Ring Theory
4.8
Definition: A ring \(R\) is a set together with two binary operation together with "\(+\)" and "\(\times\)", obeying:
- \((R,+)\) is an Abelian group.
- \(\times\) is associative: \((a\times b)\times c=a\times(b\times c)\) for \(\forall a,b,c\in R\).
- The distributive laws hold in \(R\): \((a+b)\times c=a\times c+b\times c\) and \(a\times(b+c)=a\times c+b\times c\).
The ring is called commutative if multiplication is commutative.
The ring is said to have an identity if there is an element \(1\in R\) with \(1\times a=a\times 1=a\) for \(\forall a\in R\).
The additive identity of \(R\) will always be denoted as \(0\) and the additive inverse of an element \(a\) will be denoted as an \(-a\).
A ring with an identity \(1\) where \(1\ne 0\) is called a division ring (or skew field) if every non-zero element \(a\in R\) has a multiplicative inverse, i.e., there exists \(b\in R\) such that \(ab=ba=1\). A commutative division ring is called a field.
Examples:
- Trivial rings: \(\forall a,b\in R,a\times b=0\).
- Zero ring: \(R=\{0\}\).
- The Hamilton Quaterions: elements in the form \(a+bi+cj+dk\) where \(a,b,c,d\in\mathbb{R}\) and \(i^2=j^2=k^2=1,ij=-ji=k,jk=-kj=i,ki=-ik=j\).
- Ring of functions: Let \(X\ne\varnothing\) and \(A\) be any ring, \(f:X\to A\), \((f+g)(x)=f(x)\), \((fg)(x)=f(x)g(x)\).
Property 1. Let \(R\) be a ring, then
- \(0a=a0=0\) for \(\forall a\in R\).
- \((-a)b=a(-b)=-(ab)\) for \(\forall a,b\in R\).
- \((-a)(-b)=ab\).
- If \(R\) has identity, then \(-a=(-1)a\).
Definition. Let \(R\) be a ring
- A non-zero element \(a\in R\) is called a zero-divider if there exists \(b\in R,b\ne 0\) such that \(ab=0\) or \(ba=0\).
- If \(1\in R\), an element \(u\) of \(R\) is called a unit in \(R\) if \(\exists v\in R\), such that \(uv=vu=1\). The set of all unit is denoted by \(R^{\times}\).
4.15
Definition. A commutative ring with identity \(1\ne 0\) is called an integral domain if it has no zero divisors.
Property 2. Assume \(a,b,c\in R\), and \(a\) is not a zero divisor. If \(ab=ac\), then either \(a=0\) or \(b=c\). In particular, if \(a,b,c\) are any elements in an integral domain and \(ab=ac\), then either \(a=0\) or \(b=c\).
Proof. If \(ab=ac\), then \(a(b-c)=0\) so either \(a=0\) or \(b=c\).
Corollary 3. Any finite integral domain is a field.
Definition. A subring of the ring \(R\) is a subgroup of \(R\) that is closed under multiplication and addition.
Quadric Integer Rings
Let \(D\) be a square-free integer. It is immediately from the addition and multiplication that the subset \(Z(D)=\{a+b\sqrt{D}|a,b\in\mathbb{Z}\}\) forms a subring of the quadric field \(Q(\sqrt{D})\) defined earlier.
If \(D\equiv 1\pmod{4}\) then the slightly larger subset \(Z(\dfrac{1+\sqrt{D}}{2})=Z(D)=\{a+b\dfrac{1+\sqrt{D}}{2}|a,b\in\mathbb{Z}\}\) is also a subring.
Polynomial Rings
Let \(x\) be an indeterminate, \(R\) be a commutative ring, then \(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\) with \(n\in\mathbb{Z}\), \(n\ge 0\) and each \(a_i\in R\) is called a polynomial in \(x\) with coefficients \(a_i\) in \(R\). If \(a_n\ne 0\), the polynomial is said to be of degree \(n\), \(a_nx^n\) is called the leading term, and \(a_n\) is called the leading coefficient. The set of all such polynomials is called the ring of polynomials in the variable \(x\) in \(R\) and will be denoted \(R[x]\). The polynomial is called monic if \(a_n=1\).
The ring in which the coefficients are taken makes a substantial difference in the behavior of polynomials. For example, \(x^2+1\) is not a perfect square in \(Z[x]\), but is a perfect square in \(Z/2Z[x]\).
Property 4. Let \(R\) be an integral domain and let \(p(x),q(x)\) be non-zero elements of \(R[x]\). Then
- The degree of \(p(x)q(x)\) equals to the sum of the degree of \(p(x)\) and the degree of \(q(x)\).
- The units of \(R[x]\) are just the units of \(R\).
- \(R[x]\) is an integral domain.
Matrix Rings
Fix an arbitrary ring \(R\) and let \(n\in\mathbb{Z}^+\). Let \(M_n(R)\) be the set of all \(n\times n\) matrices with entries from \(R\). \(M_n(R)\) is a ring under the usual addition and multiplication of matrices.
When \(n\ge 2\), \(M_n(R)\) is not commutative even if \(R\) is commutative.
Group Rings
Fix a commutative ring \(R\) with identity \(1\ne 0\) and let \(G=\{g_1,g_2,\cdots,g_n\}\) be any finite group with group operation written multiplicatively. Define the group ring, \(RG\), of \(G\) with coefficients in \(R\) to be the set of all formal sums: \(a_1g_1+a_2g_2+\cdots+a_ng_n\), \(a_i\in R\), \(1\le i\le n\).
The ring \(RG\) is commutative if and only if \(G\) is commutative.
Ring homomorphisms and Quotient Rings
Definition. Let \(R\) and \(S\) be rings:
- A ring homomorphism is a map \(\varphi:R\to S\) satisfying:
- \(\varphi(a+b)=\varphi(a)+\varphi(b)\) for \(\forall a,b\in R\).
- \(\varphi(ab)=\varphi(a)\varphi(b)\) for \(\forall a,b\in R\).
- The kernel of the ring homomorphism \(\varphi\), \(\ker\varphi=\varphi^{-1}(0)\).
- A bijective ring homomorphism is called an isomorphism.
Example:
- The map \(\varphi:\mathbb{Z}\to Z/2Z\) defined by sending even integers to \(0\) and odd integers to \(1\) is a ring homomorphism.
- For \(n\in\mathbb{Z}\), the map \(\varphi_n:\mathbb{Z}\to\mathbb{Z}\) defined by \(\varphi_n(x)=nx\) are not in general a ring homomorphism.
Property 5. Let \(R\) and \(S\) and let \(\varphi:R\to S\) be a homomorphism.
- The image of \(\varphi\) is a subring of \(S\).
- The kernel of \(\varphi\) is a subring of \(R\). Furthermore, if \(a\in\ker\varphi\) then \(ra\) and \(ar\) are inside \(\ker\varphi\) for \(\forall r\in R\).
Proof: Note that both \(\text{im}(\varphi)\) and \(\ker\varphi\) are empty.
- If \(s_1,s_2\in\text{im}\varphi\), then \(\exists r_1,r_2\in R\), such that \(\varphi(r_1)=s_1,\varphi(r_2)=s_2\). \(s_1-s_2=\varphi(r_1- r_2)\in\text{im}\varphi\) and \(s_1s_2=\varphi(r_1r_2)\in\text{im}\varphi\). Thus \(\text{im}\varphi\) is a subring of \(S\).
- If \(a,b\in\ker\varphi\), then \(\varphi(a)=\varphi(b)=0\). Hence \(\varphi(a-b)=\varphi(ab)=0\), so \(a-b,ab\in\ker\varphi\), i.e., \(\ker\varphi\) is a subring of \(R\). Similarly for \(\forall r\in R\), we have \(\varphi(ra)=\varphi(r)\varphi(a)=0\), \(\varphi(ar)=\varphi(a)\varphi(r)=0\), so \(ra,ar\in\ker\varphi\).
Definition. Let \(R\) be a ring, \(I\) be a subset of \(R\) and \(r\in R\).
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\(rI=\{ra|a\in I\}\) and \(Ir=\{ar|a\in I\}\).
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A subset \(I\) of \(R\) is a left ideal of \(R\) if
- \(I\) is a subring of \(R\)
- \(I\) is closed under left multiplication by elements from \(R\), i.e., \(rI\subseteq I\), for \(\forall r\in R\).
Similarly, \(I\) is a right ideal if (i) holds and \(Ir\subseteq I\) for \(\forall r\in R\).
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A subset \(I\) that is both a left ideal and a right ideal is called an ideal of \(R\).
Property 6. Let \(R\) be a ring and \(I\) be an ideal of \(R\). Then the (additive) quotient group \(R/I\) is a ring under the binary operations: \((r+I)+(s+I)=(r+s)+I\) and \((r+I)(s+I)=(rs)+I\) for \(\forall r,s\in R\). Conversely, if \(I\) is any subring such that the above operation are well-defined, then \(I\) is an ideal of \(R\).
Definition. When \(I\) is an ideal of \(R\) the ring \(R/I\) with the operations in the previous proposition is called the quotient ring of \(R\) by \(I\).
Theorem 7.
- (the First Isomorphism Theorem for Rings) If \(\varphi:R\to S\) is a homomorphism of rings, then \(\ker\varphi\) is an ideal of \(R\), the image of \(\varphi\) is a subring of \(S\) and \(R/\ker\varphi\) is isomorphic as a ring to \(\varphi(R)\): \(R/\ker\varphi\cong\varphi(R)\).
- If \(I\) is any ideal of \(R\), then the map \(R\to R/I\) defined by \(r\to r+I\) is a surjective ring homomorphism, with kernel \(I\). Thus every ideal is the kernel of a ring homomorphism and vice versa.
Theorem 8.
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(the Second Isomorphism Theorem for Rings). Let \(A\) be a subring and \(B\) be an ideal of \(R\), then \(A+B=\{a+b|a\in A,b\in B\}\) is a subring of \(R\), \(A\cap B\) is an ideal of \(A\) and \((A+B)/B=A/(A\cap B)\).
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(the Third Isomorphism Theorem for Rings). Let \(I,J\) be ideals of \(R\) with \(I\subseteq J\). Then \(J/I\) is an ideal of \(R/I\) and \((R/I)/(J/I)\cong R/J\).
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(the Fourth or Lattice Isomorphism Theorem for rings). Let \(I\) be an ideal of \(R\), the correspondence \(A\leftrightarrow A/I\) is an inclusion preserving bijection between the set of subrings \(A\) of \(R\) that contain \(I\) and the set of subrings of \(R/I\). Furthermore, \(A\) (a subring containing \(I\)) is an ideal of \(R\) if and only if \(A/I\) is a subring of \(R/I\).
Definition. Let \(I,J\) be ideals of \(R\).
- Define the sum of \(I\) and \(J\) by \(I+J=\{a+b|a\in I,b\in J\}\).
- Define the product of \(I\) and \(J\) by \(IJ\) equals to all finite sums of \(ab\) with \(a\in I,b\in J\).
It is easy to see that \(I+J\) is the smallest ideal of \(R\) containing both \(I\) and \(J\), and that \(I+J\) is an ideal contained in \(I\cap J\).
Definition. Let \(R\) be a ring with \(1\ne 0\), \(A\subseteq R\),
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The ideal generated by \(A\): the smallest ideal of \(R\) containing \(A\), denoted by \((A)\).
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\(RA=\{r_1a_1+r_2a_2+\cdots+r_na_n|r_i\in R,a_i\in A,n\in\mathbb{Z}^+\}\).
\(AR=\{a_1r_1+a_2r_2+\cdots+a_nr_n|r_i\in R,a_i\in A,n\in\mathbb{Z}^+\}\)
\(RAR=\{r_1a_1r'_1+r_2a_2r'_2+\cdots+r_na_nr'_n|r_i,r'_i\in R,a_i\in A,n\in\mathbb{Z}^+\}\).
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An ideal generated by a single element is called a principal ideal.
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An ideal generated by a finite set is called a finitely generated ideal.
Since the intersection of any non-empty collection of ideals of \(R\) is also an ideal, we have \((A)=\cap_{A\subseteq I}I\) such that \(I\) is an ideal of \(R\).
The left ideal generated by \(A\) is the intersection of all left ideals of \(R\) that contain \(A\). \(RA\) is precisely the left ideal generated by \(A\) and \(RAR\) is the ideal generated by \(A\).
If \(R\) is commutative then \(RA=AR=RAR=(A)\).
Example:
- The trivial ideal \(0\) and the ideal \(R\) are both principal: \(0=(0)\) and \(R=(1)\).
- In \(\mathbb{Z}\) we have \(n\mathbb{Z}=\mathbb{Z}n=(n)=(-n)\) for all integers of \(n\). These are all the ideals of \(\mathbb{Z}\), so every ideal of \(Z\) is principal. Furthermore, the ideal generated by two nonzero integers \(n\) and \(m\) is the principal ideal generated by their greatest common divisor.
Property 9. Let \(R\) be a ring with \(1\ne 0\) and \(I\) be an ideal of \(R\).
- \(I=R\) if and only if \(I\) contains a unit.
- Assume \(R\) is commutative. Then \(R\) is a field if and only if its only ideal are \(0\) and \(R\).
Corollary 10. If \(R\) is a field then any non-zero ring homomorphism from \(R\) to another ring is an injection.
Proof: The kernel of a homomorphism is an ideal. The kernel of a non-zero homomorphism is a proper ideal hence \(0\).
4.29
Definition. An ideal \(M\) in an arbitrary ring \(S\) is called a maximal ideal if \(M\ne S\) and the only ideals containing \(M\) are \(M\) and \(S\).
Property 11. In a ring with identity every proper ideal is contained in a maximal ideal.
Property 12. Assume \(R\) is commutative. The ideal \(M\) is a maximal ideal if and only if the quotient ring \(R/M\) is a field.
Proof: The ideal \(M\) is maximal if and only if there are no ideals \(I\) with \(M\subset I\subset R\). By the Lattice Isomorphism Theorem, the ideals of \(R\) containing \(M\) correspond bijectively with the ideals of \(R/M\), so \(M\) is maximal if and only if the only ideals of \(R/M\) are \(0\) and \(R/M\). By Property 9, we see that \(M\) is maximal if and only if \(R/M\) is a field.
Example:
- Let \(n\) be a non-negative integer, the ideal \(nZ\) of \(Z\) is a maximal ideal if and only if \(Z/nZ\) is a field, if and only if \(n\) is a prime number.
- The ideal \((x)\) is not a maximal ideal in \(\mathbb{Z}[x]\) because \((x)\subset(2,x)\subset \mathbb{Z}[x]\). In fact, \(\mathbb{Z}[x]/(x)\cong\mathbb{Z}\) which is not a field.
Definition. Assume \(R\) is commutative. An ideal \(P\) is called a prime ideal if \(P\ne R\) and whenever \(ab\in P\) for \(a,b\in R\), at least one of \(a\) and \(b\) are inside \(P\).
Property 13. Assume \(R\) is commutative. Then ideal \(P\) is a prime ideal in \(R\) if and only if the quotient ring \(R/P\) is an integral domain.
Proof: The ideal \(P\) is prime if and only if \(P\ne R\) and whenever \(ab\in P\), then either \(a\in P\) or \(b\in P\). Use the bar notation for elements of \(R/P\): \(\bar{r}=r+P\). Note that \(r\in P\) if and only if \(\bar{r}\) is zero in \(R/P\). Thus in the terminology of quotients \(P\) is a prime ideal if and only if \(\bar{R}\) and whenever \(\bar{ab}=0\) either \(\bar{a}=0\) or \(\bar{b}=0\), i.e., \(R/P\) is an integral domain.
Corollary 14. Let \(R\) be commutative. Every maximal ideal of \(R\) is a prime ideal.
Proof: If \(M\) is a maximal ideal then \(R/M\) is a field by Property 12. A field is an integral domain so the corollary follows from Property 13.
Example:
- The principal ideals generated by primes in \(\mathbb{Z}\) are both prime and maximal ideals. The zero ideal in \(\mathbb{Z}\) is prime but not maximal.
- The ideal \((x)\) is a prime ideal in \(\mathbb{Z}[x]\) since \(\mathbb{Z}[x]/(x)\cong\mathbb{Z}\). This ideal is not a maximal ideal. The ideal \(0\) is a prime ideal in \(\mathbb{Z}[x]\) but is not a maximal ideal.
Ring of Fractions.
Theorem 15. Let \(R\) be a commutative ring. Let \(D\) be any non-empty subset of \(R\) that does not contain \(0\) or any zero divisors and is closed under multiplication. Then there exists a commutative ring \(Q\) with \(1\) such that \(Q\) contains \(R\) as a subring and every element of \(D\) is a unit in \(Q\). The ring \(Q\) has the following properties:
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Every element of \(Q\) is of the form \(rd^{-1}\) for some \(r\in R\) and \(d\in D\). In particular, if \(D=R-\{0\}\) then \(Q\) is a field.
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(uniqueness of \(Q\)) The ring \(Q\) is the "smallest" ring containing \(R\) in which all elements of \(D\) become units, in the sense that: any ring containing an isomorphic copy of \(R\) in which all the elements of \(D\) become units must also contain an isomorphic copy of \(R\).
Definition. Let \(R,D,Q\) be as in Theorem 15:
- The ring \(Q\) is called the ring of fractions of \(R\) with respect to \(D\) and is denoted as \(D^{-1}R\).
- If \(R\) is an integral domain and \(D=R-\{0\}\), \(Q\) is called the field of fractions or quotient field of \(R\).
The Chinese Remainder Theorem
Definition. The ring direct product of \(R_1,R_2\), denoted by \(R_1\times R_2\), is defined as:
- Addition: \((r_1,r_2)+(s_1,s_2)=(r_1+s_1,r_2+s_2)\).
- Multiplication: \((r_1,r_2)(s_1,s_2)=(r_1s_1,r_2s_2)\).
Definition. The ideals \(A\) and \(B\) are said to be comaximal if \(A+B=R\).
Theorem 17. (The Chinese Remainder Theorem) Let \(A_1,A_2,\cdots,A_k\) be ideals in \(R\), the map \(R\to(R/A_1)\times (R/A_2)\times\cdots(R/A_k)\) defined by \(r\to(r+A_1,r+A_2,\cdots,r+A_k)\) is a ring homomorphism with kernel \(A_1\cap A_2\cap\cdots\cap A_k\). If for each \(i,j\in\{1,2,3,\cdots,k\}\) with \(i\ne j\) the ideals \(A_i\) and \(A_j\) are comaximal, then this map is surjective and \(A_1\cap A_2\cap\cdots\cap A_k=A_1A_2\cdots A_k\) so \(R/(A_1A_2\cdots A_k)=R/(A_1\cap A_2\cap\cdots\cap A_k)\cong (R/A_1)\times (R/A_2)\times\cdots(R/A_k)\).
Proof: We first prove this for \(k=2\), the general case will follow by induction. Let \(A=A_1,B=A_2\). Consider the map \(\varphi:R\to(R/A)\times(R/B)\) defined by \(\varphi(r)=(r\bmod A,r\bmod B)\). \(\varphi\) is a ring homomorphism since it is just the natural projection of \(R\) into \(R/A\) and \(R/B\) for the two components. \(\ker\varphi\) consists of all the elements \(r\in R\) that are in \(A\) and \(B\), i.e., \(A\cap B\). When \(A+B=R\), \(\exists x\in A,y\in B\) such that \(x+y=1\). Thus \(\varphi(x)=(0,1),\varphi(y)=(1,0)\). If now \((r_1\bmod A,r_2\bmod B)\) is an arbitrary element in \((R/A)\times(R/B)\), then the element \(r_2x+r_1y\) maps to this element since \(\varphi(r_2x+r_1y)=\varphi(r_2)\varphi(x)+\varphi(r_1)\varphi(y)=(0,r_2\bmod B)+(r_1\bmod A,0)=(r_1\bmod A,r_2\bmod B)\). This shows that \(\varphi\) is surjective. Finally, the ideal \(AB\) is always contained in \(A\cap B\). If \(A\) and \(B\) are comaximal and \(x,y\) are as above, then for any \(c\in A\cap B\), \(c=c·1=cx+cy\in AB\), so \(AB=A\cap B\).
The general case follows easily by induction from the case of two ideals using \(A=A_1\) and \(B=A_2A_3\cdots A_k\) once we show that \(A_1\) and \(A_2A_3\cdots A_k\) are comaximal. By hypothesis, for each \(i\in\{2,3,\cdots,k\}\) there are \(x_i\in A_1\) and \(y_i\in A_i\) such that \(x_i+y_i=1\). Since \(x_i+y_i\equiv y_i\bmod A_1\), it follows that \(1=(x_2+y_2)(x_3+y_3)\cdots(x_k+y_k)\) is an element in \(A_1+(A_2A_3\cdots A_k)\), this completes the proof.
Corollary 18. Let \(n\) be a positive integer and let \(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\) be its factorization into powers of distinct primes. Then \(Z/nZ\cong (Z/p_1^{\alpha_1}Z)\times(Z/p_2^{\alpha_2}Z)\times\cdots\times (Z/p_k^{\alpha_k}Z)\). \(\varphi(n)=p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)\cdots p_k^{\alpha_k-1}(p_k-1)\).
Euclidean Domains
Definition: Any function \(N:R\to\mathbb{Z}^+\cup\{0\}\) with \(N(0)=0\) is called a norm on the integral domain \(R\). If \(N(a)>0\) for \(\forall a\ne 0\), then \(N\) is called a positive norm. The integral domain \(R\) is said to be a Euclidean domain if there exists \(N\) such that for \(\forall a,b\in R,b\ne 0\), \(\exists p,q\in R\) such that \(a=qb+r\) with \(r=0\) or \(N(r)<N(b)\). The element \(q\) is called the quotient and \(r\) the remainder.
Euclidean Algorithm: \(a=q_0b+r_0,b=q_1r_0+r_1,r_0=q_2r_1+r_2,r_1=q_3r_2+r_3,\cdots\).
Examples:
- Fields are trivial examples of Euclidean domains since \(\forall a,b\), \(b\ne 0\), we have \(a=ab^{-1}·b+0\).
- The integer ring \(\mathbb{Z}\) are Euclidean domains with norm given by \(N(a)=|a|\), the usual absolute value.
- If \(F\) is a field, then \(F[x]\) is a Euclidean Domain.
Property 1. Every ideal of a Euclidean domain is principal. More precisely, if \(I\) is any non-zero ideal in the Euclidean domain \(R\) then \(I=(d)\), where \(d\) is any non-zero element of \(I\) of minimum norm.
Proof: If \(I\) is zero ideal, then nothing is to prove. So we suppose \(I\) is not the zero ideal. Let \(d\in I\) with minimum norm, obviously \((d)\in I\). Inversely, let \(a\in I\) and use the division algorithm to write \(a=qd+r\) with \(r=0\) or \(N(r)<N(d)\). Then \(r=a-qd\in I\). By the minimality of the norm of \(d\), we see that \(r\) must be \(0\). Thus \(a=qd\in(d)\), showing \(I=(d)\).
Example: Let \(R=\mathbb{Z}[x]\). Since the ideal \((2,x)\) is not principal, so \(\mathbb{Z}[x]\) is not a Euclidean domain for any choice of norm, even though \(\mathbb{Q}[x]\) is.
Definition. Let \(R\) be a commutative ring and \(a,b\in R\), \(b\ne 0\).
- \(a\) is said to be a multiple of \(b\) if there exists \(x\in R\) such that \(a=xb\). In this case \(b\) is said to divide \(a\) or be a divisor of \(a\), written \(b\mid a\).
- A greatest common divisor of \(a\) and \(b\) is a non-zero element \(d\) such that \(d\mid a,d\mid b\), and if \(d'\mid a,d'\mid b\), then \(d'\mid d\). A greatest divisor of \(a\) and \(b\) will be denoted as \(\gcd(a,b)\), or \((a,b)\) if \(a,b\) is clear in the context.
Since \(b\mid a\leftrightarrow(a)\sube(b)\), the properties of greatest common divisor translated into the language of ideals therefore becomes:
If \(I\) is the ideal generated by \(a\) and \(b\), then \(d\) is a greatest common divisor of \(a\) and \(b\) if:
- \(I\) is contained in \((d)\).
- If \((d')\) is any principal ideal containing \(I\) then \((d)\sube(d')\)
Property 2. If \(a\) and \(b\) are non-zero elements in the commutative ring \(R\) such that the ideal generated by \(a\) and \(b\) is a principal ideal \((d)\), then \(d=\gcd(a,b)\).
An integral domain in which every ideal \((a,b)\) is principal is called a Bezout Domain.
5.6
Property 3. Let \(R\) be an integral domain. If \(d\) and \(d'\) generate the same principal ideal, i.e., \((d)=(d')\), then \(d'=ud\) for some unit \(u\). In particular, if \(d\) and \(d'\) are both greatest common divisors of \(a\) and \(b\), then \(d=ud'\) for some unit \(u\).
Theorem 4. Let \(R\) be an Euclidean Domain and let \(a,b\) be non-zero elements of \(R\) and \(d=r_n\) be the last non-zero remainder in the Euclidean Algorithm for \(a\) and \(b\). Then
- \(d=\gcd(a,b)\).
- \((d)=(a,b)\). In particular, \(\exists x,y\in R\) such that \(d=ax+by\).
Since the equation \(ax+by=n\) is simply another way of stating that \(n\) is the element of the ideal generated by \(a\) and \(b\). Hence the equation is solvable in integers \(x\) and \(y\) if and only if \(N\) is divisible by \(\gcd(a,b)\).
Principal Ideal Domain (P.I.D)
Definition. A principal ideal domain is an integral domain in which every ideal is principal.
Of course, every Euclidean domain is always a principal ideal domain.
Example:
- \(\mathbb{Z}\) is a principal ideal domain, but \(\mathbb{Z}[x]\) is not.
- \(\mathbb{Z}[\sqrt{-5}]\) is not a principal ideal domain, since \((3,1+\sqrt{-5})\) is not principal.
It is not true that every P.I.D is a Euclidean domain. For example, \(\mathbb{Z}[\dfrac{1+\sqrt{-19}}{2}]\) is a P.I.D, but not a Euclidean domain.
Property 6. Let \(R\) be a P.I.D and \(a,b\) are non-zero elements of \(R\). Let \(d\) be a generator for \((a,b)=(d)\), then
- \(d\) is a greatest common divisor of \(a\) and \(b\).
- \(\exists x,y\in R\), such that \(d=ax+by\).
- \(d\) is unique up to some unit in \(R\).
Property 7. Every non-zero prime ideal in a P.I.D is a maximal ideal.
Proof: Let \((p)\) be a non-zero prime ideal in the P.I.D \(R\), and \(I=(m)\) be any ideal containing \((p)\). We must show that \(I=(p)\) or \(I=R\). Since \(p\in (m)\), \(p=rm\) for some \(r\in R\). Since \((p)\) is a prime ideal and \(rm\in (p)\), either \(r\) or \(m\) must lie in \((p)\). If \(m\in(p)\) then \((p)=(m)=I\). If \(r\in(p)\), write \(r=ps\), then \(p=rm=psm\), so \(sm=1\), \(m\) is a unit so \(I=R\).
Corollary 8. If \(R\) is any commutative ring such that \(R[x]\) is a P.I.D, then \(R\) is a field.
Proof: Assume \(R[x]\) is a P.I.D. Since \(R\) is a subring of \(R[x]\), then \(R\) must be an integral domain. The ideal \((x)\) is a non-zero prime ideal in \(R[x]\) because \(R[x]/(x)\) is isomorphic to the integral domain \(R\). Thus by Property 7, \((x)\) is a maximal ideal, \(R\cong R[x]/(x)\) is a field.
Definition. Let \(R\) be an integral domain:
- Suppose \(r\in R\) is non-zero and not a unit. Then \(r\) is called irreducible in \(R\) whenever \(r=ab\) with \(a,b\in R\), at least one of \(a,b\) must be a unit in \(R\). Otherwise, \(r\) is reducible.
- The non-zero element \(p\in R\) is called prime if \((p)\) is prime.
- Two element \(a\) and \(b\) of \(R\) differing by a unit are said to be associate in \(R\) (i.e. \(a=ub\))
Example: In \(\mathbb{Z}\), the irreducible elements are prime numbers, and two integers \(a\) and \(b\) are associates of each other if and only if \(a=\pm b\).
Property 10. In a integral domain a prime element is always irreducible.
Proof: Suppose \((p)\) is a prime ideal and \(p=ab\), then \(ab=p\in(p)\), so one of \(a,b\), say \(a\), is in \((p)\). Thus \(a=pr\) for some \(r\in R\). So \(p=ab=prb\), so \(rb=1\), and \(b\) is a unit. This shows that \(p\) is irreducible.
However, an irreducible element may not a prime element. Counter example: Let \(R=\mathbb{Z}[\sqrt{-5}]\), then \(3\) is irreducible but not prime since \(9=(2+\sqrt{-5})(2-\sqrt{-5})\) but \(9\in(3)\).
Yet, if \(R\) is a P.I.D, the notations of prime and irreducible are the same.
Property 11. In a P.I.D, a non-zero element is prime if and only if it is irreducible.
Proof:
Prime->irreducible: Proven in Property 10.
Irreducible->prime: Let \(M\) be any ideal containing \((p\)), then \(M=(m)\) is a principle ideal. Since \(p\in (m)\), \(p=rm\) for some \(r\in R\). But \(p\) is irreducible, so either \(r\) or \(m\) is a unit. So either \((p)=(m)\) or \((m)=1\). Thus \((p)\) is a maximal ideal, hence prime.
Unique Factorization Domain
Definition. A U.F.D is an integral domain \(R\) in which every non-zero element \(r\in R\), which is not a unit obeys:
- \(r=p_1p_2\cdots p_m\), where \(p_1,p_2,\cdots,p_n\) are irreducible elements inside \(R\).
- The decomposition \((i)\) is unique up to associates.
Example:
- A field is always a U.F.D.
- \(\mathbb{Z}[\sqrt{-5}]\) is not a U.F.D since \(6=2·3=(1+\sqrt{-5})(1-\sqrt{-5})\).
Property 12. In a U.F.D, a non-zero element is prime if and only if it is irreducible.
Proof:
Prime->irreducible: Proven in Property 10.
Irreducible->prime: Let \(p\) be irreducible in \(R\) and assume \(p\mid ab\) for some \(a,b\in R\), we must prove \(p\mid a\) or \(p\mid b\). Since \(p\mid ab\), \(ab=pc\) for some \(c\in R\). Written \(a,b\) as product as irreducibles. We see from the decomposition of \(ab\), \(p\) must be associate to one of the irreducible occuring either in \(a\) or \(b\). So \(p\) divides either \(a\) or \(b\).
Theorem 14. A P.I.D is always a U.F.D.
Corollary 15. (Fundamental Theorem of Arithmetic): \(\mathbb{Z}\) is a U.F.D.
Fields \(\sube\) E.D \(\sube\) P.I.D \(\sube\) U.F.D \(\sube\) Integral domains \(\sube\) Rings \(\sube\) Groups.
All containments propers:
- \(\mathbb{Z}\) is a Euclidean domain, but not a field.
- \(\mathbb{Z}[\dfrac{(1+\sqrt{-19})}{2}]\) is a P.I.D, but not a Euclidean domain.
- \(\mathbb{Z}[x]\) is a U.F.D, but not a P.I.D.
- \(\mathbb{Z}[-\sqrt{5}]\) is an integral domain, but not a U.F.D.
Polynomial rings
Property 2. Let \(I\) be an ideal of the ring \(R\) and \((I)=I[x]\), then \(R[x]/(I)\cong(R/I)[x]\). In particular, if \(I\) is a prime ideal of \(R\), then \((I)\) is a prime ideal of \(R[x]\).
Proof: There is a natural map \(\varphi:R[x]\to (R/I)[x]\) given by reducing each of the coefficients modulo \(I\). Clearly \(I\) is a surjective homomorphism, and \(\ker\varphi=I[x]=(I)\). By the first the first Isomorphism Theorem, \(R[x]/(I)\cong(R/I)[x]\). In addition, if \(I\) is a prime ideal in \(R\), then \(R/I\) is an integral domain, hence also \((R/I)[x]\), \((I)\) is a prime ideal of \(R[x]\).
Definition. The polynomial ring in the variables \(x_1,x_2,\cdots,x_n\) with coefficients in \(R\), denoted \(R[x_1,x_2,\cdots,x_n]\), is defined inductively by \(R[x_1,x_2,\cdots,x_n]=R[x_1,x_2,\cdots,x_{n-1}][x_n]\), i.e. a finite sum of non-zero monomial terms: \(ax_1^{d_1}x_2^{d_2}\cdots x_n^{d_n}\).
Example: \(\mathbb{Z}[x,y]\) consists of all finite sums of monomial terms of the form \(ax^iy^j\): for example, \(p(x,y)=2x^3+xy-y^2,q(x,y)=-3xy+2y^2+x^2y^3\) are both elements of \(\mathbb{Z}[x,y]\) of degree \(3\) and \(5\).
Polynomial Rings over fields
Theorem 3. Let \(F\) be a field, then \(F[x]\) is a Euclidean Domain. Specifically, if \(a(x)\) and \(b(x)\in F[x]\) with \(b(x)\) non-zero, then there exists unique \(q(x)\) and \(r(x)\in F(x)\) such that \(a(x)=q(x)b(x)+r(x)\) with \(r(x)=0\) or \(\deg r(x)<\deg b(x)\).
Proof: If \(a(x)=0\), then take \(q(x)=r(x)=0\). Thus we may assume \(a(x)\ne 0\) and prove the existence of \(q(x)\) and \(r(x)\) by induction on \(n=\deg a(x)\). If \(n<m\), then take \(q(x)=0\) and \(r(x)=a(x)\). Assume \(n\ge m\), then write \(a(x)=a_nx^n+a^{n-1}x^{n-1}+\cdots+a_1x+a_0\), \(b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0\). Then \(a'(x)=a(x)-\dfrac{a_n}{b_m}x^{n-m}b(x)\), is of degree less than \(n\). By induction, there exists \(q'(x),r(x)\) with \(a'(x)=q'(x)b(x)+r(x)\) with \(r(x)=0\) or \(\deg r(x)<\deg b(x)\). Then letting \(q(x)=q'(x)+\dfrac{a_n}{b_m}x^{n-m}\), we have \(a(x)=q(x)b(x)+r(x)\) with \(r(x)=0\) or \(\deg r(x)<\deg b(x)\). As for uniqueness, suppose \(q_1(x)\) and \(r_1(x)\) also satisfy \(a(x)=q_1(x)b(x)+r_1(x)\), then \(r(x)-r_1(x)=b(x)(q(x)-q_1(x))\) has degree less than \(m=\deg b(x)\). So \(q_1(x)-q(x)=0\), \(r(x)=r_1(x)\).
Corollary 4. If \(F\) is a field, then \(F[x]\) is a P.I.D and U.F.D.
Property 5. (Gauss Lemma) Let \(R\) be a U.F.D with field of fractions \(F\) and let \(p(x)\in R[x]\). If \(p(x)\) is reducible in \(F[x]\), then \(p(x)\) is reducible in \(R[x]\).
Corollary 6. Let \(R\) be a U.F.D. \(F\) be its field of fractions, \(p(x)\in R[x]\). Suppose the greatest common divisor of the coefficients of \(p(x)\) is \(1\), then \(p(x)\) is irreducible in \(R[x]\) if and only if it is irreducible in \(F[x]\).
Proof. By Gauss's Lemma, if \(p(x)\) is reducible in \(F[x]\), then it is reducible in \(R[x]\). Conversely, since the greatest common divisor of the coefficients of \(p(x)\) is \(1\), so if \(p(x)\) is reducible in \(R[x]\), then \(p(x)=a(x)b(x)\) where \(a(x),b(x)\) are not constant polynomial in \(R[x]\). Thus the same factorization shows that \(p(x)\) is reducible in \(F[x]\).
Theorem 7. \(R\) is a U.F.D if and only if \(R[x]\) is a U.F.D.
Corollary 8. If \(R\) is a U.F.D, then a polynomial ring in an arbitrary number of variables with coefficients in \(R\) is also a U.F.D.
5.13
Irreducibility criteria
Property: Let \(F\) be a field and \(p(x)\in F[x]\), then \(p(x)\) has a factor of degree \(1\) if and only if \(p(x)\) has a root in \(F\), i.e, there exists \(\alpha\in F\) s.t. \(p(x)=0\).
Proof: If \(p(x)\) has a factor of degree \(1\), i.e., \((x-\alpha)\), then \(p(\alpha)=0\). Conversely, suppose \(p(\alpha)=0\), by the Euclidean Algorithm, \(p(x)=q(x)(x-\alpha)+r\), where \(r\) is a constant. Since \(p(\alpha)=r=0\), \(p(x)=q(x)(x-\alpha)\).
Property 10. A polynomial of degree \(2\) or \(3\) over a field \(F\) is reducible if and only if it has a root in \(F\).
Property 11. Let \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\in \mathbb{Z}[x]\), if \(\dfrac{r}{s}\in\mathbb{Q}\) (\(r,s\) are relatively prime) is a root of \(p(x)\), then \(r\mid a_0\) and \(s\mid a_n\). In particular, if \(p(x)\) is a monic polynomial, and \(p(d)\ne 0\) for all integers \(d\) dividing \(a_0\), then \(p(x)\) has no root in \(\mathbb{Q}\).
Proof. By hypothesis, \(p(\dfrac{r}{s})=0=a_n(\dfrac{r}{s})^n+a_{n-1}(\dfrac{r}{s})^{n-1}+\cdots+a_0\), so \(0=a_nr^n+a_{n-1}r^{n-1}s+\cdots+a_0s^n\). Thus \(a_nr^n=s(-a_{n-1}r^{n-1}-\cdots-a_0s^{n-1})\), since \((r,s)=1\), \(s\mid a_n\). Similarly \(a_0s^n=r(-a_nr^{n-1}-a_{n-1}r^{n-1}s-\cdots-a_1s^{n-1})\), \(r\mid a_0\).
Examples:
- \(x^3+3x-1\) in \(\mathbb{Z}[x]\) is irreducible.
- \(x^2-p\) and \(x^3-p\) in \(\mathbb{Q}[x]\) is irreducible.
- \(x^2+1\) in \(\mathbb{Z}/2\mathbb{Z}\) is reducible since it has a root \(x=1\).
- \(x^2+x+1\) in \(\mathbb{Z}/2\mathbb{Z}\) is irreducible.
Property 12. Let \(I\) be a proper ideal in \(R\) and \(p(x)\in R[x]\) be a non-constant monic polynomial. If the image of \(p(x)\) in \((R/I)[x]\) cannot be factored in \((R/I)[x]\) into two polynomials of smaller degree, then \(p(x)\) is irreducible in \(R[x]\).
Proof. Suppose \(p(x)\) cannot be factored in \((R/I)[x]\) but \(p(x)\) is reducible in \(R[x]\), then there exists monic, non-constant polynomials \(a(x),b(x)\) in \(R[x]\) such that \(a(x)+b(x)\). Reducing the coefficients modulo \(I\) gives a factorization in \(R/I[x]\) with non-constant factors.
Property 13. (Eisenstein's Criterion) Let \(P\) be a prime ideal of the integral domain \(R\) and \(f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0\) be a polynomial in \(R[x]\) (\(n\ge 1\)) Suppose \(a_{n-1},\cdots,a_1,a_0\) are all elements of \(P\), but \(a_0\notin P^2\), then \(f(x)\) is irreducible in \(R[x]\).
Proof. Assume \(f(x)\) were reducible, say \(f(x)=a(x)b(x)\) in \(R[x]\), where \(a(x),b(x)\) are non-constant polynomials. Reducing this equation modulo \(P\), we have \(x^n=\overline{a(x)}\cdot\overline{b(x)}\) in \((R/P)[x]\). Since \(P\) is a prime ideal, \(R/P\) is an integral domain, it follows that both \(\overline{a}(x)\) and \(\overline{b}(x)\) have zero constant term, \(a_0\in P^2\), which contradicts with the assumption.
Example:
- \(p(x)=x^4+10x+5\) in \(\mathbb{Z}[x]\) is irreducible by Eisenstein's Criterion.
- \(I_p(x)=x^{p-1}+x^{p-2}+\cdots+1\) is irreducible in \(\mathbb{Z}[x]\) by Eisenstein's Criterion.
Polynomial Rings over Fields
Property 15. The maximal ideals in \(F[x]\) are the ideals \((f(x))\) generated by irreducible polynomials \(f(x)\). In particular, \(F[x]/(f(x))\) is a field if and only if \(f(x)\) is irreducible.
Property 16. Let \(g(x)\) be a non-constant monic element of \(F[x]\) and let \(g(x)=f_1(x)^{n_1}f_2(x)^{n_2}\cdots f_k(x)^{n_k}\) be its factorization into irreducibles, where the \(f_i(x)\) are distinct. Then we have \(F[x]/(g(x))=(F[x]/(f_1(x)^{n_1}))\times(F[x]/(f_2(x)^{n_2}))\times\cdots\times (F[x]/(f_k(x)^{n_k}))\).
Property 17. If \(f(x)\) has roots \(\alpha_1,\alpha_2,\cdots,\alpha_k\) in \(F\), then \(f(x)\) has \((x-\alpha_1)\cdots(x-\alpha_k)\) factor. In particular, a polynomial of degree \(n\) over a field has at most \(n\) roots.
Field Theory
5.13s
Definition. The characteristic of a field \(F\), denoted by \(\text{ch}(F)\) is the smallest positive integer \(p\) such that \(p·1\) if such \(p\) exist. Otherwise \(\text{ch}(F)=0\).
\(\text{ch}(F)\) is either \(0\) or a prime number.
Definition. The prime subfield of a field \(F\) is the subfield of \(F\) generated by the identity \(1_F\) of \(F\). It is isomorphic to either \(\mathbb{Q}\) (if \(\text{ch}(F)=0\)) or \(\mathbb{F}_p\) (if \(\text{ch}(F)=p\))
Definition. If \(K\) is a field containing the subfield \(F\), then \(K\) is said to be an extension field (or simply an extension) of \(F\), denoted by \(K/F\).
If \(K/F\) is any extension fields, then the multiplication in \(K\) makes \(K\) into a vector space over \(F\). In particular, every field \(F\) can be considered as a vector field over its prime field.
Definition. The degree (or relative degree of index) of \(K/F\), denoted by \([K:F]\), is the dimension of \(K\) as a vector space over \(F\) (i.e., \([K:F]=\dim_FK)\). The extension is said to be finite if \([K:F]\) is finite and infinite otherwise.
Property 2. Let \(\varphi:F\to F'\) be a homomorphism of fields. Then \(\varphi\) is either identically \(0\) or is injective, so that the image of \(\varphi\) is either \(0\) or isomorphic to \(F\).
Theorem 3. Let \(F\) be a field and \(p(x)\in F[x]\) be an irreducible polynomial. Then there exists a field \(K\) containing an isomorphic copy of \(F\) in which \(p(x)\) has a root. Identifying \(F\) with this isomorphic copy shows that there exists an extension of \(F\) in which \(p(x)\) has a root.
Proof: Consider the quotient \(K=F[x]/(p(x))\). Since by assumption \(p(x)\) is an irreducible polynomial in the P.I.D \(F[x]\). The ideal \((p(x))\) is maximal hence \(K\) is a field indeed. The canonical projection \(\pi\) of \(F[x]\) to the quotient \(F[x]/(p(x))\) restricted to \(F\sub F[x]\) gives a homomorphism \(\varphi=\pi_F:F\to K\) which is not identically \(0\) since it maps \(1_F\) to \(1_K\). Hence \(\varphi(F)\cong F\) is an isomorphic copy of \(F\) in \(K\). We identify \(F\) with its isomorphic image in \(K\) and view \(F\) as a subfield of \(K\). If \(\overline{x}=\pi(x)\) denote the image of \(x\) in \(K\), then \(p(\overline{x})=\overline{p(x)}=p(x)\bmod p(x)=0\). So \(K\) does contain a root \(\overline{x}\) of the polynomial \(p(x)\).
Theorem 4. Let \(p(x)\in F[x]\) be an irreducible polynomial of degree \(n\) over the field \(F\) and let \(K=F[x]/(p(x))\). Let \(\theta=x\bmod p(x)\). Then the elements \(1,\theta,\theta^2,\cdots,\theta^{n-1}\) are a basis for \(k\) as a vector space over \(F\). So the degree of the extension is \(n\), i.e., \([K:F]=n\). Hence \(K=\{a_0+a_1\theta+a_2\theta^2+\cdots+a_{n-1}\theta^{n-1}|a_0,a_1,\cdots,a_{n-1}\in F\}\).
Proof: Let \(a(x)\in F[x]\), since \(F[x]\) is a E.D, we have \(a(x)=q(x)p(x)+r(x)\), \(q(x),r(x)\in F[x]\) with \(\deg r(x)<n\). Since \(q(x)p(x)\in(p(x))\), it follows that \(a(x)\equiv r(x)\bmod p(x)\), which shows that every residue class in \(F[x]/(p(x))\) is represented by a polynomial of degree less than \(n\). Hence the images \(1,\theta,\theta^2,\cdots,\theta^{n-1}\) of \(1,x,\cdots,x^{n-1}\) in the quotient span the quotient as a vector space over \(F\). It remains to see that these elements are linearly independent, so form a basis of \(K\). If they are not linearly independent in \(K\), then there exists a linear combination \(b_0+b_1\theta+\cdots+b_{n-1}\theta^{n-1}=0\) in \(K\) with \(b_0,b_1,\cdots,b_{n-1}\in F\), not all \(0\). This is equivalent to \(b_0+b_1x+b_2x^2+\cdots+b_{n-1}x^{n-1}\equiv 0\pmod{p(x)}\), which is impossible.
5.20
Definition. Let \(K/F\) and \(\alpha,\beta,\cdots\in K\), then the smallest subfield of \(K\) containing both \(F\) and \(\alpha,\beta\), denoted as \(F(\alpha,\beta,\cdots)\) is called the field generated by \(\alpha,\beta,\cdots\).
If the field \(K\) is generated by a single element \(\alpha\) over \(F\), \(K=F(\alpha)\), then \(K\) is said to be a simple extension of \(F\) and the element \(\alpha\) is called a primitive element for the extension.
Theorem 6. Let \(F\) be a field, \(p(x)\in F[x]\) be an irreducible polynomial. Suppose \(K/F\) containing a root \(\alpha\) of \(p(x)\): \(p(\alpha)=0\). Then \(F(\alpha)\cong F[x]/(p(x))\).
Proof: There exists a natural homomorphism \(\varphi:F[x]\to F(\alpha)\), as \(\varphi(a(x))=a(\alpha)\). Since \(p(\alpha)=0\), \(p(x)\in\ker\varphi\), so we obtain an induced homomorphism from \(F[x]/(p(x))\to F(\alpha)\). Since \(p(x)\) is irreducible, \(F[x]/(p(x))\) is a field, and \(\varphi\) is not the zero map, hence \(\varphi\) is an isomorphism of the field on the left to its image. Since this image is then a subfield of \(F[\alpha]\) containing \(F\) and \(\alpha\), hence equal to \(F(\alpha)\).
Theorem 8. Let \(\varphi:F\to F'\) be an isomorphism of fields. Let \(p(x)\in F[x]\) be an irreducible polynomial and \(p'(x)\in F'(x)\) be the irreducible polynomial obtained by applying \(\varphi\) to the coefficients of \(p(x)\). Let \(\alpha\) be a root of \(p(x)\), and \(\beta\) be a root of \(p'(x)\). Then there exists an isomorphism \(\sigma\) from \(F(\alpha)\) to \(F'(\beta)\) which maps \(\alpha\) to \(\beta\) and extending \(\varphi\), i.e. \(\sigma|_F=\varphi\).
Proof. Clearly, \(\varphi\) induces a natural isomorphism from \(F[x]\) to \(F'[x]\), which maps the maximal ideal \((p(x))\) to the maximal ideal \((p'(x))\). Taking quotients by these ideals, we obtain an isomorphism: \(F[x]/(p(x))\to F'[x]/(p'(x))\). By Theorem 6, \(F[x]/(p(x))\cong F(\alpha)\) and \((p'(x))\cong F(\beta)\). Composing these isomorphisms, we obtain \(\sigma\).
Algebraic Extensions: \(K/F\).
Definition. \(\alpha\in K\) is said to be algebraic over \(F\) if \(\alpha\) is a root of some non-zero polynomial \(f(x)\in F[x]\). Otherwise \(\alpha\) is transcendental over \(F\). \(K/F\) is said to be algebraic if every element of \(K\) is algebraic over \(F\).
Property 9. Let \(\alpha\) be algebraic over \(F\), then there exists a unique monic irreducible polynomial \(m_{\alpha,F}\in F[x]\) which has \(\alpha\) as a root. A polynomial \(f(x)\in F[x]\) has a root if and only if \(m_{\alpha,F}(x)\) divides \(f(x)\) in \(F[x]\).
Corollary. If \(L/F\) and \(\alpha\) is algebraic over both \(F\) and \(L\), then \(m_{\alpha,L}(x)\) divides \(m_{\alpha,F}(x)\) in \(L[x]\).
The polynomial \(m_{\alpha,F}(x)\) is called the minimal polynomial for \(\alpha\) over \(F\). The degree of \(m_{\alpha}(x)\) is called the degree of \(\alpha\).
Property 12. The element \(\alpha\) is algebraic over \(F\) if and only if the simple extension \(F(\alpha)/F\) is finite.
Proof: If \(\alpha\) is algebraic, then \(m_{\alpha,F}(x)\) is of finite degree. Conversely, suppose \([F(\alpha):F]=n\) be finite, then the \(n+1\) elements, \(1,\alpha,\alpha^2,\cdots,\alpha^n\) of \(F(\alpha)\) are linearly dependent over \(F\). Say \(b_0+b_1\alpha+b_2\alpha^2+\cdots+b_n\alpha^n=0\) with \(b_0,b_1,\cdots,b_n\in F\) not all \(0\). Proven.
Corollary 13. If the extension \(K/F\) is finite, then it is algebraic.
Theorem 14. Let \(F\sube K\sube L\) be fields. Then \([L:F]=[L:K][K:F]\).
Corollary 15. Suppose \(L/F\) is a finite extension and let \(K\) be any subfield of \(L\) containing \(F\), \(F\sube K\sube L\). Then \([K:F]\) divides \([L:F]\).
Example: \([\mathbb{Q}(\sqrt[6]{2}),\mathbb{Q}]=6\), \([\mathbb{R}(\sqrt[6]{2}),\mathbb{R}]=1\), \([\mathbb{Q}(\sqrt[6]{2}),\mathbb{Q}(\sqrt{2})]=3\).
Theorem 17. \(K/F\) is finite if and only if \(K\) is generated by a finite number of algebraic element.
Theorem 20. \(K\) is algebraic over \(F\) and \(L\) is algebraic over \(K\), then \(L\) is algebraic over \(F\).
Definition. Let \(K_1,K_2\) be two subfields of a field \(K\), then the composite field of \(K_1\) and \(K_2\), denoted by \(K_1K_2\), is the smallest subfield of \(K\) containing both \(K_1\) and \(K_2\).
Property 21. Let \(K_1,K_2\) be two finite extensions of a field \(F\) containing \(K\). Then \([K_1K_2:F]\le[K_1:F][K_2:F]\) with equality if and only if an \(F\)-basis for one of the fields remains linearly independent over the other field. If \(\alpha_1,\alpha_2,\cdots,\alpha_n\) and \(\beta_1,\beta_2,\cdots,\beta_m\) are bases for \(K_1\) and \(K_2\) over \(F\) respectively, then \(\alpha_i\beta_j\) span \(K_1K_2\) over \(F\).
Splitting fields
Definition. \(K/F\) is called a splitting field for the polynomial \(f(x)\in F[x]\) if \(f(x)\) factors completely into linear factors (or splits completely) in \(K[x]\), and \(f(x)\) doesn't factor completely into linearly factors over any proper subfield of \(K\) containing \(F\).
Theorem 25. For any field \(F\), if \(f(x)\in F[x]\), then there exists an extension \(K\) of \(F\) which is a splitting field for \(f(x)\).
Example:
- The splitting field for \(x^2-2\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt{2})\).
- The splitting field for \((x^2-2)(x^2-3)\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt{2},\sqrt{3})\).
- The splitting field for \(x^4+4\) over \(\mathbb{Q}\) is \(\mathbb{Q}(i)\).
The splitting field has degree at most \(n!\).
Theorem 27. Let \(\varphi:F\to F'\) be an isomorphism, \(f(x)\in F[x]\), \(f'(x)\in F'[x]\). Let \(E\) be the splitting field for \(f(x)\) over \(F\), and \(E'\) be a splitting field for \(f'(x)\) over \(F'\). Then \(\varphi\) extends to an isomorphism \(\sigma:E\cong E'\)
5.27
Definition. The field \(\bar{F}\) is called an algebraic closure of \(F\) if \(\bar{F}\) is algebraic over \(F\) and if every polynomial \(f(x)\in F[x]\) splits completely over \(\bar{F}\). A field \(K\) is algebraic closed if every polynomial inside \(K[x]\) has a root in \(K\).
Theorem (Fundamental Theorem of algebra). The field \(\mathbb{C}\) is algebraic closed.
Separable extension
Definition. A polynomial over \(F\) is called separable if it has no multiple roots. Otherwise it is inseparable.
Property 33. A polynomial \(f(x)\) has a multiple root \(\alpha\) if and only if \(\alpha\) is also a root of \(D_xf(x)\). In particular, \(f(x)\) is separable if and only it is relatively prime to its derivative: \((f(x),D_xf(x))=1\).
Example: \(f(x)=x^{p^n}-x\) over \(F_p\) is separable.
Corollary 34. Every irreducible polynomial over a field of characteristic \(0\) (for example, \(\mathbb{Q}\)), is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials.
Property 35. Let \(F\) be a field of characteristic \(p\), then for \(\forall a,b\in F\), \((a+b)^p=a^p+b^p\), and \((ab)^p=a^pb^p\). Put another way, \(\varphi(a)=a^p\) is an injective field homomorphism from \(F\) to \(F\).
Corollary 36. Suppose \(F\) is a finite field of characteristic \(p\), then every element of \(F\) is a \(p\)-th power in \(F\).
Property 37. Every irreducible polynomial over a finite field is separable.
Definition. A field \(K\) of characteristic \(p\) is called perfect if every element of \(K\) is a \(p\)-th power in \(K\), i.e. \(K=K^p\). Any field of characteristic \(p\) is also called perfect.
Definition. The field \(K\) is said to be separable (or separable algebraic) over \(F\) if every element of \(K\) is the root of a separable polynomial over \(F\).
Cyclotomic polynomial & extensions
Definition. The \(n\)-th cyclotomic polynomial \(\Phi\) is the polynomial whose roots are the primitive \(n\)-th roots of unity. \(\Phi_n(x)=\prod\limits_{1\le a<n,(a,n)=1}(x-\xi_n^a)\).
\(x^n-1=\prod\limits_{d\mid n}\Phi_n(x)\).
Galois Theory
Definition. An isomorphism \(\varphi\) of \(K\) with itself is called an automorphism of \(K\). \(\sigma\in\text{Aut}(K)\) is said to fix an element \(\alpha\in K\) if \(\sigma(\alpha)=\alpha\).
Definition. Let \(K/F\) be an extension of fields. Let \(\text{Aut}(K/F)\) be the collection of automorphisms of \(K\) which fix \(F\).
Property 2. Let \(K/F\) and \(\alpha\in K\) be algebraic over \(F\). Then for any \(\sigma\in\text{Aut}(K/F)\), \(\sigma(\alpha)\) is a root of the minimum polynomial for \(\alpha\) over \(F\).
Proof. Suppose \(\alpha\) is a root of \(f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in F[x]\), for any \(\sigma\in\text{Aut}(K/F)\), \(\sigma(\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_1\alpha+a_0)=(\sigma(\alpha))^n+a_{n-1}(\sigma(\alpha))^{n-1}+\cdots+a_1\sigma(\alpha)+a_0=0\).
Example:
- Let \(K=\mathbb{Q}(\sqrt{2})\). If \(\tau\in\text{Aut}(\mathbb{Q}(\sqrt{2}))=\text{Aut}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})\), then \(\tau(\sqrt{2})=\pm\sqrt{2}\) since these are the two roots of \(m_{\sqrt{2},\mathbb{Q}}=x^2-2\). So \(\text{Aut}(K)=\{1,\sigma\}\), where \(\sigma(a+b\sqrt{2})=a-b\sqrt{2}\).
- Let \(K=\mathbb{Q}(\sqrt[3]{2})\), \(\text{Aut}(K)=\{1\}\).
Property 3. Let \(H\le\text{Aut}(K)\), then the collection \(F\) of elements of \(K\) fixed by all elements of \(H\) is a subfield of \(K\).
Proof: Let \(h\in H,a,b\in F\), then \(h(a\pm b)=h(a)\pm h(b)=a\pm b\), \(h(ab)=h(a)h(b)=ab\), \(h(a^{-1})=a^{-1}\).
The subfield of \(K\) fixed by all the elements of \(H\) is called the fixed field of \(H\).
- If \(F_1\sube F_2\sube K\), then \(\text{Aut}(K/F_2)\le\text{Aut}(K/F_1)\).
- If \(H_1\le H_2\le\text{Aut}(K)\) are two subgroups of automorphisms with associated fixed fields \(F_1\) and \(F_2\) respectively, then \(F_2\sube F_1\).
Property 5. Let \(E\) be the splitting field over \(F\) of the polynomial \(f(x)\in F[x]\), then \(|\text{Aut}(E/F)|\le[E:F]\) with equality if \(f(x)\) is separable over \(F\).
Definition. Let \(K/F\) be a finite extension, the \(K\) is said to be Galois over \(F\) and \(K/F\) is a Galois extension if \(\text{Aut}(K/F)=[K:F]\). If \(K/F\) is Galois the group \(\text{Aut}(K/F)\) is called the Galois group of \(K/F\), denoted as \(\text{Gal}(K/F)\).
Corollary. If \(K\) is the splitting field over \(F\) of a separable polynomial \(f(x)\), then \(K/F\) is Galois.
Example:
- \(\mathbb{Q}(\sqrt{2})/\mathbb{Q}\) is Galois but \(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}\) is not.
- \(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}\) is Galois.
- The splitting field of \(x^3-2\) over \(\mathbb{Q}\) is Galois of degree \(6\). The roots of \(x^3-2\) are \(\sqrt[3]{2},\rho\sqrt[3]{2},\rho^2\sqrt[3]{2}\), where \(\rho=\dfrac{-1+\sqrt{-3}}{2}\). \(\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\rho)/\mathbb{Q})=\{1,\sigma,\tau,\sigma^2,\tau\sigma,\tau\sigma^2\}\), where \(\sigma(\sqrt[3]{2})=\rho\sqrt[3]{2},\sigma(\rho)=\rho\) and \(\tau(\sqrt[3]{2})=\sqrt[3]{2},\tau(\rho)=\rho^2\).
The Fundamental Theorem of Galois Theory
Definition. A character \(\chi\) of a group \(G\) with values in a field \(L\) is a homomorphism from \(G\) to the \(L^{\times}\). \(\chi:G\to L^{\times}\). The characters \(\chi_1,\chi_2,\cdots,\chi_n\) of \(G\) are said to be linearly independent over \(L\) if they are linearly independent as functions of \(G\), i.e., there is no non-trivial relation \(a_1\chi_1+a_2\chi_2+\cdots+a_n\chi_n=0\).
Theorem 7. If \(\chi_1,\chi_2,\cdots,\chi_n\) are distinct characters of \(G\) with values in \(L\) then they are linearly independent.
Theorem 9. Let \(G=\{\sigma_1=1,\sigma_2,\cdots,\sigma_n\}\) be a subgroup of automorphism of a field \(K\) and \(F\) be the fixed field, then \([K:F]=|G|\).
Corollary 10. Let \(K/F\) be any finite extension. Then \(|\text{Aut}(K/F)|\le[K:F]\) with equality if and only if \(F\) is the fixed field of \(\text{Aut}(K/F)\).
标签：mathbb,域论,varphi,field,ideal,Let,环论,抽象代数,alpha From： https://www.cnblogs.com/tzcwk/p/18215808/abstract-algebra2