# Is $\mathbb{Z}_p$ a field? (p prime)

I was wondering if $\mathbb{Z}_p$ ($p$ prime) was a field, because in some notes I read there's written that $\mathbb{F}_p = \mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$ is a "prime subfield"

But I was wondering about the non-invertible $0$ element inside $\mathbb{Z}_p$

• 0 is always 'non-invertible'. That's a standard part of the field axioms. Jun 21, 2016 at 12:49
• Are you also worried about the non-invertible element $0$ inside $\mathbb R$?
– 5xum
Jun 21, 2016 at 13:10
• @5xum ... Good point. I think I'm just muddling between fields and groups and their definitions...
– ela
Jun 21, 2016 at 13:41
• @AlessioMartorana It's a common beginner's mistake: it's tempting to think "but $p$ isn't invertible in $\mathbb Z_p$" while glossing over $0$ because that obviously doesn't have an inverse; but $p$ and $0$ are one and the same. Jun 21, 2016 at 16:09

$\mathbb{Z}/p\mathbb{Z}$ is a field, it is a ring in which every nonzero element is invertible (use Bezout's theorem).

When you read that $\mathbb{Z}/p\mathbb{Z}$ is a prime subfield, it means that it doesn't have any other subfield than itself. Besides, $\mathbb{Z}/p\mathbb{Z}$ is the prime subfield of any field of characteristic $p$, it means that it is the smallest subfield of any field of characteristic $p$.

$\newcommand{Z}{\mathbb{Z}}$Yes, $\Z_p$ is a field. What this means is that $\Z_p$ is a (commutative) ring and all non-zero elements have inverses. So it is not a problem that $0$ does not have a (multiplicative) inverse.

How does $\Z_p$ look like? Without too many details, $\Z_p = \{0, 1, \dots p-1\}$. Addition and multiplication is modulo $p$. So, for example, $p = 0$.

In fact, $\Z_n$ is a field exactly when (if and only if) $n$ is a prime.

Another example of a field is the real numbers. Note also here that $0$ does not have a multiplicative inverse. But all other elements do have inverses.

For any commutative ring $A$ with unit $1_A$, we have a ring homomorphism $\varphi\colon\mathbf Z\to A$ de fined by $\varphi(m)=m\cdot1_A$. The kernel of this homomorphism is an ideal of $\mathbf Z$, generated by a unique element $n\in\mathbf N$, and $\varphi$ factors as $$\mathbf Z\longrightarrow\mathbf Z/n\mathbf Z\hookrightarrow A.$$ Now, if $A$ is an integral domain, the ideal $n\mathbf Z$ is a prime ideal, so $n$ is either $0$ or a prime number, called the characteristic of the domain. This is the case in particular if $A$ is a field $F$.

If $F$ has characteristic $0$, $\varphi$ extends to a field homomorphism from the field of fractions $\mathbf Q$ of $\mathbf Z$ to $F$, so that $F$ (and all its subfields) contains a subfield isomorphic to $\mathbf Q$.

If $F$ has characteristic $p>0$, $\varphi$ induces a field homomorphism from the field $\mathbf F_p=\mathbf Z/p\mathbf Z$ to $F$, so $F$ and all its subfields contain a subfield isomorphic to $\mathbf F_p$.

In all cases this smallest subfield of $F$ is called its prime subfield.

Yes, $\mathbb{Z}_p$ is a field.

The axiom on multiplicative inverses for a field states that each nonzero element must have a multiplicative inverse. Note that only the nonzero elements need to have multiplicative inverses.