## Category Archives: Algebra

### An example of non-free projective module over integral domain

An ideal $I$  of an integral domain $R$  is a free $R$-module if and only if it is generated by one element. Indeed, given any two non-zero elements $x,y\in I$, $x\cdot y+(-y)\cdot x=0$. So any two elements are linear dependent. Hence, if $I$ were free module, it must generated by one element, or $I$ is a principle ideal. Our strategy is to find an ideal $I$ which is not principle and show that $I$ is projective.

Let $R=\mathbb{Z}[-21]=\{a+b\sqrt{-21}|a,b\in\mathbb{Z}\}$. Obviously $R$ is an integral domain. Let $I$ be the ideal generated by 5 and $2+\sqrt{-21}$. $I=\{5a+(2+\sqrt{-21})b|a,b\in R\}$.

$\mathbf{Claim:}$ $I$ is not a principal ideal.
$\mathbf{Proof:}$
Define $\phi:R\to \mathbb Z$ by $\phi(a+b\sqrt{-21})=a^2+21b^2$, then it is straightforward to verify $\phi(r_1r_2)=\phi(r_1)\phi(r_2)$, for all $r_1,r_2\in R$. If $r\in R$ is a unit, then $\phi(r)=1$, which means $a=\pm 1$ and $b=0$. So the only units in $R$ are $\pm 1$.

Now suppose $I$ is a principle ideal, then $\exists\,u$ such that $I=Ru$. Then $\exists\, r\in R$ such that $5=ru$, which means $25=\phi(r)\phi(u)$. Since $x^2+21y^2=5$ has no solution in $\mathbb Z$, $\phi(r)$ or $\phi(u)$ can not be 5. So we must have $\phi(r)=25$ and $\phi(u)=1$ or $\phi(r)=1$ and $\phi(u)=25$. The second case is impossible, because $\phi(r)=1$ means $r=\pm 1$. Then $I=5\cdot R$. We can obtain $2+\sqrt{-21}=5(a+b\sqrt{-21})$, this is impossible.

The only case we left is $\phi(u)=1$, which means $u=\pm 1$. Then $I=R$. We have $R\cong \mathbb Z[x]/(x^2+21)$, $I$ corresponds to the ideal $(5,2+x)/(x^2+21)$. If $I=R$, then $(5,2+x)=\mathbb Z[x]$. This fact leads to $1=f(x)\cdot 5+g(x)\cdot (2+x)$, for some $f(x)$, $g(x)\in \mathbb Z[x]$. If we set $x=-2$, then $1=f(-2)\cdot 5$ in $\mathbb Z$, this is impossible.

In conclusion, $I$ can not be principle. $\text{Q.E.D. }\hfill\square$

$\mathbf{Claim:}$ $I$ is a projective module.

$\mathbf{Proof:}$Define $\psi:R^2\to I$ by $\psi(a,b)=5a+(2+\sqrt{-21})b$. If we can find $i:I\to R^2$ such that $\psi i=id_I$, then $I$ is a direct summand of $R^2$, which is a free $R$-module. Consequently, $I$ is a projective module.

Let $x=5a+(2+\sqrt{-21})b\in I$, $a,b\in R$. There exist unique $u,v\in R$ such that
$(3-4\sqrt{-21})x=5u$
$(16+2\sqrt{-21})x=5v$
A straightforward calculation shows that, $u=3a+18b-(4a+b)\sqrt{-21}$, $v=16a-2b+(2a+4b)\sqrt{-21}$. Define $i(x)=(u,v)\in R^2$. It is easy to verify that $i$ is an $R$-module homomorphism. Moreover,
$\psi(u,v)=u\cdot 5+v\cdot(2+\sqrt{-21})$
$=5(3a+18b-(4a+b)\sqrt{-21})+[16a-2b+(2a+4b)\sqrt{-21}](2+\sqrt{-21})= 5a+(2+\sqrt{-21})b.$

This means $\psi i(x)=x$. Thus $I$ is a projective module. $\text{Q.E.D. }\hfill\square$

As discussed in the beginning of this section, since $I$ is not principle, we get a non-free projective module.

### A brief review of projective modules

Projective modules were introduced by Henri Cartan and Samuel Eilenberg in 1956.

In general one can define a $R$-module $P$ to be projective by any one of the following:
(1) Any short exact sequence $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$ splits;
(2) $P$ is a direct summand of a free module, which $\exists\, Q$ such that $P\oplus Q$ is a free $R$-module;
(3) Given an epimorphism $\displaystyle f:M\to N$ and $p:P\to N$, then $\exists\, g:P\to M$ such that $p=fg$, that is

(4) hom$(P,\textendash)$ is an exact functor.

These four definitions show that projective module has very rich properties and is related to other types of modules. (2) means projective module is a generalization of free modules. (3) is called the lifting property of projective module, it is also a generalization of free module, because if $P$ is a free module, there exists a unique $g:P\to M$ such that the following diagram commutes

In (3) we don’t require $g$ to be unique, so $P$ does not have the universal property, consequently not necessarily a free object in category sense. Dual to definition (3), injective module is introduced
1)Given a monomorphism $f:M\to N$ and a module homomorphism $q:M\to Q$, then $\exists\, g:N\to Q$ such that the following diagram commutes

2) hom$(\textendash, Q)$ is an exact (contravariant) functor;
3) Any short exact sequence $0\rightarrow Q\rightarrow M\rightarrow N\rightarrow 0$ splits.

Since we know free modules are all projective ones, two natural questions emerge naturally.
(1)whether there is a projective module but not free;
(2)when the projective module is free.
For the first question, there is an easy answer. Let $R=\mathbb{Z}/6\mathbb{Z}$ regarded as a module for $R$-module, then $R$ is a free module. By the Chinese remainder theorem, $R\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$. Thus $\mathbb{Z}/2\mathbb{Z}$ is a direct summand of free $R$-module. $\mathbb{Z}/2\mathbb{Z}$ is projective from the definition, but not free $R$-module because actually it is a torsion $R$-module. Carefully investigating this example, the ring $R$ is decomposable, $R\cong R_1\oplus R_2$, then it is easy to find a non-free projective module, in particular $R_1$. What if we require the $R$ is not decomposable, say $R$ is an integral domain in particular? Is it still possible to find a non-free projective module? Our main result of this paper is devoted to this purpose. It turns out to be much more difficult to give such examples. Before finding an example, we should have a rough understanding on which kind of ring projective modules are always free, which is related to the answer of question (2).

The study of (2) can be summarized as following:
$\mathbf{Fact 1}$ If $R$ is a PID (the commutative principal ideal domains), all projective modules are free.
This is because any submodule of free module over $R$ is free. Projective is direct summand of free module, hence a submodule of free module.
$\mathbf{Fact 2}$ $R$ is a local ring, any finitely generated projective module over $R$ is free.
This result is proved by Irving Kaplansky in 1958. It is easy to prove this result for finitely generated projective modules, but the general case is difficult.
The following theorem is a well known result of H. Bass.
$\mathbf{Fact 3}$ If $R$ is a connected (i.e.without nontrivial idempotents) Noetherian ring. Then any infinitely generated projective $R$-module is free.
The following theorem is known as Serre’s Conjecture, which was solved by Quillen, Daniel and Suslin, Andrei A. independently in 1976.
$\mathbf{Fact 4}$ If $R=D[x_1,x_2\cdots,x_n]$, where $D$ is a PID, then every projective module over $R$ is free.
Note that Bass’s theorem is a special case of Serre’s Conjecture.

### Albert radicals of norm and Hilbert’s Satz 90

$\mathbf{Problem(Albert):}$ Let $E$ be a cyclic extension of dimension $n$ over $F$ and let $\eta$ be a generator of $\text{Gal }E/F$. Let $r|n$, $n=rm$ and suppose $c$ is a non-zero element of $F$ such that $c^r=N_{E/F}(u)$ for some $u\in E$. Show that there exists a $v$ in the (unique) subfield $K$ of $E/F$ of dimensionality $m$ such that $c=N_{K/F}(v)$.

$\mathbf{Proof:}$ $G=\text{Gal }E/F$ is a cyclic group, then it  has a unique subgroup $H=\{\eta^m,\eta^{2m},\cdots,\eta^{rm}=1\}$ of index $r$. By the Galois corresponding theorem, there exists a unique subfield $K=\text{Inv }H$ such that $\text{Gal K/F}\cong G/H$. $\text{Gal }K/F=\{\eta^1|_K,\eta^2|_K\cdots,\eta^m|_K\}$. $K$ has dimensionality $m$ over $F$.

Consider $w=c^{-1}u\eta(u)\eta^2(u)\cdots \eta^{m-1}(u)$, then $\displaystyle\eta(w)=\frac{\eta^m(u)}{u}w$. We also have

$\displaystyle N_{E/K}(w)=\eta^m(w)\eta^{2m}(w)\cdots \eta^{rm}(w)=c^{-r}\eta(u)\eta^2(u)\cdots\eta^n(u)=1$,

by Hilbert’s Satz 90, $\exists \, l\in E$ such that $\displaystyle w=\frac{\eta^m(l)}{l}$.

Let $\displaystyle v=\frac{ul}{\eta(l)}$, then $v\in K$, because

$\displaystyle \eta^m(v)=\frac{\eta^m(u)\eta^m(l)}{\eta^{m+1}(l)}=\frac{\eta^m(u)\eta(l)}{\eta^{m+1}(l)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{wl}{\eta(l)}=u\frac{l}{\eta(l)}=v$

Surprisingly we have

$\displaystyle N_{K/F}(v)=v\eta(v)\cdots \eta^{m-1}(v)=u\eta(u)\cdots\eta^{m-1}(u)\frac{l}{\eta^m(l)}=cw\frac{l}{\eta^m(l)}=c$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacabson p300. This problem puzzled me for three weeks. Finally it turns out to be very easy.

### When the norm map is surjective

$\mathbf{Problem:}$ Suppose $F$ is finite field with $q=p^m$ elements, $E$ an Galois extension of $F$ such that $[E:F]=n$. Prove that $N_{E/F}:E^\ast\to F^\ast$ is surjective.

$\mathbf{Proof:}$ From theorem 4.26 we know that $\text{Gal }E/F$ is cyclic over $F$ generated by $\eta:a\to a^q$.

$\displaystyle N_{E/F}(a)=aa^qa^{q^2}\cdots a^{q^{n-1}}=a^{\frac{q^n-1}{q-1}}$.

$N_{E/F}:E^\ast\to F^\ast$ is a homomorphism. It is well known that  any finite subgroup of the multiplicative group of field is cyclic, which means $E^\ast$ and $F\ast$ is cyclic.

Since $|E\ast|=q^n-1$ and $|F^\ast|=q-1$, and $N_{E/F}(a)=a^m$, where $m={\frac{q^n-1}{q-1}}$, then $|\text{Ker } N|=(q^n-1,m)=m$ and $|\text{Im }N|=\frac{q^n-1}{(q^n-1,m)}=q-1$. Thus $N$ is surjective.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p300.

### Criterion for embedding in cyclic field

$\mathbf{Problem:}$ Assume $F$ has $p$ distinct pth roots of 1. $p$ a prime, and $E/F$ is cyclic of dimensional $p^f$. Let $z$ be a primitive $p$th root of 1. Show that if $E/F$ can be imbedded in a cyclic field $K/F$ of dimension $p^{f+1}$, then $z=N_{E/F}(u)$ for some $u\in E$.

$\mathbf{Proof:}$ Suppose $\sigma$ is the generating isomorphism of cyclic Galois group $K/F$. Then $\displaystyle E=\text{Inv }\sigma^{p^f}$. $N_{K/E}(z)=z^p=1$, so by Hilbert satz 90, there exists $a\in K$ such that $\displaystyle z=\sigma^{p^f}(a)a^{-1}$ .
$\displaystyle \sigma^{p^f}(a^{-1}\sigma(a))=\sigma^{p^f}(a^{-1})\sigma^{p^f+1}(a)=(z a)^{-1}\sigma(z a)=a^{-1}\sigma(a)$.
So $a^{-1}\sigma(a)=a_0\in E$, then
$N_{E/F}(a_0)=a^{-1}\sigma(a)\sigma(a^{-1})\sigma^2(a)\cdots \sigma^{p^f-1}(a)\sigma^{p^f}(a)=a^{-1}\sigma^{p^f}(a)=z$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p300.

### Galois group of C(t,u) over C(t^n,u^n) and application to cosnx

$\mathbf{Problem:}$ Let $E$ be an extension of $\mathbb{C}$ such that $E=\mathbb{C}(t,u)$ where $t$ is transcedental over $\mathbb{C}$ and $u$ satisfies the equation $u^2+t^2=1$ over $\mathbb{C}(t)$. Find the Galois group of $\mathbb{C}(t,u)$ over $\mathbb{C}(t^n,u^n)$ for any $n\in \mathbb{N}$ Show that

$\displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n], i=\sqrt{-1}$

is contained in $\mathbb{C}(t^n,u^n)$. Use this to prove that the function cos$nx$ is expressible rationally with complex coefficients in terms of $\cos^nx$ and $\sin^n x$. Does this hold for sin $nx$.
$\mathbf{Proof:}$ $\forall\, \eta\in \text{Gal }\mathbb{C}(t,u)/\mathbb{C}(t^n,u^n)$, we must have

$\displaystyle\eta(t^n)=\eta^n(t)=t^n$    (1)
$\displaystyle\eta(u^n)=\eta^n(u)=u^n$    (2)
$\displaystyle\eta^2(t)+\eta^2(u)=1$      (3)

This means $(\eta(t)/t)^n=1$. So there exist $w_1$, a n-th root of 1, such that $\eta(t)=w_1t$. Similarly $\exists\, w_2$ such that $\eta(u)=w_2u$. (3) becomes

$w^2_1t^2+w^2_2u^2=1$. Combing with $t^2+u^2=1$, we get $w^2_1=w^2_2=1$. Since $w^n_1=w^n_2=1$, we know that when $n$ is even $w_1=w_2=1$ and when $n$ is odd $w_1=\pm 1$ and $w_2=\pm 1$.

In conclusion, the Galois group is trivial when $n$ is odd and Klein group when $n$ is even.

No matter $n$ is even and odd, $\displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n]$ is invariant, so $u_n$ is in $\mathbb{C}(t^n,u^n)$.

If $n$ is odd, $\displaystyle v_n=\frac{1}{2i}[(t+iu)^n-(t-iu)^n]$ is in $\mathbb{C}(t^n,u^n)$.  If $n$ is even, $v_n$ is not in the ground field, because $\eta(t)=t, \eta(u)=-u$ can not fix $v_n$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ At first I am surprised by the result I got. I can not believe that the galois is trivial when $n$ is odd.  So I conculted to Zhuohui. Thank him for his kind help. I drew the following picture after discussion with him.

case n=4

case n=3

### Inverse Galois Problem

$\mathbf{Problem:}$ Use the fact that any finite group $G$ is isomorphic to a subgroup of $S_n$ to prove that given any finite group $G$ there exists fields $F$ and $E/F$ such that

$\displaystyle \text{Gal }E/F\cong G$

$\mathbf{Proof:}$ Any field $K$, $x_1,x_2,\cdots,x_n$ are indeterminate. Let $E=K(x_1,x_2,\cdots,x_n)$. $D$ is the field of symmetric rational functions in $latex x_1,x_2,\cdots,x_n$ Or if we let $p_1,p_2,\cdots,p_n$ are basic symmetric polynomials, namely

$\displaystyle p_1=\sum x_i$, $\displaystyle p_2=\sum _{i>j} x_ix_j$, $\cdots \displaystyle p_n=x_1x_2\cdots x_n$,

then $D=K(p_1,p_2,\cdots,p_n)$.

As we all know, $\text{Gal }E/D=S_n$. Since $G\leq S_n$, by the Galois corresponding theorem, there exists an intermediate field $F$ such that $\text{Gal }E/F\cong G$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ This is the easiest inverse Galois problem since one does not specify the ground field. If you restrict $F=\mathbb{Q}$, things are much trickier.

### Polynomial of prime degree and solvable transitive subgroup

$L$ is the group of transformations of $\mathbb{Z}_p$ of the form $x\to ax+b$, $a\neq 0$.
$H$ is all the translations $x\to x+b$.

$\mathbf{Lemma 1:}$ $|L|=p(p-1)$ and $|H|=p$, $H\lhd L$.

$\mathbf{Lemma 2:}$ $G$ is a group of transformation in $\mathbb{Z}_p$ containing the group $H$ of translations as normal subgroup. Show that $G$ is a subgroup of $L$, this means $L$ is the normalizer of $H$.
$\mathbf{Proof:}$ Suppose $\tau:x\to x+1$ and let $\eta\in G$, $\eta \tau\eta^{-1}$ has the form $x\to x+k$. Hence $\eta(x+1)=\eta\tau(x)=\eta(x)+k$. If $\eta(0)=b$, then $\eta(x)=kx+b$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Lemma 3:}$ Any solvable transitive subgroup of $S_p$, $p$ a prime, is equivalent to a subgroup of $L$ containing the group of translations.

$\mathbf{Proof:}$ Suppose $G$ is this group. Then $G$ has a composition series
$\displaystyle G\rhd H_1\rhd H_2\rhd \cdots \rhd H_{n-1}\rhd H_n=1$
whose factors are all cyclic groups of prime order.
By lemma 4, we know that $H_{n-1}$ is also a solvable transitive subgroup of $S_p$. Since $H_{n-1}$ is cyclic of prime order, then it must be generated by some p-cycle, otherwise it can not be transitive.
Apply some conjugation to $G$ in $S_p$, we can assume $H_{n-1}$ is generated by $(12\cdots p)$. So $H_{n-1}$ is a translation group. We can identify the element of $G$ as the transformation of $\mathbb{Z}_p$. By lemma 2, we know that $G$ is equivalent to some subgroup of $L$.
$\text{Q.E.D}\hfill \square$
$\mathbf{Lemma 4:}$ Let $H$ be a normal nontrivial subgroup of a transitive subgroup $G$ of $S_n$ of transformations of $\{1,2,\cdots,n\}$. Show that all $H-$orbits have the same cardinality. Hence show that if $n=p$ is a prime, then $H$ is transitive.

$\mathbf{Theorem(Galois):}$ Let $f(x)\in F[x]$ be irreducible of prime degree over $F$ of characteristic 0, $E$ a splitting field over $F$ of $f(x)$. Show that $f(x)$ is solvable by radicals over $F$ if and only if $E=F(r_i,r_j)$ for any two roots $r_i,r_j$ of $f(x)$.

$\mathbf{Proof:}$ Since $F$ has characteristic 0, $f(x)$ is solvable by radicals if and only if $G=\text{Gal }E/F$ is a solvable subgroup of symmetric group $S_p$. $f(x)$ is irreducible if and only if $G$ is transitive.
(1) if $f$ is solvable by radicals, then $G$ is a solvable transitive subgroup of $S_p$.
By lemma 3, we know $G$ is equivalent to a subgroup of $L$ containing $H$. WLOG, assume $H\leq G\leq L$. The action of $G$ on roots looks like $r_i\to r_{ai+b}$, $a\neq 0$.
Consider $\text{Gal }E/F(r_i,r_j)$, $r_i\neq r_j$. $\forall\,\eta\in \text{Gal }E/F(r_i,r_j)$, $\exists\, a,b$ unique such that $i=ai+b$ and $j=aj+b$. Then $(a-1)(i-j)=0$. Since $i\neq j$, we get $a=1$ and $b=0$, which means $\eta=id$. So $E=F(r_i,r_j)$.
(2) If $E=F(r_i,r_j)$, then $p| |\text{Gal }E/F(r_i,r_j)|$. This Galois group contains a normal Sylow p-subgorup which is isomorphic to $H$. By lemma 2, we know $G$ is equivalent to a subgroup of $L$ which is solvable. So $G$ is solvable, $f(x)$ is solvable by radicals.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p262.

### Algebraically closed and Primitive element theorem

$\mathbf{Problem:}$ Let $F$ be a perfect field and $F\subset E$ an algebraic field extension, such that every non-constant polynomial $f(x)\in F[x]$ has a root in $E$. Show that $E$ is algebraically closed.

$\mathbf{Proof:}$ Suppose $f(x)\in F[x]$ is a non-constant polynomial. Let $K$ be a splitting field of $f$.
Since $F$ is perfect, $K$ is a separable extension of $F$. Primitive element theorem implies that any finite separable extension is simple extension. That is $\exists\,\alpha\in K$ such that $K\in F(\alpha)$. Suppose $\alpha$ has a minimal polynomial $g(x)$ over $F$. By assumption, $g(x)$ has a root $\gamma\in E$. Since there exists an isomorphism $\eta:F(\alpha)/ F\to F(\gamma)/ F\subset E$ by $\eta(\alpha)=\gamma$, then $E$ contains a splitting field of $f(x)$ over $F$.
$\forall\, f\in E[x]$, adjoin a root $\xi$ of $f(x)$, we get $E(\xi)/E$. Since $E(\xi)/E$ and $E/F$ are algebraic. $E(\xi)/F$ is also algebraic. Hence, $\xi$ is a root of some non-constant polynomial $h(x)\in F[x]$. By the previous proof, $h(x)$ splits over $E$. We conclude that $\xi\in E$ and $E(\xi)=E$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### The discriminant of a cubic equation and its relation with roots

$\mathbf{Problem:}$ Let $F=\mathbb{R}$ and let $f(x)=x^3-a_1x^2+a_2x-a_3$ with discriminant $d$. Show that $f(x)$ has multiple roots, three distinct real roots, or one real root and two non-real roots according as $d=0$, $d>0$, $d<0$.

$\mathbf{Proof:}$ $f$ has multiple roots if and only if $d=0$.

Supoose $d\neq 0$. Since cubic equation always has at least one  root in $\mathbb{R}$, assume $f(x)=(x-r)g(x)$, $r\in \mathbb{R}$. Then Galois group $G_f$ of $f$ is equal to that of $g(x)$ over $\mathbb{R}$, which is the symmetric group $S_2$ or the alternating group $A_2=1$.

$d>0$ if and only if $G_f$ is contained in symmetric group $A_3$. Since $G_f$ has only two cases. $d>0$ if and only if $G_f$ is $A_2=1$ which means $f$ has three distinct roots in $\mathbb{R}$

Correspondingly, we get $d<0$ if and only if   one real root and two non-real roots.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p260.