quick ring questions

Oct 2007
347
17
Hello, this questions came up in my exam and I was wondering if I got it right!

TRUE or FALSE:
Every field of 16 elements is commutative

I said TRUE since if we take a field minus the 0 element, then we have a group of units of order 15, which is cyclic and hence abelian by Sylow theory! So if we put the zero element back in clearly that commutes with everything too. Is this correct?

I also said that R[X] was a subring of C[X] , ie. real polynomials are a subring of complex polynomials, is this correct? Thanks for any advice
 
May 2009
1,176
412
Hello, this questions came up in my exam and I was wondering if I got it right!

TRUE or FALSE:
Every field of 16 elements is commutative

I said TRUE since if we take a field minus the 0 element, then we have a group of units of order 15, which is cyclic and hence abelian by Sylow theory! So if we put the zero element back in clearly that commutes with everything too. Is this correct?

I also said that R[X] was a subring of C[X] , ie. real polynomials are a subring of complex polynomials, is this correct? Thanks for any advice
Do you mean field, or do you mean division ring? A field is commutative by definition...However, you are correct. The ring of units has order 15, and there is only one group of order 15, \(\displaystyle C_{15}\).

Yes, \(\displaystyle \mathbb{R}[X]\) is a subring of \(\displaystyle \mathbb{C}[X]\) as every real number is a complex number.
 
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Oct 2007
347
17
ahh.. I may have got it wrong the question actually says any RING with 16 elements is commutative, so the group of units is not necessarily 15- is there a counterexample for this?
 

NonCommAlg

MHF Hall of Honor
May 2008
2,295
1,663
ahh.. I may have got it wrong the question actually says any RING with 16 elements is commutative, so the group of units is not necessarily 15- is there a counterexample for this?
yes, the ring of \(\displaystyle 2 \times 2\) matrices with entries from \(\displaystyle \mathbb{Z}/2\mathbb{Z}\) has 16 elements and it's not commutative.