LEC ALG3 4

Quotients of polynomial rings

If $k$ is a field, then $k[x]$ is what is called a Euclidean domain. A Euclidean domain is an integral domain with a “degree-like” function $N$ into $\mathbb{N}$ that makes division with remainder possible. The Euclidean algorithm can be executed in a Euclidean domain to obtain the “gcd” of two elements, and the extended Euclidean algorithm can be executed to express this gcd as a linear combination of the two elements. In general, a polynomial ring over an arbitrary ring is not a Euclidean domain. However, you can still perform Euclidean division by monic polynomials in any polynomial ring, due to some special properties¹:

Lemma 109.1(Lemma).

Let $R$ be a nonzero ring. Let $f(x)\in R[x]$ be a monic polynomial.

1. $f(x)$ is a non-zero-divisor.
2. $\text{deg}(f(x)q(x))=\text{deg}f(x)+\text{deg}q(x)$ for all polynomials $q(x)\in R[x]$.

Thus, if $f(x)\in R[x]$ is monic and $g(x)\in R[x]$ is another polynomial, there exist $q(x),r(x)\in R[x]$ such that

$$g(x)=f(x)q(x)+r(x)$$

and $\text{deg}r(x)<\text{deg}f(x)$. Again, this can performed over any ring when dividing by monic polynomials. Further, quotients and remainders are uniquely determined by $g(x)$ and $f(x)$:

Lemma 109.2(Aluffi (2009) III.4.5).

Let $f(x)$ be a monic polynomial, and assume

$$f(x)q_1(x)+r_1(x)=f(x)q_2(x)+r_2(x)$$

with both $r_1(x)$ and $r_2(x)$ polynomials of degree less than $\text{deg}f(x)$. Then $q_1(x)=q_2(x)$ and $r_1(x)=r_2(x)$.

We will now restrict ourselves to commutative rings. We have shown that, if $f(x)$ is monic, then for every $g(x)\in R[x]$, there exists a unique polynomial $r(x)$ of degree $<\text{deg}f(x)$ and such that

$$g(x)+(f(x))=r(x)+(f(x)).$$

Polynomials of degree $<d$ may be seen as elements of a direct sum $R^{\oplus d}$.

Proposition 109.3(Aluffi (2009) III.4.6).

Let $R$ be a commutative ring, and let $f(x)\in R[x]$ be a monic polynomial of degree $d$. Then the function

$$\phi :R[x]\to R^{\oplus d}$$

defined by sending $g(x)\in R[x]$ to the remainder of the division of $g(x)$ by $f(x)$ induces an isomorphism of abelian groups

$$\frac{R[x]}{(f(x))}\cong R^{\oplus d}.$$

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Example 109.4(Example).

If $f(x)=x-a$ is monic of degree $1$ for some $a\in R$, Prp 3 gives us an isomorphism

$$\frac{R[x]}{(x-a)}\cong R$$

of abelian groups $g(x)\mapsto g(a)$. It is easy to verify that this is also a ring homomorphism. Thus, for all $a,b\in R$, we have

$$\frac{R[x]}{(x-a)}\cong R\cong \frac{R[x]}{(x-b)}$$

as rings.

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Prime and maximal ideals

All rings in this section are commutative.

Definition 109.5(Definition).

Let $I\ne (1)$ be an ideal of a commutative ring $R$.

1. $I$ is a prime ideal if $R/I$ is an integral domain.
2. $I$ is a maximal ideal if $R/I$ is a field.

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Since a finite integral domain is a field, if $R/I$ is finite, then $I$ is prime iff $I$ is maximal.

Example 109.6(Example).

For all $a\in R$, the ideal $(x-a)$ is prime in $R[x]$ iff $R$ is an integral domain; it is maximal iff $R$ is also a field (Exm 4).

The ideal $(2,x)$ is maximal in $\mathbb{Z}[x]$ since

$$\begin{array}{r}\frac{\mathbb{Z}[x]}{(2,x)}{\cong }_1\frac{\mathbb{Z}[x]/(x)}{(2+(x))}{\cong }_2\frac{\mathbb{Z}}{(2)}=\mathbb{Z}/2\mathbb{Z},\end{array}$$

is a field, where ${\cong }_1$ uses this result and ${\cong }_2$ uses the isomorphism obtained in Exm 4.

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Lemma 109.7(Lemma).

A commutative ring $R$ is a field iff its only ideals are $(0)$ and $(1)$.

Proof.

Let $I\ne (0)$ be an ideal of a field $R$ and let $u$ be a non-zero element of $I$. If $v$ is the inverse of $u$, $uv=1\in I$, which implies $I=(1)$.

Conversely, assume the only ideals of a commutative ring $R$ are $(0)$ and $(1)$. Let $u\in R$ be non-invertible. Then, $(u)$ is a proper ideal of $R$, a contradiction. Thus, $R$ must be a division ring, and hence a field.□

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Proposition 109.8(Proposition).

Let $I\ne (1)$ be an ideal of a commutative ring $R$. Then

1. $I$ is prime iff for all $a,b\in R$, $ab\in I⟹(a\in I\text{ or }b\in I)$;

2. $I$ is maximal iff for all ideals $J$ of $R$, $I\subseteq J⟹(I=J\text{ or }J=R)$.

Proof.

The ring $R/I$ is an integral domain iff for all $\overline{a},\overline{b}\in R/I$, $\overline{a}\cdot \overline{b}=0⟹(\overline{a}=0\text{ or }\overline{b}=0)$. This condition translates immediately into the given condition in $R$.

The maximality condition follows from the correspondence between ideals of $R/I$ and ideals of $R$ containing $I$ and Lem 7.□

Proposition 109.9(Aluffi (2009) III.4.13).

Let $R$ be a PID, and let $I$ be a nonzero ideal in $R$. Then $I$ is prime iff it is maximal.

Proof.

Maximal ideals are prime in every ring. Let$I=(a)$ be a prime ideal in $R$, with $(a)\ne 0$, and assume $I\subseteq J$ for an ideal $J$ of $R$. As $R$ is a PID, $J=(b)$ for some $b\in R$. Since $I=(a)\subseteq (b)=J$, we have $a=bc$ for some $c\in R$. Since $I$ is prime, $b\in I$ or $c\in I$. If $b\in I$, $(b)\subseteq (a)$, and $I=J$. If $c\in I$, we have $c=da$ for some $d\in R$. But then, $a=bda$, so $db=1$ ($R$ is an integral domain). Thus, $b$ is a unit, and $(b)=J=R$.□

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Theorem 109.10(Conrad (24 C.E.) 3.3).

The intersection of all prime ideals in a nonzero commutative ring is the nilradical of the ring.

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Power series rings

The set of all formal power series in $x$ with coefficients in a commutative ring $R$ constitute another ring $R[[x]]$, called the ring of formal power series in the variable $x$ over $R$.

As a set, $R[[x]]$ can be constructed as the set $R^{\mathbb{N}}$. Addition is defined componentwise, and multiplication is defined to be the Cauchy product:

$$(a_n{)}_{n\in \mathbb{N}}\times (b_n{)}_{n\in \mathbb{N}}={\left(\sum _{k=0}^n{a}_k{b}_{n-k}\right)}_{n\in \mathbb{N}}.$$

With these operations, $R^{\mathbb{N}}$ becomes a commutative ring with zero $(0,0,0,\dots )$ and identity $(1,0,0,\dots )$.

Lemma 109.11(Lemma).

Let $R$ be a commutative ring. A formal power series $f(x)={\sum }_{n\ge 0}{a}_n{x}^n$ is a unit in $R[[x]]$ iff its constant coefficient $a_0$ is a unit in $R$.

Proof.

Suppose $a_0$ is a unit in $R$. Then, we can write $f(x)=a_0^{-1}(1+a_1{a}_0^{-1}x+a_2{a}_0^{-1}{x}^2+\dots )$, the inverse of which can be constructed recursively. Conversely, if $a_0$ is not a unit in $R$ and $g(x)={\sum }_{n\ge 0}{b}_n{x}^n$ is the inverse of $f(x)$, we have $a_0{b}_0=1$, a contradiction.□

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Theorem 109.12(Theorem).

If $K$ is a field, then $K[[x]]$ is a PID.

Proof.

Define the order of a nonzero series

$$\mathrm{ord}(f)=\min n:a_n\ne 0.$$

Properties: $\mathrm{ord}(0)=\infty$, and $\mathrm{ord}(fg)=\mathrm{ord}(f)+\mathrm{ord}(g)$.

Now let $I$ be a nonzero ideal in $K[[x]]$. Choose $f\in I$ with minimal order $n:=\mathrm{ord}(f)$. Then $f=x^{n}u$ where $u$ has order $0$, so $u$ is a unit by Lem 11. Hence $x^n=u^{-1}f\in I$. Now for any $g\in I$, $\mathrm{ord}(g)\ge n$ by minimality, so $g=X^{n}h$ for some $h\in K[[X]]$. Thus

$$I\subseteq (X^n).$$

Since $X^n\in I$, we also have $(X^n)\subseteq I$. So $I=(X^n)$.□

Side note: By Lem 11, the set of non-units is precisely the the ideal $(x)$. By ^a2d371, $(x)$ is the unique maximal ideal of $K[[x]]$.

Footnotes

1. Note that this does not mean the Euclidean algorithm works: the remainder may not be monic! ↩

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

Conrad, K. (24 C.E.). ZORN’S LEMMA AND SOME APPLICATIONS. https://kconrad.math.uconn.edu/blurbs/zorn1.pdf