Exercise:

2. Set $ r$ to a zero-factor ring with only a limited number of elements but at least two elements. Prove that $ r$ is a body.

To prove this, you only need to describe $ \ {R ^ *; \ cdot \} $ to form a group. because $ r$ is a ring, $ \ {R ^ *; \ cdot \} $ forms a finite group. In addition, $ r$ has no zero factor, so $ \ {R ^ *; \ cdot \} $ satisfies the left and right cancellation rules, and thus $ \ {R ^ *; \ cdot \} $ is a group. that is, $ \ {R ^ *; +, \ cdot \} $ is the body.

3. set $ r$ to a ring. If $ A _ {1}, A _ {2}, \ cdots, A _ {n} \ In r$ exists, and each $ A _ {I} \ neq0 $ makes

$ A _ {1} A _ {2} \ cdots A _ {n} = 0 $

Proof: $ r$ has a zero factor.

The mathematical induction is used. When $ n = 2 $, the conclusion is obvious. If $ n = K $ is true, consider the situation of $ k + 1 $. If

$ A _ {1} A _ {2} \ cdots A _ {k} A _ {k + 1} = 0, A _ {I} \ neq0 $

If $ A _ {1} A _ {2} = 0 $, $ r$ has a zero factor $ A _ {1}, A _ {2} $, and the conclusion is true.

If $ A _ {1} A _ {2} \ neq0 $, remember $ B = A _ {1} A _ {2} $, then

$ Ba _ {3} \ cdots A _ {k + 1} = 0 $

Assume that $ r$ has a zero factor.

In summary, according to the principle of mathematical induction, $ r$ has a zero factor.

5. Set $ r$ as the ring, $ A \ In r$. proof:

$ <A >=\ left \{\ sum _ {I = 1} ^ {n} X _ {I} AY _ {I} + RA + As + Na | r, s, X _ {I}, Y _ {I} \ In R, n \ In \ mathbb Z \ right \} $

It is easy to verify that the upper right side set is $ S $. let's explain that $ S $ is the ideal minimum of $ r$. choose $ r$'s ideal $ I $ that includes $ A $, and set $ X _ {I}, Y _ {I}, r, s \ In r according to the formula in the question, n \ In \ mathbb Z $, which can be understood according to the ideal definition.

$ X _ {I} AY _ {I}; As; RA; Na \ In I $

Then \ begin {Align *} \ sum _ {I = 1} ^ {n} X _ {I} AY _ {I} + RA + As + Na & \ In I \ \ rightarrow S & \ subset I \ end {Align *}

So $ <A> = S $.

6. Set $ r$ to zero-factor ring, and $ I $ to $ r$. Do you need to know if the quotient ring $ R/I $ is zero-factor ring?

The answer is not certain. For example, if $ r = \ mathbb Z, I = 6 \ mathbb Z $, then

$ R/I = \ mathbb Z _ {6} $

In $ \ mathbb Z _ {6} $

$ \ Overline {2} \ cdot \ overline {3 }=\ overline {0} $

There is a zero factor.

7. Set $ \ mathbb p $ as the number field. Proof: Ring $ M _ {n} (\ mathbb p) $ is only an ordinary ideal.

Prove that $ r = M _ {n} (\ mathbb p) $, and take the non-zero ideal $ I $ of $ r$. We will describe $ I = r$, for this reason, you only need to describe the RMB in $ r$.

$ E = e $

You can. $ e $ is the $ N $ level unit array on the number field $ \ mathbb p $. any non-zero element in $ I $ A = \ left (A _ {IJ} \ right) _ {n \ times n} $ is not general. We can set

$ A _ {IJ} =\left \{\ begin {array} {CC} 1 & I = j = 1 \\ 0 & Other \ end {array} \ right. $

That is, $ A = E _ {11} $. This is because we can take some primary phalanx columns in $ r$, multiply them left and right by $ A $, and convert them into a standard type.

$ \ Left (\ begin {array} {CC} e _ {r} & \ & 0 \ end {array} \ right), r \ geq1 $

Then, the $ E _ {11} $ matrix in $ r$ makes $ E _ {11} \ In I $, then, use the matrix in $ r$ to perform proper column/column swapping for $ E _ {11} $.

$ E _ {II} \ In I, I = 1, 2, \ cdots, N $

In this way, the metadata $ e = E _ {11} + E _ {22} + \ cdots + E _ {NN} \ In I $

To traverse all the elements in $ r$, according to $ Re = r \ In I $

$ R = I $. That is to say, the ring $ r$ only has an ordinary ideal.

8. Set $ r$ as the ring. If $ r$ is used as the addition group, it indicates that $ r$ is the switching ring.

Prove that the elements in the question $ r$ are like $ Na, n \ In \ mathbb Z $. in this way, $ ka, La \ In r$, obviously $ (KA) \ cdot (LA) = KLA ^ 2 = (LA) \ cdot (KA) $

Thus $ r$ is a switching ring.

Additional questions:

2. Set $ r$ as the ring, which is called $ x \ In r$ as a reversible element. If $ Y \ In r$ exists, make $ xy = YX = 1 $

Set $ a, B \ In r$ to prove that $1-AB $ is reversible only when $ 1-ba $ is reversible.

This proves that we first deduce the form and note that

\ Begin {Align *} \ frac {1} {1-AB} & = 1 + AB + (AB) ^ 2 + \ cdots \ & = 1 + A \ left (1 + BA + (BA) ^ 2 + \ cdots \ right) B \ & = 1 + \ frac {AB} {1-ba} \ end {Align *}

That is, $ (1-AB) ^ {-1} = 1 + a (1-ba) ^ {-1} B $. the rest is only the result of mechanical validation. therefore, the conclusion is true.

3. set $ r$ to ring, $ A \ In r$. if $ m \ In \ mathbb N $ causes $ A ^ m = 0 $, $ A $ is a zero-power element. proof: If $ r$ is an exchange ring, the entire power of $ r$ constitutes the ideal of $ r$.

A set of all idempotent elements whose $ I $ is $ r$. first, it is not difficult to prove that if $ A $ is a power zero, then $-A $ is also a power zero, and the power zero indexes of both are the same. and if $ A ^ m = 0 $

$ A ^ m = a ^ {m + 1} = \ cdots = 0 $

Therefore, $ a, B \ In I $, and the power zero index is $ K _ {1}, K _ {2} $, respectively, for a sufficiently large $ m> K _ {1} + K _ {2} $, consider \ begin {Align *} (a-B) in the switching ring $ r$) ^ {m} & = \ sum _ {I = 0} ^ {m} \ binom {n} {I} a ^ I (-B) ^ {M-I} = 0 \ end {Align *}

The above formula is zero because of any $ I = 0, 1, \ cdots, M $, $ M-I \ geq K _ {2} $ or $ I \ geq K _ {1} $ is required. therefore, $ a-B \ In I $. for any $ r \ In r$

$ (RA) ^ {K _ {1 }}= R ^ {K _ {1} a ^ {K _ {1 }}= 0 $

So $ Ra \ In I $. In summary, we can see that $ I $ is the ideal of $ r$. (because it is a switching ring, you only need to verify one side)

4. Set $ r$ as the ring, $ A \ In r$. If $ A \ neq0 $ and $ A ^ 2 = A $, $ A $ is called idempotent. proof:

(1) If all non-zero elements of the ring $ r$ are idempotent elements, $ r$ must be an exchange ring;

(2) If $ r$ is a zero-factor ring with an idempotence, $ r$ only has a unique idempotence and $ r$ is a ring.

Proof (1) any non-zero $ A $ of $ r$, then $ A ^ 2 = A $

Obviously $-A \ neq0 $, and $-A $ is also an idempotent element, that is, $ (-a) ^ 2 = a ^ 2 = A =-A $

In this way, for any $ A \ NEQ B \ In r$, obviously $ A + B \ neq0 $, which is also an idempotent element

\ Begin {Align *} A + B & = (a + B) = a ^ 2 + B ^ 2 + AB + BA \\\ rightarrow AB & =-BA = BA \ end {Align *}

So $ r$ is a switching ring.

(2) notes $ E (Ea-a) = EA-Ea = 0 $

$ R$ has no zero factor, so $ Ea = A $. Similarly, $ AE = A $. therefore, $ e $ is $ r$. it is known from the uniqueness of the primary element that $ e $ is also the only idempotent element.

6. Set $ r$ to a zero-factor ring. $ e $ is $ r$'s left (right) operator about multiplication. Proof: $ e $ must be $ r$'s operator.

If you specify a non-zero dollar in $ r$ $ A, B $, \ begin {Align *} AB-AB & = 0 \ rightarrow (EA) B-A (EB) & = 0 \ rightarrow (Ea-AE) B & = 0 \ end {Align *}

$ R$ does not have a zero factor. Therefore, $ AE = Ea = A $ indicates that $ e $ is $ r$.