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I'm struggling with this a lot. I think it all boils down to my basic understanding of what the ring $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ is. As much as I know is that it is a ring of 10 elements, but I have no idea how I would go about finding the units in this.

The way I view it is:

$\mathbb Z/2\mathbb Z = \{0,1\},\qquad \mathbb Z/5\mathbb Z= \{0,1,2,3,4\}$.

So would $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z= \{0\times0, 0\times1, 0\times2, 0\times3, 0\times4, 1\times0, 1\times1, 1\times2, 1\times3, 1\times4\} = \{0,0,0,0,0,0,1,2,3,4\}$?

This seems totally incorrect to me, so it would be great if someone could clear this up for me.

I understand a unit is an element $u$ such that $uv=vu$, but in this I don't really even know what the elements are... Would the unit be $0$ and $1$?

Any help would be great. Thanks

B. Mehta
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ASynosure
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  • You've tagged this ring theory and mentioned the word unit - as far as I'm aware units are only defined for rings, but the question only uses groups. Do you mean to look at $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/5\mathbb{Z}$ as rings, or are you using the word unit differently? – B. Mehta Apr 16 '18 at 19:16
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    Also, the product of groups/rings isn't defined by multiplying all the elements together - instead, you want to form pairs. So, the elements of the product should be ${(0,0),(0,1),\dotsc,(0,4),(1,0),(1,1),\dotsc}$ – B. Mehta Apr 16 '18 at 19:18
  • Z/2ZxZ/5Z is a ring, as far as I know. So I think what you've said first there is right. – ASynosure Apr 16 '18 at 19:19
  • In that case, do you mean to define units as elements $u$ with $uv = vu = 1$, where $1$ denotes the identity in the product ring? – B. Mehta Apr 16 '18 at 19:21
  • Yes, that's it. Sorry for the lack of info. – ASynosure Apr 16 '18 at 19:22
  • Indeed you have an isomorphism with $\mathbb Z_{10}$ – Mark Bennet Apr 16 '18 at 19:30
  • See also https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor – lhf Apr 16 '18 at 22:56

4 Answers4

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I'll clarify again the product of rings: For $R, S$ rings, the ring $R \times S$ is defined to be $$R \times S := \{(r,s) \mid r \in R, s \in S\}$$ with addition and multiplication defined pointwise, additive identity $(0_R, 0_S)$ and multiplicative identity $(1_R, 1_S)$ and additive inverses also defined pointwise.

So, the product $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ looks like $$\{(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(1,1),(1,2),(1,3),(1,4)\}$$ with $10$ elements as you expected.

Now, for $u=(u_1,u_2)$ to be a unit, we need to find a $v = (v_1, v_2)$ such that $uv = 1$. Looking at the definition of the product ring, this means $u_1 v_1 = 1$ and $u_2 v_2 = 1$, i.e. $u_1$ and $u_2$ are units in their respective rings. In $\mathbb{Z}/2\mathbb{Z}$, the only unit is $1$ and in $\mathbb{Z}/5\mathbb{Z}$, the non-zero elements $1,2,3,4$ are all units. So, $u = (1,1),(1,2),(1,3),(1,4)$ are the units of $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$.

B. Mehta
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The group of units here is given by $$ U(\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z)=U(\mathbb Z/2\mathbb Z)\times U(\mathbb Z/5\mathbb Z)\cong 1\times C_4\cong C_4. $$ Here we use that $U(\mathbb Z/p\mathbb Z)\cong C_{p-1}$ is cyclic of order $p-1$.

Dietrich Burde
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The product-ring $\;\mathbf Z/2\mathbf Z \times \mathbf Z/5\mathbf Z$ is made up of all pairs of congruence classes of integers modulo $2$ and modulo $5$ respectively. The units are pairs of such classes, which are units in $\mathbf Z/2\mathbf Z$ and $\mathbf Z/5\mathbf Z$ respectively.

As $2$ and $5$ are primes, $\mathbf Z/2\mathbf Z$ and $\mathbf Z/5\mathbf Z$ are fields, so the units are made up of pairs of congruence classes which are non-zero modulo $2$ and modulo $5$: $$\bigl\{(1_2,1_5),(1_2,2_5),(1_2,3_5),,(1_2,4_5)\bigr\}$$

Note: by the Chinese remainder theorem, it is equivalent to find the units in the isomorphic ring $\;\mathbf Z/10\mathbf Z$, which are made up of the classes of elements $<10$ which are coprime to $10$, i.e. $$\bigl\{1_{10},3_{10} ,7_{10},9_{10}\bigr\}.$$

Bernard
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Just make the multiplication table.

The ring $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ is made up of all ordered pairs $(p,q)$, where $p \in \mathbb Z/2\mathbb Z$ and $q \in \mathbb Z/5\mathbb Z$.

Multiplication is defined in the natural way: $(p,q) \times (p',q') = (p\times p',q\times q')$.

The units are all those elements with multiplicative inverses.

You want pairs, for which there is another pair with $(p,q) \times (p',q') = (1,1)$, remembering to reduce the first component modulo 2, and the second component modulo 5, e.g.

$$(1,4) \times (1,3) = (1\times 1, 4 \times 3) = (1,12) = (1,2)$$

Fly by Night
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