# Question on Subring test and Subfield test

The question goes like this:

$$P = \{x - 2^{1/3}y - 2^{2/3}z : x, y, z \in \mathbb{Z}\}$$

Part(i) prove that $$P$$ is a subring of the ring of real numbers.

Part(ii) prove that $$P$$ is a subfield of the field of real numbers.

My attempt:

I use the subring test for part (i), I had already shown that addition and multiplication binary operations are both closed in $$P$$.

Now I am confused by the remaining criteria(s) I need to show for completing the subring test.

Some abstract algebra books said that I only need to show that subtraction and multiplication is closed in $$P$$.

While some youtuber said I need to show the additive identity of the ring of real number is in $$P$$.

While still some others said that I have to show that $$P$$ is an abelian subgroup and show both associate and distributive laws exists in $$P$$.

So I don't know exactly what I need to show for subring test.

Thank you very much!

• Hi, I've edited your post to clean up some of the formatting. Please check to make sure that I haven't inadvertantly changed the meaning of anything in the process. – user3482749 Nov 29 '20 at 10:04
• Your editing is correct, thank you for your help! As I had a hard time typing all the symbols correctly here. – Ramesh Karl Nov 29 '20 at 10:05
• The common subnring test is discussed in this question. Is there something not clear about it? – Bill Dubuque Nov 29 '20 at 10:18

To show that $$P$$ is a subring, we need to show that it is a ring. That is: we need to show that:

1. Addition in $$P$$ is associative.
2. Addition in $$P$$ is commutative.
3. Multiplication in $$P$$ is associative.
4. Multiplication is left and right distributive over addition.
5. For all $$x, y \in P$$, there is $$x + y \in P$$.
6. For all $$x, y \in P$$, there is $$xy \in P$$.
7. $$0 \in P$$
8. For all $$x \in P$$, there is $$-x \in P$$.
9. $$1 \in P$$ (see note).

Note: opinions vary on whether rings must have multiplicative identities. I like them to, so I've included it.

I've listed them in that slightly odd order for a reason: the first four are all immediate: they're all true in all of $$\mathbb{R}$$, and don't have any existential quantification going on, so they're true in $$P$$ as well.

So we actually only have those last five things to check, and we can, if we like, just go through and check them all separately. However, we can be lazy:

If $$P$$ is closed under subtraction and nonempty, then we have some $$x \in P$$, so we have $$0 = x - x \in P$$ (so we don't need to check $$0 \in P$$ separately), and also $$0 - x = -x \in P$$ (so we don't need to check that $$P$$ has additive inverses separately), and finally for any $$y \in P$$, we have $$y - (-x) = y + x = x + y \in P$$ (so we don't need to check closure under addition separately. However, $$0$$ is very often the easiest thing to show that $$P$$ contains, so very often we'll do that, then subtraction, and have all of the addition axioms sorted.

So to finish the job off, we only need to check that $$1 \in P$$ (if we like our rings to have identities) and $$xy \in P$$ for all $$x, y \in P$$.

However, we can be even lazier: we can look ahead, and check that it's a subfield, which will then immediately imply that it's a subring.

We could check that it's a subfield by just doing the above, then checking the extra field axioms (commutativity of multiplication is immediate as with the first four above, we now definitely need $$1 \in P$$ if we didn't already, we need to check multiplicative inverses, and we need $$1 \neq 0$$, but that's also immediate from $$\mathbb{R}$$).

We could, again, just check both of the new things that need checking. However, we can now do the same thing as we did with addition above, and just check that $$P$$ is closed under non-zero division and that $$P$$ contains something other than $$0$$, and get the rest for free (we then have $$x \in P \setminus \{0\}$$, and so $$1 = xx^{-1}$$ and $$x^{-1} = 1x^{-1}$$, and finally $$xy = y(x^{-1})^{-1}$$ for all $$y \in P$$).

• I can't find the multiplicative inverses for some elements in P, i.e. there is no b in P that satisfies ab = 1 = ba for i.e. a = 1 - 2 (2)^(1/3) - 3(2)^(2/3). As there isn't multiplicative inverses for ALL non-zero elements in P, so P isn't a subfield for the field of real number? – Ramesh Karl Nov 30 '20 at 5:00
• If you can prove that there is no such $b$, then yes, that follows. – user3482749 Nov 30 '20 at 17:55
• thank you very much! – Ramesh Karl Dec 1 '20 at 0:19