# Ring of $\mathbb{Z}_2$-valued functions

Let $R =\{f:\{1,2,3,4,\cdots ,10\}\to \mathbb{Z}_2 \}$ be the set of all $\mathbb{Z}_2$ -valued functions on the set $\{1,2,3......10\}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?

1) $R$ has a unique maximal ideal.

2) Every prime ideal of $R$ is also maximal.

3 ) The number of proper ideals of $R$ is $511$.

4 ) Every element of $R$ is idempotent.

R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .

• what is "Z2" is it "$\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z} = \mathbb{F}_2$" the field with 2 elements? or $\mathbb{Z}^2$ a lattice? or something completely different? – mdave16 Jun 19 '17 at 12:17
• If you want $\mathbb{Z}^2$ just change \$\mathbb{Z}_2\$ to \$\mathbb{Z}^2\$. – kingW3 Jun 19 '17 at 12:20
• Yes you are right Z2 is field with 2 elements F2 – Gilll Jun 19 '17 at 12:24
• R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless – Gilll Jun 19 '17 at 12:29

You shouldn't have any trouble showing that your ring is isomorphic to $\mathbb Z_2^{10}$ (the product ring of $10$ copies of $\mathbb Z_2$.) The idea is that you send $\phi :\{1,2,\ldots, 10\}\to\mathbb Z_2$ to $(\phi(1),\phi(2),\ldots,\phi(10))\in \mathbb Z_2^{10}$.

Of course $\mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.

It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.

The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.

The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.

4) is true because every element of $\mathbb{Z}_2=\{0,1\}$ is idempotent, so $\forall \Phi \in R, \forall k\in \{1,...,10\}, \Phi^2(k)= \Phi(k) * \Phi(k) = \Phi(k)$.

Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $a\in A$, Define: $I_{a}=\{\ f:A \to \mathbb{Z}_{2}|f(a) =0 \}\$. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring. (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.

Now about the number of ideals. observe first that if $B\subseteq A$ we can define $I_{B}= \{\ f:A \to \mathbb{Z}_{2}|f(b) =0$ $\forall$ b $\in B \}\$.

Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.

Now how many such proper ideal we have already.

we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a \in A$ , there exists $f_{a} \in I$ such that $f_{a}(a)=1$.

Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} \in I$. Observe $\sum_{a\in A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.

So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.