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• 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.

• Well 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.

• Well 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 havezero 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 termzero 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 termzero divisorsmakes sense.

• In the integers mod 12, 3 × 4 = 0, so both 3 and 4 divide 0.

• They really arezero divisors.”

• Let’s return to the equation X² + 5X + 6 = 0,

• except this time well 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 wordintegralinintegral domain.”

• There is, however, a term for an arbitrary ring with no zero divisors: a DOMAIN.

• But youll 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)

• Theyre 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)

Suppose you want to solve the equation X² - 4 = 0.

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B1 mod integral equation factoring domain ring

# Integral Domains (Abstract Algebra)

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林宜悉 posted on 2020/03/06
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