(Z_n, +, other multiplication) can be a field?

Question

Is it possible for every $N=p^n, n\in\mathbb{N}^*, p$ prime, to define an operation $\star:\mathbb{Z}_N\times\mathbb{Z}_N\to\mathbb{Z}_N$ such that the structure:

$$(\mathbb{Z}_N, +, \star)$$ will be a field, where the adition is the usual one in the group $\mathbb{Z}_N$?

Answer 1

edit: Taking into account the comments on this answer, I have to add one point: I am assuming in my answer that $1$ is the multiplicative neutral element for the multiplication $\star$. That doesn't need to be true in general. Given the multiplication $\star$ with neutral element $a$, we first show that $1$ is a unit.
For that, use that $$a = a\star a = (\sum_{i=1}^a 1)^2 = \sum_{i=1}^{a^2} 1^2 = 1 \star \sum_{i=1}^{a^2} 1.$$ Therefore, let $1^{-1}$ be the inverse of $1$ with respect to $\star$. Then we can define a new multiplication $\star'$ by $$x \star' y := x \star y \star 1^{-1}.$$ This will give us a multiplication with $1$ as neutral element and by mapping $1 \mapsto a$, both rings are isomorphic (or you get a contradiction with the axioms, e.g. in the case $gcd(a,N) > 1$).
Therefore, we can w.l.o.g. assume that the neutral element is indeed $1$.

Sorry that I forgot about that before, it is not that trivial to get that we can assume $1$ as neutral element, but it is very easy to just assume it without thinking enough about it.

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The comments are most likely due to the fact that all finite fields are known and classified, and from that we can easily see that $\mathbb{Z}_N$ with its natural addition will never be one of them, unless $N = p$ is a prime of course (but there we already know which multiplication to take).
As I think that you didn't study this classification yet, let's look for a more elemental proof that this does not work:
Assume that $N$ is not a prime. Then there are numbers $1 < a,b < N$ such that $N = ab$. When we multiply these resp. their residues in $\mathbb{Z}_N$, we can use the distributive law to look at $$a \star b = (\sum_{i=1}^a 1)\star (\sum_{i=1}^b 1) = \sum_{i=1}^{ab} 1\star 1.$$ Of course you might want to add further steps here, but be sure to only use the axioms for a multiplication, e.g. distributive law, associative law, etc.
Now in the last sum, we sum $N = ab$ ones, so this product needs to be equal to $N$ - no matter how exactly $\star$ was defined. But that is a problem (why?).

Note that in this way, you can even show that the classical multiplication is the only one that makes $\mathbb{Z}_N$ with its normal addition into a ring.

Answer 2

As stated in the comments, this is not possible. But you can show even more:

Let $n \in \mathbb N$. Up to isomorphism, there is a unique commutative ring with identity with underlying cyclic abelian group $\mathbb Z_n := \mathbb Z/n\mathbb Z$.

Proof. Since the underlying abelian group has elements of order $n$, the characteristic of the ring is at least $n$. In particular the multiplicative identity $e$ has additive order at least $n$ and subsequently it is equal to $n$. Thus $e$ is co-prime to $n$ and there is an automorphism of the abelian group $\mathbb Z_n$ which sends $e$ to $1$, hence our ring will be isomorphic to the usual ring structure on $\mathbb Z_n$.

Answer 3

As several people have said, the answer is no. But there's an interesting reason why. Here, I'll let $\mathbb{F}_p$ denote the "standard" field $\mathbb{Z} / p \mathbb{Z}$ for prime $p$. Let $K$ be a finite field.

Firstly, every field has what's called a characteristic, where $\operatorname{char}(K)$ is the least $n \geq 1$ such that $\sum_{j = 1}^{n} 1_K = 0_K$, where we say that $K$ has characteristic $0$ if no such $n$ exists. But $K$ is finite, so by a counting argument it must have positive characteristic. Moreover, it can be shown that every field of positive characteristic in fact has prime characteristic. So let $p = \operatorname{char} (K)$.

Now I claim that $\mathbb{F}_p$ is a subfield of $K$. Let $\phi : \mathbb{Z} \to K$ be the ring homomorphism $\phi(1_\mathbb{Z}) = 1_K$, which has kernel $p \mathbb{Z}$. Thus by the universal property of quotient morphisms, there exists a unique $\Phi : \mathbb{F}_p \to K$ such that $\phi = \Phi \circ \pi$, where $\pi : \mathbb{Z} \to \mathbb{F}_p$ is the familiar projection. Moreover, we can observe that $\Phi$ is an injection, so we have our embedding of $\mathbb{F}_p$ as a subgroup of $K$. We call $\mathbb{F}_p = \mathbb{F}_{ \operatorname{char} (K) }$ the characteristic subgroup of $K$.

Now, our insightful step is to say that we can treat $K$ as a vector space over $\mathbb{F}_p$, where "scalar multiplication" is just multiplication by $\mathbb{F}_p$. We can construct a basis for the vector space $K$, call it $\{ \theta_1 , \ldots , \theta_n \}$ for some $n$. But this means that $K$ has $p^n$ elements.

Some things that come out from this argument:

1. It's actually impossible to give a field structure to any set of $N$ elements if $N$ isn't an integer power of a prime.

2. If $K$ is a field of order $p$ for a prime $p$, then $K$ is isomorphic to $\mathbb{F}_p$.

3. Since every finite integral domain is a field, we can replace every instance of the word "field" in the last two with the word "integral domain".

In fact, a strengthening of 2 is possible, but not shown here: For every positive prime $p$, positive integer $n$, there exists a finite field $\mathbb{F}_{p^n}$ of order $p^n$, and if $K$ is another field of order $p^n$, then $K$ is isomorphic to $\mathbb{F}_{p^n}$. In other words, there exists (up to isomorphism) exactly one field of order $p^n$.