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So, how exactly do greatest common divisors factor into PIDs? Well, given two elements $a,b\in R$ we know that the ideal $(a,b)$ generated by the two is principal, and so equal to $(d)$ for some $d\in R$. Well, unsurprisingly, as is easily seen in $\mathbb{R}$, one has that $d$ is a greatest common divisor of $a,b$. Indeed:

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Theorem: Let $R$ be an integral domain and suppose that $d\in R$ are such that $(d)=(a,b)$. Then, $d$ is a greatest common divisor of $a,b$.

Proof: Since $a,b\in (d)$ we have by definition that $d\mid a,b$. Suppose now that $c\mid a,b$ then we know that $(c)\supseteq(a,b)$ and so $(c)\supseteq(d)$. But, by definition this implies that $c\mid d$. $\blacksquare$

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So, think about how this jives with our usual notion of greatest common divisors in $\mathbb{Z}$. We usually call the greatest common divisor of $a,b\in\mathbb{Z}$ the largest (positive) common divisor of $a,b$. In other words $\text{gcd}(a,b)=\max\left\{x\in\mathbb{N}:x\mid a,b\right\}$. Now,  from the previous theorem we know that if $d=\text{gcd}(a,b)$ and $c\mid a,b$ then $(c)\supseteq (d)$ and so, in particular, we have that $d\in(c)$. But, by previous theorem we know $(c)$ is generated by an element of least degree in $(c)$. In particular, $d(c)\leqslant d(x)$ for every $x\in (c)$ and so, in particular, $d(c)\leqslant d(d)$. From this it’s not hard to conclude that, agreeing with our idea of greatest common divisors in $\mathbb{Z}$, one has the following:

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Theorem: Let $(R,d)$ be a Euclidean domain with $d$ satisfying the $d$-inequality. Then if $a,b\in R$ one has that $d$ is a greatest common divisor of $a,b$ if and only if $d\mid a,b$ and whenever $c\mid a,b$ one has that $d(c)\leqslant d(d)$.

Proof: Suppose first that $d$ is a greatest common divisor of $a,b$ and let $c\mid a,b$. As mentioned in the previous paragraph we have that $(c)\supseteq (d)$ and so $d\in (c)$ and so $d(c)\leqslant d(d)$. We shall prove the other direction in our next post. $\blacksquare$

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Regardless, another common fact (known colloquially as Bezout’s identity) which holds true in $\mathbb{Z}$ holds more generally in PIDs:

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Theorem: Let $R$ be a PID. Then, if $a,b\in R$ one has that $1$ is a greatest common divisor of $a,b$ if and only if $ax+by=1$ for some $x,y\in R$.

Proof: Suppose first that $a,b$ have $1$ as a greatest common divisor. We know then that $(a,b)=(1)=R$ and so, in fact, for any $r\in R$ there exists $x,y\in R$ such that $ax+by=r$. Conversely, suppose that $ax+by=1$ for some $x,y\in R$ then evidently $(a,b)=R=(1)$ and so $1$ is a greatest common divisor of $a,b$. $\blacksquare$

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If $a,b\in R$ have $1$ as a greatest common divisor we say that $a,b$ are coprime. In fact, we see that the above theorem may be stated as follows:

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Theorem: Let $R$ be a PID and $a,b\in R$. Then, $a,b$ are coprime if and only if $(a),(b)$ are comaximal

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And so the CRT tells us that if $a,b$ are coprime then $R/(ab)=R/((a)(b))\cong R/(a)\times R/(b)$.

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As a last structural result we mention the following:

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Theorem: Let $R$ be a PID, then $R/\mathfrak{a}$ is a PID if and only if $\mathfrak{a}\in\text{Spec}(R)$.

Proof: The necessity for $\mathfrak{a}$ being prime is clear since we need $R/\mathfrak{a}$ to be an integral domain. Suppose now that $\mathfrak{a}$ is prime, then $R/\mathfrak{a}$ is an integral domain. Now, if $\mathfrak{b}$ is an ideal in $R/\mathfrak{a}$ then we know by the lattice isomorphism theorem that $\mathfrak{b}=\mathfrak{c}/\mathfrak{a}$ for some $\mathfrak{c}\supseteq\mathfrak{a}$. But, by assumption that $R$ is a PID we know that $\mathfrak{c}=(c)$ for some $c\in R$ and it’s not hard to see then that $\mathfrak{b}=\mathfrak{c}/\mathfrak{a}=\left(c+\mathfrak{a}\right)$. $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.