"Abstract algebra" 06-Ideal with straight and

1. Homomorphism and ideals

The homomorphism theorem and the normal subgroup play an important role in the analysis of the structure of the group, we can do the same discussion on the ring. If the ring \ (r_1\) to another system \ (r_2\) has mapping \ (F:r_1\mapsto r_2\), satisfies the formula (1), such a mapping is called the homomorphic mapping . If the map is full, it is called \ (r_1,r_2\) homomorphism , recorded as \ (R_1\sim r_2\). Easy to prove \ (r_2\) is also a ring, and \ (r_1\) of 0 yuan, negative, unit, inverse, exchangeable and other properties will map to \ (r_2\), but the 0 factor is not necessarily maintained.

\[f (a+b) =f (a) +f (b); \quad f (AB) =f (a) f (b) \tag{1}\]

　　• verification: \ (Z_m\sim z_n\) the necessary and sufficient conditions are \ (N\mid m\).

As already known in the group, any homomorphism mapping corresponds to a normal subgroup (homomorphic nucleus), and the same-ring homomorphism can be equivalent to the study of homomorphic nuclei. Like the group, the homomorphism nucleus of the ring homomorphism is the image of the 0 element in the r_2\. It is easy to prove that the homomorphism nucleus is a sub-ring, just like the particularity of the normal subgroup, it is not a normal sub-ring. Considering the zeroing of 0 elements, the homomorphism kernel must satisfy the following conditions. Generally, the addition subgroup in the Ring \ (r\) (n\) If the following right one is satisfied, it is called the Ring of the Left (right) ideal, both are satisfied with the call ideal , recorded as \ (N\trianglelefteq r\), it is easy to prove that the ideal (including the ideal) is a sub-ring.

\[n\in n,\: R\in r\quad\rightarrow\quad rn\in n,\: nr\in n\tag{2}\]

By definition, the ideal is the subgroup of the addition group, so it is the normal subgroup under addition. It is easy to prove that the homomorphism mapping in the addition group to the normal subgroup is also the homomorphism map (multiplication closed) in the ring, so that each homomorphism map of the ring corresponds to the ideal one by one of the ring, and the ideal serves as the normal subgroup. Like normal subgroups, ideals are not transitive, i.e. ideal ideals are not necessarily ideal. It is easy to prove that the ideal intersection is ideal, and any sub-rings of the loop are ideal. To the general ring \ (r\), obviously \ (ra\) and \ (ar\) respectively is its right and left ideal.

Ideal is a special sub-ring, each ring \ (r\) have \ (\{0\}\) and \ (r\) Two ordinary ideals, other ideal shouting truth ideal, there is no real ideal ring called single ring . From the definition of ideal, to any \ (n\in n\) have \ (Nr\subseteq n\), compared to the group, the structure is "collapsed", from this association to the single ring and "good" ring must have what relationship. A good ring, of course, refers to the multiplication of groups to form the addition of the ring and the domain, if they have non-0 ideal \ (n\), by \ (aa^{-1}=1\) know \ (1\in n\), thus \ (n=r\), that is, except the ring and the domain must be a single ring.

For any ring \ (r\), because \ (ra\) is its left ideal, if \ (r\) does not have a non-trivial left ideal, then \ (ra\) is \ (r\) or \ (\{0\}\). If present \ (ra=\{0\}\), it is easy to prove \ (a\) The build ring is ideal, so that the build ring is \ (r\), it is a 0-multiplicative ring. Conversely, if \ (ra=r\) is always established, that is, the first equation \ (ya=b\) always has a solution, so \ (r\) is a de-ring. In the above discussion, if the ring does not have a non-trivial left ideal (or right ideal), it is either a zero-multiplication ring or a de-loop.

　　• if \ (H\trianglelefteq n\trianglelefteq r\) and \ (n\) There are unit elements, verification \ (H\trianglelefteq r\);

　　• Verification: Only a limited number of ideal integral rings are domains. (Hint: investigate all left ideals \ (ra\))

As already known from the previous discussion, the ring \ (r\) ideal \ (n\) of all accompanying sets form a ring, which is isomorphic to the homogeneous mappings of the ideal nucleus, known as factor rings , which is recorded as a \ (r/n\). As in group theory, this conclusion is called the homomorphism theorem of the ring, which is the basic tool of the analytic ring structure. The homomorphism theorem of a ring can also get its three isomorphism theorems, which are very similar to the isomorphism theorems of the group, and do not explain much.

(1) first isomorphism theorem : \ (R/\text{ker}\:f\cong f (R) \);

(2) second isomorphism theorem : \ (N\trianglelefteq r,\:h\leqslant r\quad\rightarrow\quad (h+n)/n\cong h/(h\cap N) \);

(3) third isomorphism theorem : \ (H,n\trianglelefteq r,\:n\subseteq h\quad\rightarrow\quad r/h\cong (r/n)/(h/n) \).

　　• discuss the quotient of the Gaussian ring in the main ideal \ (\langle m+ni\rangle\), proving that it has a \ (m^2+n^2\) element and lists the representative elements. (Hint: Start with a large class of imaginary numbers, then discuss the integer class)

2. Special Ideal 2.1 main ideal

For any subset of rings, we can use it to generate the smallest rings and ideals. It is easy to prove that the addition subgroup generated by element \ (a\) is a cyclic loop, so it is the generation sub-ring of \ (a\). The ideal that is generated by element \ (a\) is called a master ideal (Principal Ideal), which is recorded as \ (\langle a\rangle\), below to see the structure of the main ideal. First the master ideal must contain \ (a\) generated addition group \ (\{na\}\), which requires that it is ideal to include \ (ra,ar\), in addition to the closed they have a unified format \ (ax+by+na\). Next, according to the closeness of the multiplication, which must also include \ (rar\), its uniform format is expanded to \ (ax+by+na+\sum{x_kay_k}\). Now you can prove that all elements of this form constitute an ideal, so it is the main ideal that \ (a\) generates.

\[\langle A\rangle=\{ax+by+na+\sum_{k=1}^{m}{x_kay_k}\}\tag{3}\]

In summary, each element of the master ideal is given the form of a formula (3), where the \ (m,n\) integer (the number of construction steps is limited). In special cases, there will be simpler expressions, please self-derivation. For example, if multiplication is commutative, the form becomes \ (ax+na\). When there is a unit element, the expression can be unified to \ (\sum\limits_{k=1}^{m}{x_kay_k}\). Both exchangeable and unit elements are simplified to \ (ax\). In particular, each ideal of a loop ring is the master ideal.

Now we look at the rings generated by multiple elements, and its structural form is complex, but the results are better for the ideal. First, it is easy to prove by inductive method that if \ (R_k\) is ideal, then \ (\sum{r_k}\) is also ideal. So for any subset \ (\{a_1,a_2,\cdots,a_n\}\), \ (\langle a_1\rangle+\langle a_2\rangle+\cdots+\langle a_n\rangle\) is an ideal, And obviously it is generated by \ (\{a_1,a_2,\cdots,a_n\}\) the smallest ideal, and thus has the following form.

\[\langle a_1,a_2,\cdots,a_n\rangle=\langle a_1\rangle+\langle A_2\rangle+\cdots+\langle a_n\rangle\tag{4}\]

2.2 Vegetarian ideals and great ideals

We have already mentioned that the general ring is really not "perfect", and sometimes we would like to study the whole ring, single ring, except the ring or domain. With the aid of the homomorphism theorem, it is possible to try to get the proper ideal and make the factor rings "perfect". First of all to consider the business ring \ (r/n\) is the whole ring of the scene, the whole ring first zero factor, if there is \ ((a+n) (b+n) =n\), then there must be one of \ (n\). Expand to get, if there is \ (Ab\in n\), then there must be \ (a\in n\) or \ (B\in n\). Of course, the whole ring also requires exchangeable, in a switching ring, satisfies the following conditions ideal called the ideal. It is easy to prove that the quotient group \ (r/n\) of the commutative ring is the necessary and sufficient condition of the whole ring (n\) for the vegetarian ideal.

\[ab\in n\quad\rightarrow \quad a\in n\:\vee\: b\in n\tag{5}\]

According to the third isomorphism theorem, to make \ (r/n\) A single ring, there must be no more "large" ideals than \ (n\). The exact definition is: if \ (N\ne r\), and except \ (n,r\) does not contain the ideal of \ (n\), then \ (n\) is called the great ideal of \ (r\). Obviously, \ (n\) is a great ideal and the necessary and sufficient conditions are for \ (r/n\) as a single ring. The conclusion of synthesizing the front single ring shows that if \ (r\) has a unit element, then \ (r/n\) is the necessary and sufficient condition for the removal of the ring (n\) is a great ideal, plus exchangeable conditions, the conclusion of the domain is also established.

　　• verification: \ (z\) all the vegetarian ideal for \ (\{0\}\) and \ (\langle p\rangle\);

　　• verification: \ (z\) the great ideal only \ (\langle p\rangle\).

3. Straight and decomposition 3.1 straight and

In group theory, we can see that the straight integral solution is the best method to deconstruct the group, and the thought is also applied to the ring. To ring \ (r_1,r_2,\cdots,r_n\), easy proof set \ (r=\{(a_1,a_2,\cdots,a_n) \mid a_k\in r_k\}\) also forms a ring under the following operations, \ (r\) is generally called \ (r_1,r_2,\cdots , r_n\) of the outer straight and . \ (r\) ideal \ (R ' _k=\{(0,\cdots,0,a_k,0\cdots,0) \mid a_k\in r_k\}\) is isomorphic to \ (r_k\), and \ (r=r ' _1+r ' _2+\cdots+r ' _n\), And the decomposition of each element is unique.

\[(A_1,a_2,\cdots,a_n) + (b_1,b_2,\cdots,b_n) = (a_1+b_1,a_2+b_2,\cdots,a_n+b_n) \tag{6}\]

\[(A_1,a_2,\cdots,a_n) \cdot (b_1,b_2,\cdots,b_n) = (a_1b_1,a_2b_2,\cdots,a_nb_n) \tag{7}\]

In view of the above discussion, when the ring \ (r\) has the ideal \ (r_1,r_2,\cdots,r_n\) to satisfy: (1) \ (r=r_1+r_2+\cdots+r_n\); (2) \ (r\) any element \ (A\) can be uniquely represented as \ (a=a_1+a_2+\ Cdots+a_n, (a_k\in r_k) \). then called \ (r\) for \ (r_1,r_2,\cdots,r_n\) of the inner straight and , abbreviation straight and , remember as \ (R_1\oplus r_2\oplus\cdots\oplus r_n\).

The second condition in the definition has an equivalent form that is easier to use, one is the representation of the 0 element, and the other is the independence of each straight and the item (equation (8)). The second equivalence condition illustrates the independence of the straight and the item, i.e. \ (R_i\cap r_j=\{0\}\), if there is \ (a_i\in r_i,b_j\in r_j\), then \ (A_ib_j\in r_i+r_j\), so \ (a_ib_j=0\). Further if \ (a,b\) has a straight and decomposition \ (a=a_1+\cdots+a_n,b=b_1+\cdots+b_n\), you can have a formula (9), that is, any element of the operation can be mapped to each straight and the item. Straight and decomposition is an unrelated decomposition, which decomposes large rings into unrelated small loops to study.

\[r_k\cap (r_1+\cdots+r_{k-1}+r_{k+1}+\cdots+r_n) =\{0\}\tag{8}\]

\[ab=a_1b_1+a_2b_2+\cdots+a_nb_n\tag{9}\]

3.2 Ideal with straight and

The direct and decomposition allows us to discuss the nature of the ring in a smaller ideal, and now we look at the relationship between the general ideal and the straight and the decomposition. First consider the ideals of the straight and the items \ (N\trianglelefteq r_k\), then to any \ (N\in n\), have \ (Nr=n (n_1+\cdots+n_n) =nn_k\in n\), with the same (Rn\in n\). Thus there \ (N\trianglelefteq r\), that is straight and the ideal of the item is also straight and ideal. By this conclusion it is easy to have, straight and the ideal of the term (N_k\trianglelefteq r_k\) of the straight and also \ (r\) ideal (formula (10)).

\[n_1\oplus n_2\oplus\cdots\oplus N_n\trianglelefteq r\tag{10}\]

Conversely to any ideal \ (N\trianglelefteq r\), \ (N_k=n\cap r_k\) is also ideal, then \ (n\) is \ (N_k\) The straight and? Essentially as long as proving any \ (N\in n\), it's straight and decomposed to meet \ (N_k\in n\). To make this the nature of the establishment, the need for the use of units \ (1_k\), \ (N_k=1_kn\in n\), it can be assumed that \ (r\) existence unit element, so that the inverse proposition is established, because the unit of the straight and decomposition will get \ (r_k\) unit element.

Now the question is naturally, what kind of ring has straight and decomposition? How to do straight and explode? Assuming \ (r\) is characterized by \ (n\), and there is coprime decomposition \ (n=n_1n_2\), we want \ (r\) to be decomposed into straight and items with eigenvalues of \ (n_1,n_2\) respectively. Because \ (n_1,n_2\) coprime, there is \ (sn_1+tn_2=1\), which examines the collection \ (R_1=\{sn_1a\mid a\in r\}\) and \ (R_2=\{tn_2a\mid a\in r\}\). First, it is easy to prove that they are ideal, and then because \ (a=sn_1a+tn_2a\), so there is \ (r=r_1+r_2\). Assuming \ (a\in r_1\cap r_2\), it is easy to have \ (n_1a=n_2a=0\), which in turn gets \ (a=0\), so \ (R_1\cap r_2=\{0\}\), thereby \ (R=r_1\oplus r_2\).

Finally, the characteristics of \ (r_1,r_2\) are calculated (m_1,m_2\), according to the definition of \ (r_1,r_2\) First (m_1\leqslant n_1,m_2\leqslant n_2\), and then by \ (n\) is \ (r\) feature has \ (m_1m_2\ Geqslant n\), thereby \ (m_1=n_1,m_2=n_2\). The conclusion is that if the prime factorization of \ (n\) is performed (N=p_1^{\alpha_1}\cdots p_m^{\alpha_m}\), the ring can be decomposed into a straight sum of the power characteristics (equation (11)).

\[r=r_1\oplus R_2\oplus\cdots\oplus R_m,\quad\text{char}\,r_k=p_k^{\alpha_k}\tag{11}\]

3.3 Application of straight and

First of all, to discuss the existence of the ring, it is clear that any order of exchange rings are present, such as \ (z_n,z\). Hamilton Ring gives an example of an infinite-order non-commutative ring, and we now want to know if a finite-order non-commutative ring exists? In group theory we know that any finite commutative group can be straight and decomposed by invariant factors. For the addition of ring \ (r\) there is also \ ((r,+) =\langle b_1\rangle\oplus\cdots\oplus\langle b_m\rangle\), where \ (|b_k|\mid |b_{k+1}|\). If \ (n=| r|\) does not contain a factor higher than one time, then \ (R=\langle b_1\rangle\) is a loop ring, which is exchangeable. This means that a non-commutative ring must have a squared factor \ (n=n_1^2n_2\).

Conversely to such a \ (n\), in fact, can also construct a non-commutative ring, we only need to construct a non-commutative \ (n_1^2\) Order ring, it and any \ (n_2\) Order ring straight and is the \ (n\) Order non-commutative ring. For a \ (n_1\) Order ring R, examine the two-tuple \ ((x, y) \) Set, define addition and multiplication as follows, it is easy to prove that the set constitutes a non-commutative ring under the defined addition and multiplication. The necessary and sufficient condition for the existence of the finite-order non-commutative rings is that the order of the rings contains the square factor.

\[(x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2), \quad (x_1,y_1) (x_2,y_2) = (x_2+y_2) (x_1,y_1) \tag{12}\]

Finally, we use the language of the ring to describe the " Chinese remainder theorem " and review the contents of the theorem: if \ (m_1,m_2,\cdots,m_n\) coprime, then the equation Group \ (x\equiv A_k\pmod{m_k}, (K=1,2,\cdots,n) \) There is only one solution under modulo \ (M_1m_2\cdots m_n\). Standing at the angle of the ring, \ (m_k\) The congruence class is a main ideal ring, so the study of the Ring \ (r\) ideal \ (i_1,i_2,\cdots,i_n\). M_i,m_j\ can be said to be \ (I_i\oplus i_j=r\), and the conclusion of the proof is the formula (13).

\[r/\cap i_k\cong r/i_1\times r/i_2\times\cdots\times r/i_n\tag{13}\]

First, easy authentication \ (r\to r/i_1\times r/i_2\times\cdots\times r/i_n\) is the homomorphism mapping, if it can be proved that it is full-shot, by the homomorphism fundamental theorem can be concluded. The proof method is the same as the essence in the elementary number theory, we need to construct for each dimension \ (r_k= (\cdots,0,a_k,0\cdots) \). This condition is equivalent to \ (r_k\in a_k+i_k\) and \ (R_k\in (\prod{i_i})/i_k\), or \ (r=i_k+ (\prod{i_i})/i_k\). If the ring has a unit element, the equation can be easily pushed from \ (r=i_i+i_j\), so the conclusion is proven.

"Abstract algebra" 06-Ideal with straight and