# A field with 729 elements [duplicate]

I am looking for an explicit description of the additive, multiplicative group structure and automorphism group of the field with 729 elements.

Sorry for not being clear. I have no clue what does "explicit description" mean, but I am guessing that the underlying group might be cyclic, but I do not know how to give a proof.

For its automorphism group, What I have tried: 1). Comparing this with the Galois group of this field over $$F_3$$ which is cyclic. 2). I also know that Aut(Z_3） = Z_2, which is, in general, true for or prime fields. That is why I am guessing that the underlying additive group should be cyclic, then by the same method, I can show that the automorphism group is a cyclic group of order 728 probably.

## marked as duplicate by Somos, Travis, Eevee Trainer, Lord Shark the Unknown abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 21 at 6:16

• Please be explicit by what you mean by 'explicit' ...Namely, not the basic structures of the groups, but generators? – peter a g Mar 21 at 0:58
• Also what have you tried, and what is the context for this question? – jgon Mar 21 at 1:07
• The additive group will not be cyclic; the field will have characteristic $3$, i.e. $x + x + x = 0$ for any $x$ in the field. The multiplicative group will be cyclic, as is always the case for finite fields. – Theo Bendit Mar 21 at 1:45

A field $$F$$ with $$729$$ elements is a vector space of dimension $$6$$ over $$\mathbb F_3$$ because $$729=3^6$$. Therefore, the additive group of $$F$$ is isomorphic to $$\mathbb F_3^6$$, the cartesian product of six copies of $$\mathbb F_3$$.
The multiplicative group of every finite field is cyclic. Therefore, the multiplicative group of $$F$$ is cyclic of order $$728$$.
$$F$$ is a simple extension of $$\mathbb F_3$$ generated by any root $$\theta$$ of any polynomial of degree $$6$$ that is irreducible mod $$3$$. The simplest one if $$x^6+x+2$$. An automorphism of $$F$$ must send $$\theta$$ to another root of the same polynomial. Therefore, the automorphism group of $$F$$ has order $$6$$. You just have to decide whether it is $$C_6$$ or $$S_3$$.
Alternatively, $$F$$ is a splitting filed (of $$x^{729}-x$$ for instance) and so is a Galois extension of $$\mathbb F_3$$. Since its only proper subfields are $$\mathbb F_{3^2}$$ and $$\mathbb F_{3^3}$$, the Galois group must be $$C_6$$.
• Why must the automorphism group be of order $6$? – zach Mar 21 at 2:31
• @zach, because there are $6$ roots of $x^6+x+2$ in $F$ to choose. – lhf Mar 21 at 10:44