字幕表 動画を再生する 英語字幕をプリント Suppose you want to solve the equation X² - 4 = 0. In regular algebra you would factor the left-hand side giving you (x-2)(x+2)=0. Here the product of two terms is zero, so one of the terms must be zero. If x–2=0, then X = 2. And if x +2 = 0, then X = -2. So this equation has two solutions: 2 and -2. This seems straightforward, right? But to solve this equation we used a BIG assumption: that if the product of two terms is zero, then one of the terms must be zero. In Abstract Algebra, this isn’t always true. Let’s see an example where this technique of solving equations does NOT work. We’ll solve the equation x² + 5x + 6 = 0 in the integers mod 12. If you factor the left-hand side, you get (x+2)*(x+3) is congruent to 0 mod 12. If you set each term to 0 mod 12, you get x=10 or x=9. Are these all the solutions? Since there are only 12 numbers in the ring of integers mod 12, let’s go ahead and check all possibilities. We’ll plug in 0 through 11 into the polynomial x² + 5x + 6 and see what we get. If you plug in 0 you get 6, so x=0 is not a solution. If you plug in 1 you get 12 which is congruent to 0 mod 12, so 1 IS a solution. This is interesting. Here’s a solution to the equation that we did NOT find by factoring. Plugging in 2 gives us 8. 3 gives you 6. And let’s quickly fill out the table for the remaining numbers. We see this equation has not two, but FOUR solutions: 1, 6, 9 and 10. Factoring only identified the solutions 9 and 10. So what went wrong? To see the problem, let’s take a closer look at the equation. After factoring, we have (x+2)*(x+3) is congruent to 0 mod 12. Setting each term to zero gave us two of the solutions: 9 and 10. But why did we miss the other two? To see why, look what happens when you plug in 1. This gives you 3 × 4, which is 0 mod 12. And if you plug in 6, you get 8 × 9 which is also 0 mod 12. We now see the culprit. Working mod 12, it’s possible to multiply two non-zero numbers together and get 0. We say the integers mod 12 have “zero divisors,” and it’s the zero divisors which make solving equations more difficult. With real and complex numbers, this doesn’t happen. These fields do not have zero divisors. That’s why in regular algebra, you learn to solve many equations by factoring and setting the terms to zero. But in abstract algebra, this technique is not good enough. I’d like to take a moment and talk about the term “zero divisors.” This name was chosen because division can be defined in terms of multiplication. For example, when working with integers, we say B divides A if B × C = A for some integer C. From this definition, the term “zero divisors” makes sense. In the integers mod 12, 3 × 4 = 0, so both 3 and 4 divide 0. They really are “zero divisors.” Let’s return to the equation X² + 5X + 6 = 0, except this time we’ll work mod 11 instead of mod 12. The integers mod 11 do NOT have any zero divisors. This is because 11 is a prime number. The only way the product of two numbers is a multiple of 11 is if one of the numbers is divisible by 11. And none of the integers 1 through 10 is a multiple of 11, so this ring does not have any zero divisors. Like before, we can factor the left-hand side giving us (X + 2)*(X + 3) is congruent to 0 mod 11. Setting each term to zero gives us x = 9 and x = 8. Let’s see if there are any other solutions by plugging in all the integers mod 11. When x = 0, x² + 5x + 6 is equal to 6. Plugging in x = 1 gives us 1. x = 2 gives us 9. Let’s go ahead and fill out the rest of the table. We see that the only solutions are 8 and 9, the two solutions we found by factoring. This is a bit of good news. When a ring R does NOT have any zero divisors, the traditional technique of solving an equation by factoring and setting the terms to zero DOES work. For this reason, there’s a term for such rings: integral domains. Here’s the complete definition: an integral domain is a commutative ring R with a multiplicative identity 1 and NO zero divisors. A natural question is why do we require an integral domain to be commutative? After all, it seems that the most important thing, at least for solving equations, is there are no zero divisors. One reason, is when the idea of an integral domain was developed, much of the focus was on generalizing the integers for use in number theory. Hence the word “integral” in “integral domain.” There is, however, a term for an arbitrary ring with no zero divisors: a DOMAIN. But you’ll encounter integral domains more frequently. Integral domains have another useful property: the cancellation property. Suppose you want to find the solutions to the equation 2X = 6Y. Your instinct might be to first simplify this by dividing both sides by 2, giving you x = 3y. But when working with a ring, you may not be able to divide. So a different approach would be to factor both sides, then cancel the 2s, giving you x = 3y. Factoring and cancelling does not require division. But can you safely cancel in any ring? Unfortunately, no. Look at the equation 3x = 6y in the ring of integers mod 12. This equation has multiple solutions, but one solution is x = 6 and y = 1. This is because 18 is congruent to 6 mod 12. But look what happens if you factor out 3 then cancel. We get x = 2y. Now if you plug in x=6 and y=1 you get 6 is congruent to 2 mod 12, which is false. So in the original equation, 6 and 1 is a solution, but after cancelling, it is not. So, when can you safely cancel?? To find out, suppose a*x = a*y in some ring R, and 'a' is not 0. If we get everything on one side, we can factor out 'a' using the distributive property. If the ring R does NOT have any zero divisors, then we know that x – y must be 0 because 'a' isn’t 0. This implies that x = y. In other words, we can cancel 'a' from both sides. So another benefit of an integral domain is the cancellation property. To recap, an integral domain is a commutative ring R with 1, that has no zero divisors. The absence of zero divisors means you can use the cancellation property. It also means you can solve an equation by factoring and setting the terms to zero. Here’s a puzzle for you to discuss: how many different quadratic equations are there mod 12? And do any of them have exactly two solutions? And what’s the deal with integral domains? (laughter) They’re neither an integral, nor a function's domain? (laughter) I mean, come on, you know? And how about those viewers who haven’t subscribed? I mean it’s free, it’s easy? What’s stopping them?? (laughter)

B1 中級 積分領域 (抽象代数) (Integral Domains (Abstract Algebra)) 7 0 林宜悉 に公開 2021 年 01 月 14 日 シェア シェア 保存 報告 動画の中の単語