4

Prove that there is no integral domain of order 4.

Note: Trying with a contradiction.But without any results!

UNM
  • 1,205
  • What exactly were you asked to prove? I can think of an integral domain of order $4$... – Alex Wertheim Aug 11 '13 at 02:49
  • Two excellent answers have been posted on this. However, I encourage you to take this problem one step further. See if you can't prove that every finite integral domain is a field - and hence, every finite integral domain has prime power order. Since there is a field of every prime power order, that really classifies a lot! – Alex Wertheim Aug 11 '13 at 02:54

3 Answers3

13

That's because the statement is false; there is an integral domain of order $4$, namely $$(\mathbb{Z}/2\mathbb{Z})[x]/(x^2+x+1).$$ This is the only one (up to isomorphism). More generally, for a natural number $n$, there is an integral domain with $n$ elements if and only if $n$ is a prime power. This is discussed in an older answer of mine.

Community
  • 1
Zev Chonoles
  • 132,937
  • 1
    Our lecturer helped us in solving that " There is no integral domain of order 6"! And he gave this as a homework! – UNM Aug 11 '13 at 02:52
11

The answers by Zev and T. Bongers are excellent but let me note:

Exercise 1: Prove that a finite integral domain is a field.

Exercise 2: Prove that every finite field has prime power order.

Exercise 3: Prove that there is a field of every possible prime power order. (Zev has linked to another one of his excellent answers; have a look.)

So, if you give me a number and can tell me whether or not it's a prime power, then I can tell you whether or not there's an integral domain of that order!

I hope this helps!

Amitesh Datta
  • 21,429
9

There is an integral domain (in fact a field) of order $4$; namely, the field $\mathbb{F}_{2^2}$.