Every finite Integral Domain is a field Proof |Maths |Mad Teacher
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Every finite Integral Domain is a field Proof |Maths |Mad Teacher

Hello everybody in this video we are going to prove that every finite integral domain is a field So if you’re interested in that then just keep watching the video Yeah, please subscribe it won’t hurt So, to start the proof we suppose a random finite integral domain and use its features to prove that it is a field But before that let’s revisit the definition of integral domain real quick So integral domain is basically a commutative ring with identity 1 is not equal to 0 and containing no zero divisors Here 1 is not equal to 0 means that multiplicative identity, that is 1 is different from additive identity that is 0 so we have two different identities in this ring also The integral domain has no zero divisors Whatsoever. If you don’t know, what a zero divisor is? I have a whole other video explaining its concept It’s pretty simple. So just check it out in the description box below Ok, let’s compare the features of an integral domain and a field so integral domain is a ring It is commutative under multiplication and includes multiplicative identity as well. The same features lie in the field too Except one additional feature that all the elements have their multiplicative inverses in the ring Except 0 or the additive identity as 0 has no multiplicative inverse So this is the only feature we need to proof that a finite integral domain has Ok, now consider that d our finite integral domain has these elements an additive identity 0 a multiplicative identity 1 and some other random elements we are calling a 1 a 2 up to so on a n So we will prove that for any random element other than additive identity 0 let’s say a that belongs to D. We have another element that is also not the additive identity in D. Let’s say B gives multiplicative identity 1 on multiplication Showing that they are inverses of each other We can find the inverse of a by multiplying it with all the elements of D so we got these elements a1 aa1 aa2 up to so on aan and all these elements are distinct from each other in D You may ask why because if a is multiplied by Two distinct elements in D and gives the same answer then by using cancellation law in integral domains, we say that those distinct elements were basically the same Also, if you know the concept of zero divisors, we know that no pair of nonzero elements gives zero on multiplication in integral domains so all these elements a 1 a a 1 a a 2 up to so on a a n are Actually non zeros, if you don’t know the concept of zero divisors, you know what to do One other thing that we already know is The D is closed with respect to multiplication So it means that a 1 a a 1 a a 2 upto so on a a n are all distinct Elements, they are all different from each other but Also, these are the same elements of D that we considered in the beginning of the proof in some order So we conclude that either a 1 is equal to 1 which implies a equals to 1 by the way or aa i is equal to 1 where ai is any random element of D. so we have found the multiplicative inverse of a random element a which means that all the elements of D have Multiplicative inverses in D. So, At last we look into the features of D one more time. ring D is commutative under multiplication that’s right and D has multiplicative identity and now D has multiplicative inverses of each of its element except zero, sounds about right So my friends D has finally qualified to become a field. Hence, proved Please share and subscribe to my channel and leave your requests down below. I will see you all in my next video. Bye


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