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.

For part (ii) for showing subfield, I have no clue yet, please help.

Thank you very much!

  • 1
    $\begingroup$ 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. $\endgroup$ – user3482749 Nov 29 '20 at 10:04
  • $\begingroup$ Your editing is correct, thank you for your help! As I had a hard time typing all the symbols correctly here. $\endgroup$ – Ramesh Karl Nov 29 '20 at 10:05
  • $\begingroup$ The common subnring test is discussed in this question. Is there something not clear about it? $\endgroup$ – 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 $y \in P$).

  • $\begingroup$ 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? $\endgroup$ – Ramesh Karl Nov 30 '20 at 5:00
  • 1
    $\begingroup$ If you can prove that there is no such $b$, then yes, that follows. $\endgroup$ – user3482749 Nov 30 '20 at 17:55
  • $\begingroup$ thank you very much! $\endgroup$ – Ramesh Karl Dec 1 '20 at 0:19

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