Abstract Nonsense | My little journal

Some basic properties of tensor product

Posted by Doralikesmath on May 20, 2012 in Commutative Algebra | 0 comments

After spent quite a bit of time teaching commutative algebra myself, I realized that one can go nowhere without deeply understanding of basic constructions and properties of elementary objects in algebra. Thus, in this post, I’m going to summarize my knowledge about tensor product.

1. Tensor product of modules

Proposition 1.1

Given two $A$ -module $M$ and $N$ , there exists an $A$ -module denoted by $M\otimes_A N$ or simply $M\otimes N$ and an $A$ -bilinear mapping $f:M\times N \rightarrow M\otimes N$ such that given any $A$ -module $P$ with an $A$ -bilinear mapping $g: M\times N \rightarrow P$ , there exists a unique $A$ -linear map $\varphi: M\otimes N \rightarrow P$ such that $g=\varphi\circ f$ . Moreover, the pair $(M\otimes N,f )$ is unique up to isomorphism in the sense that if $(T, f')$ has the same properties then there exists a unique isomorphism $j:M\otimes N\rightarrow T$ such that $f'=j\circ f$ .

We have the following commutative diagram:

$\xymatrix{ M\times N \ar@{->}[rr]^g \ar@{->}[rd]^f & &P\\ &M\otimes N\ar@{->}[ru]^{\varphi} }$

Proof:

a. The existence:

Let $C$ be the free module $C=A^{M\times N}$ , the set of all formal linear combinations of elements of $M\times N$ with coefficients in $A$ . Let $D$ be the submodule of $C$ generated by all elements of the form

$(x+x',y)-(x,y)-(x',y)$

$(x,y+y')-(x,y)-(x,y')$

$(ax,y)-a(x,y)$

$(x,ay)-a(x,y)$

Then let $M\otimes N=C/D$ and let $x\otimes y$ denote the image of $(x,y)$ in $M\otimes N$ . Also let $f:M\times N\rightarrow M\otimes N$ be the mapping defined by $f(x,y)=x\otimes y$ . It’s easy to see that

$(x+x')\otimes y=x\otimes y+x'\otimes y$

$x\otimes (y+y')=x\otimes y+x\otimes y'$

$(ax)\otimes y=x\otimes (ay)=a(x\otimes y)$

Thus $f$ is bilinear. Now any bilinear map $g:M\times N \rightarrow P$ can be extend by linearity to an $A$ -linear map $\overline{g}: C\rightarrow P$

$\xymatrix{ C \ar@{->}[rr]^{\overline{g}} \ar@{->}[rd]^{\overline{f}} & &P\\ &M\otimes N\ar@{->}[ru]^{\varphi} }$

By our definition of $D$ , $\overline{g}$ vanishes on $D$ , i.e., $D\subset \ker \overline{g}$ . Hence, $\overline{g}$ induces a well-defined homomorphism $g':M\otimes N=C/D\rightarrow P$ such that $g'(x\otimes y)=g(x,y)$ . The mapping $g'$ is uniquely determined by this construction.

b. Uniqueness:

Suppose that $(T,f')$ also satisfies our properties. Then we have the following commutative diagram

$\xymatrix{ M\times N \ar@{->}[rr]^{f'} \ar@{->}[rdd]^f & &T\ar@<1ex>[ldd]^{\varphi'}\\ \\ &M\otimes N\ar@{->}[ruu]^{\varphi} }$

The uniqueness follow from our diagram.

$\square$

Proposition 1.2:

Let $M$ , $N$ , and $P$ be $A$ -modules. The following results hold:

$M\otimes N\cong N\otimes M$ .
$A\otimes M=M$ .
$(M\otimes N)\otimes P=M\otimes N\otimes P=M\otimes (N\otimes P).$
$(M\oplus N)\otimes P=(M\otimes P)\oplus (N\otimes P)$ .

Proof:

$\square$

Proposition 1.3:

Let $A$ and $B$ be two rings and $P$ be a two sided $A,B$ - module. Then for given $A$ -module $M$ and $B$ -module $N$ , we have

$\Hom_A(M,\Hom_B(P,N))\cong \Hom_B(M\otimes_A P, N)$ .
$(M\otimes_A P)\otimes_B N\cong M\otimes_A(P\otimes_B N)$

Proof:

$\square$

On Hensel’s Lemma (2)

Posted by Doralikesmath on May 12, 2012 in Algebraic Number Theory, Commutative Algebra | 0 comments

Most of the results in this post can be found in NT-[1]

II. Valued Fields

Definition 2.1:

Let $F$ be an arbitrary field. An absolute value on $F$ is a map $|\cdot|:F\rightarrow \RR$ with the following properties:

(V1) $|0|=0$ , $|a|>0$ for all $a\in F$ with $a\neq 0$ ;
(V2) $|ab|=|a||b|$ for all $a,b\in F$ ;
(V3) $|a+b|\le |a|+|b|$ for all $a,b\in F$ .

A non-archimedean absolute value on $F$ is a map $|\cdot|:F\rightarrow \RR$ with properties (V1), (V2), and

(V4) $|a+b|\le \max\{|a|,|b|\}$ for all $a,b\in F$ .

An absolute value is called trivial if

$|0|=0$ , and
$|a|=1$ for all $a\neq 0$ .

Definition 2.2:

Let $F$ be a field with absolute value $|\cdot|$ and let $E$ be a vector space over $F$ . A norm on $E$ is a map $\Vert\cdot\Vert: E\rightarrow \RR$ with the following properties:

$\Vert a\Vert>0$ for all $a\in E$ with $a\neq 0$ ;
$\Vert\alpha a\Vert=|\alpha|\Vert a\Vert$ for all $\alpha\in F$ and $a\in E$ ;
$\Vert a+b\Vert\le \Vert a \Vert +\Vert b\Vert$ for all $a,b\in E$ >

From now on, $F$ will denote a field with a complete non-archimedian absolute value $|\cdot|$ , $R$ and $M$ denotes the corresponding valuation ring and valuation ideal, and $k=R/M$ is the residue field.

Proposition 2.3:

Let

$f(x)=\sum_{j=0}^nc_jx^j\in F[x].$

If $c_0c_n\neq 0$ and if $|c_0|\le |c_m|$ , $|c_n|\le |c_m|$ for some $0<m<n$ , with at least one of the two inequalities is strict, then $f$ is reducible over $F$ .

Proof:

Suppose that $|c_0|<|c_m|$ and $|c_n|\le |c_m|$ . With out loss of generality, we can choose $m=\max_{0< i< n}\{|c_i|\}$ and $|c_i|<|c_m|$ for $0\le i< m$ . We can also multiple $f$ by $c_m^{-1}$ to further assume that $f\in R[x]$ , $c_m=1$ and $|c_i|<1$ for $0\le i<m$ . Hence

$\overline{f}=t^m\left(\overline{c}_nt^{n-m}+\overline{c}_{n-1}t^{n-m-1}+\cdots+1\right).$

Since the two factors of $\overline{f}$ are relatively prime, by theorem 1.1, $f$ is reducible.

If $|c_n|<|c_m|$ and $|c_0|\le |c_m|$ , consider the polynomial $g(t)=t^nf(t^{-1})$ . We have

$\overline{g}=t^{n-m}\left(\overline{c}_0t^{m}+\overline{c}_1t^{m-1}+\cdots+1\right).$

By the same argument, $g$ is reducible, which implies $f$ is also reducible.

$\square$

Below is a direct corollary of this proposition:

Corollary 2.4:

If $f(x)=a_nx^n+\cdots+a_1x+a_0\in k[x]$ is irreducible with $a_0a_n\neq 0$ , then

$|f|:=\max\{|a_0|,|a_1|,...,|a_n|\}=\max\{|a_0|,|a_n|\}.$

Proposition 2.5:

Let $E$ be a finite extension of $F$ , then the absolute value on $F$ can be extended to an absolute value on $E$ .

Proof:

Regard $E$ as a vector space of dimension $n<\infty$ over $F$ , for $a\in E$ , let $T_a:E\rightarrow E$ be the linear operator defined by $T_av=av$ for all $v\in E$ . Then $\det T_a\in F$ and we claim that

$|a|:=|\det T_a|^{1/n}$

is an absolute value on $E$ .

Clearly $|a|\ge 0$ with equality holds if and only if $a=0$ . Also, $|ab|=|a||b|$ since $T_{ab}=T_aT_b$ , so that $\det (T_{ab})=\det(T_a)\det(T_b)$ . Moreover, if $a\in F$ then $T_a=aI_n$ , so that $|a|=|\det( aI_n)|^{1/n}$ coincides with the original absolute value on $F$ . Therefore, we only need to prove that

$|a-b|\le \max\{|a|, |b|\} \hspace{5mm}\text{ for all }a,b\in F.$

Without loss of generality, we may assume that $|a|\le |b|$ . Dividing both sides of the above inequality by $b$ , we see that it suffices to show that $0<|a|\le 1$ implies $|1-a|\le 1$ . Let

$f(t)=\det (xI-T_a)=x^n+\sum_{j=0}^{n-1}a_jx^j$

be the characteristic polynomial of $T_a$ . Then $c_i\in F$ for all $i$ and $c_0=(-1)^n\det (T_a)$ . Let $g$ be the minimal polynomial of $a$ over $F$ , then $g$ is irreducible in $F[x]$ . We have then $g$ is the minimal polynomial of $T_a$ . Because the minimal and characteristics polynomials of $T_a$ share the same roots in the algebraic closure of $F$ and $\deg g\le\deg f\le n$ , we have then $f|g^n$ . Since $g$ is irreducible, $f(x)=g(x)^r$ for some $r\in\NN^*$ .

Suppose that

$g(x)=x^m+\sum_{j=0}^{m-1}b_jx^j \in F[x]$

and let $a\in E$ satisfying $|a|\le 1$ with respect to the proposed absolute value. Then $|a_0|=|\det T_a|\le 1$ and hence, since $b_0^r=a_0$ , $|b_0|\le 1$ . Again since $g$ is irreducible, by proposition 2.3, we have $|b_j|\le 1$ for all $j$ . Now since

$g(1)=1+b_{m-1}+\cdots+b_0,$

we have by the property of absolute value on $F$ ,

$|g(1)|\le \max\{|1|, |b_0|,\cdots, |b_{m-1}|\}=1,$

which implies $|\det (I-T_a)|=|f(1)|\le 1$ . This shows that $|1-a|\le 1$ . The conclusion follows.

$\square$

We will now show that such extensions of the absolute value is unique. First consider the case the absolute value on $F$ is trivial.

Lemma 2.6:

Proof:

Suppose that $x,y\in E$ , then

$\begin{align*} |x+y|^n&=\left|(x+y)^n\right|=\left|\sum_{j=0}^n\binom{n}{j}x^jy^{n-j}\right|\\ &\le \sum_{j=0}^n\left|\binom{n}{j}x^jy^{n-j}\right| \le \sum_{j=0}^n\left|x^jy^{n-j}\right|\\ &\le (n+1)\max\{|x|, |y|\}^n. \end{align*}$

Thus

$|x+y|\le \sqrt[n]{n+1}\max\{|x|, |y|\}.$

Let $n\rightarrow \infty$ , we have

$|x+y|\le \max\{|x|, |y|\}.$

$\square$

Proposition 2.7:

Suppose that $E$ is a finite extension of $F$ . Then the trivial absolute value on $F$ extends uniquely to the trivial absolute value on $E$ .

Proof:

Because the absolute value on $F$ is trivial, then $|n|\le 1$ for all positive integers $n\in E$ . The above lemma implies that the extended absolute value on $E$ is non-archimedian.

Since $n=[E:F]<\infty$ , the extension is algebraic. Suppose that $\alpha\in E$ and $|\alpha|>1$ , then

$\alpha^n+\sum_{j=0}^{n-1}c_j\alpha^j=0$

$\begin{align*} 0&=\left|\alpha^n+\sum_{j=0}^{n-1}c_j\alpha^j\right|\le \max\{|\alpha^n|, |c_{n-1}\alpha^{n-1},...,|c_0|\}\\ &\le\max\{|\alpha^j|: 0\le j\le n\}=|\alpha^n|. \end{align*}$

Therefore, the extended absolute value on $E$ is also trivial. This shows the uniqueness of such extension.

$\square$

Now consider the case when the absolute value on $F$ is nontrivial. For that, we need two following lemmas:

Lemma 2.8:

Let $E$ be a finite-dimensional vector space over $F$ . If $\Vert\cdot\Vert_1$ and $\Vert\cdot\Vert_2$ are both norms on $E$ , then there exist positive constants $\sigma, \mu$ such that

$\sigma\Vert a\Vert_1\le \Vert a\Vert_2\le \mu\Vert a\Vert_1 \text{ for every } a\in E.$

Proof:

Let $\basis=\{e_1,...,e_n\}$ be a basis for the vector space $E$ over $F$ . Then any $a\in E$ can be uniquely expressed in the form

$a=\sum_{i=1}^n\alpha_ie_i.$

We can verify that

$\Vert a\Vert=\max_{1\le i\le n}|\alpha_i|$

is a norm on $E$ . We first prove the lemma with $\Vert\cdot\Vert_2=\Vert\cdot\Vert$ . Since

$\Vert a\Vert_1\le \Vert a\Vert\left(\Vert e_1\Vert_1+\cdots+\Vert e_n\Vert_1\right)$

we can take $\sigma=\left(\Vert e_1\Vert_1+\cdots+\Vert e_n\Vert_1\right)^{-1}$ .

Now we will use induction to show the existence of $\mu$ . The case $n=1$ is obviously true. We assume that such $\mu$ exists if $\dim_F E\le n-1$ . Suppose that for $\dim_FE=n$ , there exists a sequence $a_k\in E$ such that

$\Vert a_k\Vert_1<\epsilon_k\Vert a_k\Vert,$

where $\epsilon_k>0$ and $\epsilon_k\rightarrow k$ as $k\rightarrow \infty$ . With out loss of generality, we may suppose that

$|\alpha_n^{(k)}|=\Vert a_k\Vert$

and also by replacing $a_k$ by $(\alpha_n^{(k)})^{-1}a_k$ , that $\alpha_n^{(k)}=1$ . Thus $a_k=b_k+e_n$ , where

$b_k=\sum_{i=1}^{n-1}\alpha_i^{(k)}e_i,$

and $\Vert a_l\Vert_1\rightarrow 0$ as $k\rightarrow \infty$ . The sequences $\alpha_i^{(k)}$ $(i=1,...,n-1)$ are Cauchy sequences in $F$ , since

$\Vert b_j-b_k\Vert_1\le \Vert b_j+e_n\Vert_1+\Vert b_k+e_n\Vert_1=\Vert a_j\Vert_1+\Vert a_k\Vert_1$

and, by the induction hypothesis

$|\alpha_i^{(j)}-\alpha_i^{(k)}|\le \mu_{n-1}\Vert b_j-b_k\Vert_1 \hspace{5mm} (i=1,...,n-1).$

Hence, since $F$ is complete, there exists $\alpha_i\in F$ such that $|\alpha_i^{(k)}-\alpha_i|\rightarrow 0$ for $i=1,...,n-1$ . Let

$b=\alpha_1e_1+\cdots+\alpha_{n-1}e_{n-1}.$

Since $\Vert b_k-b\Vert_1\le \sigma^{-1}_{n-1}\Vert b_k-b\Vert$ , it follows that $\Vert b_k-b\Vert_1\rightarrow 0$ . Let $a=b+e_n$ , then

$\Vert a\Vert_1\le \Vert a-a_k\Vert_1+\Vert a_k\Vert_1=\Vert b-b_k\Vert_1+\Vert b_k\Vert_1.$

Let $k\rightarrow \infty$ , we have $a=0$ , a contradiction since $a=b+e_n\neq 0$ . Thus the lemma holds for $\Vert\cdot\Vert_1$ and $\Vert\cdot\Vert$ , that is, for some positive constants $\sigma_1$ and $\mu_1$

$\sigma_1\Vert a\Vert_1\le \Vert a\Vert\le \mu_1\Vert a\Vert_1 \text{ for every } a\in E.$

Similarly, we have

$\sigma_2\Vert a\Vert_2\le \Vert a\Vert\le \mu_2\Vert a\Vert_2 \text{ for every } a\in E$

for some positive constants $\sigma_2$ and $\mu_2$ . Then let $\sigma=\frac{\sigma_1}{\mu_2}$ and let $\mu=\frac{\mu_1}{\sigma_2}$ , we have

$\sigma\Vert a\Vert_1\le \Vert a\Vert_2\le \mu\Vert a\Vert_1 \text{ for every } a\in E.$

The conclusion follows.

$\square$

Lemma 2.9:

Let $|\cdot|_1$ and $|\cdot|_2$ be two absolute values on $F$ satisfying $|a|_2<1$ for any $a\in F$ such that $|a|_1<1$ . If $|\cdot|_1$ is nontrivial then there exists a real number $\rho>0$ such that

$|a|_2=|a|_1^{\rho} \hspace{5mm}\text{ for every } a\in F.$

Proof:

By taking inverses, we see that $|a|_2>1$ for any $a\in F$ such that $|a|_1>1$ . Choose $b\in F$ with $|b|_1>1$ . For any nonzero $a\in F$ , we have $|a|_1=|b|_1^{\gamma}$ for

$\gamma=\frac{\log |a|_1}{\log |b|_1}.$

$\frac{m}{n}>\frac{\log |a|_2}{\log |b|_2}.$

Similarly, if $m'$ , $n'$ are integers with $n'>0$ such that $m'/n'< \gamma$ then

$\frac{m'}{n'}<\frac{\log |a|_2}{\log |b|_2}.$

It follows that

$\frac{\log |a|_1}{\log |b|_1}=\gamma=\frac{\log |a|_2}{\log |b|_2}.$

Therefore, if $\rho=\frac{\log |b|_2}{\log |b|_1}$ , then $\rho>0$ and $|a|_2=|a|_1^{\rho}$ . For $a=0$ , this also holds. We have the desired result.

$\square$

Proposition 2.10:
Let the field $E$ be a finite extension of $F$ . Then there is at most one extension of the absolute value on $F$ to $E$ , and $E$ is necessarily complete with respect to the extended absolute value.

Proof:

Let $\{e_1,...,e_n\}$ be a basis for the vector space $E$ over $F$ . Then any $a\in E$ can be uniquely expressed in the form

$a=\sum_{i=1}^n\alpha_ie_i.$

By lemma 2.8, for any extended absolute value, there exists $\sigma, \mu>0$ such that

$\sigma|a|\le \max_{1\le i\le n}|\alpha_i|\le \mu|a| \hspace{5mm}\text{ for any } a\in E.$

We first show that $E$ is complete. Suppose that $a_k$ is a Cauchy sequence is $E$ , and

$a_k=\sum_{j=1}^n\alpha_{k,j}e_i$

for some $\alpha_{k,j}\in F$ . Then $a_{k,j}$ is a Cauchy sequence for each $j\in [1,n]\cap \ZZ$ . Since $F$ is complete, there exists $\alpha_j\in F$ such that $\alpha_{k,j}\rightarrow \alpha_j$ when $k\rightarrow \infty$ ( $i=1,2,...,n$ ). Thus $a_k\rightarrow a$ , where $a=\alpha_1e_1+\cdots+\alpha_ne_n$ .

We will now show the uniqueness of the extension. The case of the trivial absolute value on $F$ is already proved above. Suppose now that the absolute value on $F$ is nontrivial. For a fixed $a\in F$ , consider the powers $a, a^2,...$ For each $k$ we can write

$a_k=a_{k,1}e_1+\cdots+a_{k,n}e_n.$

Since $|a|<1$ if and only if $|a_k|\rightarrow 0$ , its follows that $|a|<1$ if and only if $|a_{k,j}|\rightarrow 0$ for $j=1,2,...,n$ . This condition is independent of the absolute value on $E$ . Thus if there exists two absolute value $|\cdot|_1$ and $|\cdot|_2$ , which extend to the absolute values on $F$ , then $|a|_1<1$ if and only if $|a|_2<1$ . Hence, by lemma 2.9, there exists a positive number $\rho$ such that

$|a|_2=|a|_1^{\rho} \hspace{5mm}\text{ for every } a\in E.$

Since for some $a$ in $F$ , we must have $|a|_1=|a|_2>1$ , $\rho$ must be equal to $1$ . Therefore, $|\cdot|_1=|\cdot|_2$ .

$\square$

We finish with the following corollary:

Corollary 2.11:
Let $K$ be a finite extension of $\QQ_p$ . Then $|\cdot|_p$ extended to $K$ is a non-archimedian norm.

Proof:

The existence of the a non-archimedian norm follows from proposition 2.5 and its proof. By the above proposition, such an extension is unique. The conclusion follows.

$\square$

On Hensel’s Lemma (1)

Posted by Doralikesmath on May 11, 2012 in Algebraic Number Theory, Commutative Algebra | 0 comments

This is my final report for the ANT class I’m taking this semester. I do not claim originality of the following material, and most of the results in this part can be found in NT-[4] (See the References page).

I. Hensel’s Lemma and Its Corollaries

Theorem 1.1 (Hensel’s Lemma)

Let $(A,\m,k)$ be a local ring, and suppose that $A$ is $\m$ -adically complete. Let $F\in A[x]$ be a monic polynomial, we denote $\overline{F}$ the image of $F$ under the canonical homomorphism $A[x]\rightarrow k[x]$ . If there are monic polynomials $g,h\in k[x]$ with $\gcd(g,h)=1$ such that $\overline{F}=gh,$ then there exist monic polynomials $G,H\in A[x]$ satisfying

$F=GH,\hspace{2mm}\overline{g}=G, \hspace{2mm}\text{ and }\hspace{2mm} \overline{h}=H.$

Proof:

Suppose $G_1, H_1\in A[x]$ are such that $g=\overline{G_1}$ and $h=\overline{H_1}$ then $F\equiv G_1H_1\pmod{\m[x]}$ . Suppose by induction that we can construct monic polynomials $G_n, H_n\in A[x]$ such that $F\equiv G_nH_n\pmod{\m^n[x]}$ , and $g=\overline{G_n}$ , $h=\overline{H_n}$ . Then we can write

$F-G_nH_n=\sum_{i}\omega_iU_i(x),\text{ with } \omega_i\in \m^n \text{ and } \deg U_i<\deg F.$

Since $\gcd(g,h)=1$ , there exist $v_i, w_i\in k[x]$ such that $\overline{U_i}=gv_i+hw_i$ . Suppose that $v_i=hq_1+v_i'$ and $w_i=gq_2+w_i'$ , we have

$\begin{align*} \overline{U_i}&=g(hq_1+v_i')+h(gq_2+w_i')\\ &=h(g(q_1+q_2)+w_i')+gv_i'. \end{align*}$

Therefore, we can replace $v_i$ by its remainder modulo $h$ and adjust $w_i$ correspondingly to assume that $\deg v_i<\deg h$ . Then

$\deg hw_i=\deg(\overline{U_i}-gv_i)<\deg F \Rightarrow \deg w_i<\deg g.$

Choosing $V_i, W_i\in A[x]$ such that $\overline{V_i}=v_i$ , $\deg V_i=\deg v_i$ , $\overline{W_i}=w_i$ , $\deg W_i=\deg w_i$ , and letting

$G_{n+1}=G_n+\sum_{i}\omega_iW_i,\hspace{5mm} H_{n+1}=H_n+\sum_{i}\omega_iV_i,$

give us

$F\equiv G_{n+1}H_{n+1}\pmod{\m^{n+1}[x]}.$

Therefore, by the principal of mathematical induction, we can construct sequences of polynomials $G_n, H_n$ for $n=1,2,...$ as above. Then $\lim G_n=G$ and $\lim H_n=H$ clearly exist and satisfy $F=GH$ . Obviously, $\overline{G}=\overline{G_1}=g$ and $\overline{H}=\overline{H_1}=h$ . The conclusion follows.

$\square$

From Hensel’s lemma, we have a simple looking but surprisingly strong corollary as follows:

Corollary 1.2:

Let $(A,\m,k)$ be a complete local ring and $f\in A[x]$ a monic polynomial. Suppose that $\overline{f}(x)$ has a simple root $\alpha\in k$ , then there exists $a\in A$ such that $f(a)=0$ and $\overline{a}=\alpha$ .

Proof:

In $k[x]$ , we have $\overline{f}(x)=(x-\alpha)t(x)$ for some $t(x)\in k[x]$ not divisible by $x-\alpha$ . By Hensel’s lemma, there exist monic polynomials $g,h\in A[x]$ such that $f=gh$ , $\deg g=1$ , $\overline{g}=x-\alpha$ , and $\overline{h}=t(x)$ . Thus $g(x)=x-a$ for some $a\in A$ and $\overline{a}=\alpha$ . Therefore, $f(a)=0$ .

$\square$

We will now show some applications of corollary 1.2.

Corollary 1.3:

Let $f(x)\in \ZZ_p[x]$ and $f'(x)$ its formal derivative. If $f(x)\equiv 0\pmod p$ has a solution $a_1$ such that $f'(a_1)\not\equiv 0\pmod p$ then there exists a unique $p$ -adic integer $a$ such that $f(a)=0$ and $a\equiv a_1\pmod p$ .

Proof:

Since $f'(a_1)\neq 0$ , $a_1$ is a simple root of $\overline{f}(x)\in (\ZZ_p/\<p\>)[x]$ . By corollary 1.2, the conclusion follows.

$\square$

Example 1.4:

The polynomial $x^{p-1}-1\in \ZZ_p[x]$ splits over the residue class field $\ZZ_p/p\ZZ_p=\FF_p$ into distinct linear factors. Repeatedly applying corollary 1, we also have that it splits into linear factors in $\ZZ_p$ . Thus we obtain the result that the field $\QQ_p$ contains the $(p-1)-th$ roots of unity.

Example 1.5:

Consider $A=\CC[[z]]$ , the formal series ring of one variable over the complex numbers. Let $f(x)=x^2-(z+1)\in A[x]$ , then in $A/z\CC[[z]]$ , we have $\overline{f}=x^2-1=(x-1)(x+1)$ , which has two simple roots $1$ and $-1$ . By corollary 1.2, $f(x)=x^2-(z+1)$ factors as $(x-\alpha(z))(x-\beta(z))$ where $\alpha(z)$ and $\beta(z)\in \CC[[z]]$ . This shows the existence of the power series square roots for $z+1$ in $\CC[[z]]$ . More generally, we have the following theorem:

Theorem 1.6: (Implicit function theorem)

Let $k$ be a field and $A=k[[x_1,...,x_d]]$ . Let $p(x)=x^n+\sum_{i=0}^{n-1}a_ix^i\in A[x]$ . Suppose that the polynomial $x^n+\sum_{i=0}^{n-1}a_i(0)x^i$ has a simple root $\alpha\in k$ . Then there exists a power series $g\in A$ such that $g(0)=\alpha$ and $p(g)=0$ . In particular, if $n$ is a possible integer relatively prime to $\ch(k)$ and $f\in A$ is a power series whose constant term is an $n$ -th power in $A$ , then $f$ is an $n$ -th power in $A$ , i.e. $f=h^n$ for some $h\in A$ .

Proof:

Let $\m=\<x_1,...,x_d\>$ , then $(A,\m,k)$ is a complete local ring with respect to the $\m$ -adic topology. Also if $q=\sum_{j=0}^{\infty}a_jx^k\in A$ then $a_0=q(0)=\overline{q}$ . Then by corollary 1.2, there exists a power series $g\in A$ such that $\overline{g}=g(0)=\alpha$ and $p(g)=0$ .

For the second statement, consider the polynomial

$q(x)=x^n-f\in k[x],$

where $\gcd(n, \ch k)=1$ . We have $q'(x)=nx^{n-1}\neq 0$ because $\gcd(n, \ch k)=1$ . Therefore, $\gcd(q,q')=1$ since $f^{-1}(nq-xq')=1$ . Then $q(x)$ has a simple root. By the first statement, there exists $h\in A$ such that $q(h)=0$ , i.e. $h^n=f$ .

$\square$

Definition 1.7:

Let $(A,\m, k)$ be a complete local ring, $F\subset A$ is a field. Since $(A,\m, k)$ is local and each non-zero element in $F$ is a unit in $A$ , we have $F\cap \m=(0)$ . Thus $F=F/(0)=F/(F\cap \m)$ maps isomorphically to the image $\overline{ F}$ of $F$ in $A/\m=k$ . If the image of $F$ in $k$ is equal to $k$ , we say that $F$ is a coefficient field for $A$ .

Example 1.8:

Let $A=k[x_1,...,x_d]$ for $k$ is a field, then $\m=\<x_1,...,x_d\>$ and $A/\m=k$ . We have $k$ is a coefficient field for $A$ .

We will now attempt to prove two special cases of the Cohen’s structure theorem in which they are corollaries of Hensel’s lemma:

Corollary 1.9:

Let $(A,\m,k)$ be a complete local ring which is Hausdorff in its $m$ -adic topology. Suppose that $|k|<\infty$ , say $|k|=q=p^n$ for $p$ is a prime number and $n\in \NN^*$ . Then $x^{q-1}-1$ has distinct roots in $A$ , i.e., $A$ contains the multiplicative group $V$ of $(p^n-1)$ -th roots of unity. If in addition, $A$ contains a field, then $V\cup\{0\}$ is equal to $\FF_q$ , the field with $q$ elements, and $\FF_q$ is a coefficient field for $A$ .

Proof:

We have every non-zero element in $k$ is a root of the equation $x^{q-1}-1=0$ , they are all simple roots. By corollary 1.2, they lift to distinct elements in $A$ which form a multiplicative group $V$ . If $A$ contains a field, then since this field maps into $k$ , it must have characteristic $p$ . Now for any $a,b\in V$ , we have

$(a+b)^{p^n}=\sum_{j=0}^{p^n}\binom{p^n}{j}a^jb^{p^n-j}=a^{p^n}+b^{p^n}=a+b,$

so that either $a+b=0$ or $(a+b)^{p^n-1}-1=0$ . Therefore, we have $V\cup \{0\}=\FF_q$ . Hence, the image of $\FF_q$ in $k$ has $q$ element, which implies $\overline{\FF_q}=k$ . Therefore, $\FF_q$ is a coefficient field. The conclusions follow.

$\square$

Corollary 1.10:

Let $(A,\m,k)$ be a complete, Hausdorff local ring containing a field of characteristic zero. Then $A$ has a coefficient field.

Proof:

Let $L\subset A$ be a field with characteristic zero. Let $\mathfrak{C}$ denote the set of subfields of $A$ containing $L.$ By Zorn’s lemma, $\mathfrak{C}$ contains a maximal element $F\in \mathfrak{C}$ with respect to the inclusion, i.e. $F$ is a maximal subfield of $A$ . We will now show that $F$ is a coefficient field. Suppose not, then the image $\overline{F}$ of $F$ in $k$ is a proper subfield of $k$ . Take $\alpha\in k$ which is not in $\overline{F}$ . We have two cases as follows:

$\alpha$ is not algebraic over $\overline{F}$ . Let $a\in A$ be such that $\overline{a}=\alpha$ on $k$ . We then have $a$ is not algebraic over $F$ (in $A$ ); otherwise, $\alpha=\overline{a}$ is algebraic over $k$ . Then for all $f\in F[x]$ , we have $\overline{f}(\alpha)\neq 0$ , so that $f(a)\notin \m$ . Hence, $f(a)$ is a unit in $A$ for all $f\in F[x]$ . It follows that $F(a)$ , the rational function field in the indeterminate $a$ , is contained in $A$ , contradicting the maximality of $F$ .
$\alpha$ is algebraic over $\overline{F}$ . Let $f\in F[x]$ be such that $\overline{f}(x)$ is the minimal polynomial for $\alpha$ over $\overline{F}$ . Since $\ch(F)=0$ , $\alpha$ is a simple root. By corollary 1.2, there exists $a\in A$ such that $\overline{a}=\alpha$ and $f(a)=0$ . We also have $f$ is irreducible over $F$ , thus $F[x]/\<f(x)\cong F[a]$ is a field in $A$ properly containing $F$ , a contradiction.

Combine all of the above cases, we have $\overline{F}=k$ , so that $F$ is a coefficient field for $k$ .

$\square$

NAK’s lemma

Posted by Doralikesmath on Apr 20, 2012 in Commutative Algebra | 0 comments

The following entry is about Nakayama-Azumaza-Krull (NAK) lemma and some corollaries. About the origin of this name, see [1], pp.8.

Theorem 1:

Suppose that $M$ is an $R$ -module generated by $n$ elements, and that $\varphi\in \Hom_R(M,M)$ . Let $I$ be an ideal of $R$ such that $\varphi(M)\subset IM$ . Then there exist $a_i\in I^i$ for $i\in[1,n]\cap \ZZ$ such that

$\varphi^n+\sum_{i=1}^na_i\varphi^{n-i}=0$

as an element of $\Hom_R(M,M)$ .

Proof:

Suppose that $M=\sum_{i=1}^nr_iM$ for some $a_i\in I$ ( $1\le i\le n$ ). Then we have for each $i\in [1,n]\cap\ZZ$ ,

$\varphi(a_i)=\sum_{j=1}^na_{ji}r_j$

for some $a_{ji}\in I$ . Thus

$\sum_{j=1}^n(\delta_{j,i}\varphi-a_{j,i})r_j=0.$

Let $A=(\delta_{j,i}\varphi-a_{j,i})$ . We have then $A(r_1,...,r_n)^T=0$ , so that $\Adj(A)\times A(r_1,...,r_n)^T=0$ . Thus $\det(A)(r_1,...,r_n)^T=0$ . Since $r_1,...,r_n$ generate $M$ as an $R$ -module, they are not all zero. Therefore, $\det(A)=0$ . By expanding $\det (A)$ , we have an expression of the form

$\varphi^n+\sum_{i=1}^na_i\varphi^{n-i}=0$

for some $a_i\in I^i$ where $i\in[1,n]\cap \ZZ$ .

$\square$

Remark:

If $M$ is a free $R$ -module with finite basis and $I=R$ , we have the classical Cayley-Hamilton theorem.

Theorem 2: (NAK lemma)

Let $M$ be an $R$ -module that is generated by $n$ elements. If $I\subset R$ is an ideal such that $IM=M$ then there exists $a\in R$ such that $aM=0$ and $a\equiv 1\pmod I$ . If in addition $I\subset\rad(R)$ then $M=0$ .

Proof:

Let $\varphi$ be the identity mapping in the previous theorem, then we have and expression

$\id^n+\sum_{i=1}^na_i\id^{n-i}=\left(1+\sum_{i=1}^na_i\right)\id=0$

as an $R$ -module homomorphism. Thus let $a=1+\sum_{i=1}^na_i$ , we have $a\equiv 1\pmod I$ and $am=0$ for all $m\in M$ . Thus $aM=0$ .

Now if $I\subset \rad(R)$ , then since $a\equiv 1\pmod I$ , we have that $a$ is a unit in $R$ . This follows from the fact that $x\in \rad(R)$ if and only if $x$ is a unit in $R$ . Therefore, $M=0$ .

$\square$

Corollary 1:

Let $I$ be a ring contained in $\rad(R)$ . Suppose that $M$ is an $R$ -module and $N\subset M$ a submodule such that $M/N$ is finitely generated over $R$ . Then $M=N+IM$ implies $M=N$ .

Proof:

Suppose that $M=N+IM$ . Then in $M/N$ , we have $\overline{M}=I\overline{M}$ , thus by NAK lemma, we have $\overline{M}=0$ , which implies $M=N$ .

$\square$

Corollary 2:

Let $I$ be a ring contained in $\rad(R)$ . Suppose that $M$ is an $R$ -module. If $m_1,...,m_n\in M$ have images in $M/IM$ that generate it as an $R$ -module, then $m_1,...,m_n$ generate $M$ as an $R$ -module.

Proof:

Let $N=\sum_{i=1}^nRm_i$ . By the hypothesis, $M=N+IM$ so that $M=N$ by corollary 1.

$\square$

Corollary 3:

If $M$ and $N$ are finitely generated $R$ -modules, and $M\otimes_RN=0$ , then $\ann M+\ann N=R$ . In particular, if $R$ is local, then either $M$ or $N$ is $0$ .

Proof:

$\square$

Theorem 3:

Let $M$ be a finitely generated $R$ -module. If $f:M\rightarrow M$ is an $R$ -linear map and $f$ is surjective then $f$ is also injective. In this case, $f$ is an automorphism of $M$ .

Proof:

Since $f$ commutes with scalar multiplication by elements of $R$ , we can make $M$ an $R[x]$ -module by setting $xm=f(m)$ for all $m\in M$ . Then by the hypothesis, $xM=\<x\>M=M$ . Then by NAK lemma, there exists $y\in R[x]$ such that $(1+xy)M=0$ . Now suppose that $u\in \ker f$ , then $0=(1+xy)(u)=u+yf(u)=u$ . Therefore, $\ker f=\{0_M\}$ , which implies $f$ is an injection. The conclusions follow.

$\square$

Problem 1:

Let $R$ be a ring and $I$ is a finitely generated ideal satisfying $I^2=I$ . Prove that $I$ is generated by an idempotent element $e$ .

Solution:

Regard $I$ as an $R$ -module, by NAK lemma, there exist $e\in I$ such that $(1-e)I=0$ . Thus $I=eI=\<e\>$ . Moreover, since $I=I^2=\<e^2\>$ , we also have $e=e^2$ .