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Let $F$ either be $\mathbb R$ or $\mathbb Z/2\mathbb Z$. For each of the following subsets of $F^3$, determine whether it is a subspace. Be sure to explain your answer.

$U_1=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=0\}$; $U_2=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=4\}$;

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For 1a, what does the solution mean by "properties of addition/multiplication for any field"? It's puzzling because not every field has 2,3 as elements in it. Multiplication/addition are not defined for ANY field; the field itself has its own rule for those 2 operations, right?

For 1b, I notice that 1a uses "properties ... for any field." Why can't you do the same general approach for 1b like so.

Choose an arbitrary $(x_1,x_2,x_3) \in U_1$. We know $(0,0,0) \notin U_1$, since $0 + 2(0)+3(0) \neq 4$. Since we have used properties of addition/multiplication for any field, this statement holds over $\mathbb R$ and $\mathbb Z/2\mathbb Z$.

Here, I used the 1a approach by taking an approach for any field. But this doesn't work for $\mathbb Z/2\mathbb Z$. In short, I'm confused when I should use operations for ANY field vs a specific field; basically, I'm confused on the why the 1a approach is justified.

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Any field contains $1$, and hence contains elements $n := 1 + ... + 1$, where there are $n$ copies of $1$, for all $n$.

By "properties of addition/multiplication for any field", they mean "those properties that are true in all fields - to wit, the field axioms and their consequences.

In 1b, you're using properties that are not true in all fields: "$0 \neq 4$" is not a field axiom, nor a consequence of the field axioms.

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  • $\begingroup$ Perfect, that cleared it up. $\endgroup$
    – beginner
    Oct 27, 2020 at 23:12
  • $\begingroup$ Actually, for 1a, how do we know 0+2(0)+3(0) = 0 for an arbitrary field? Or is the arbitrary field idea only being used in part 2 of 1a? $\endgroup$
    – beginner
    Oct 27, 2020 at 23:15
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    $\begingroup$ No, that's true in any field: we know that $x0 = x(1 - 1) = x1 - x1 = x - x = 0$ for any $x$ (where each of those equals signs is an application of one of the field axioms, or two applications of the same axiom), and so $0 + 2(0) + 3(0) = 0 + 0 + 0 = 0$ (where the first equality uses the above result (three times) and the second is two applications of the additive identity axiom of a field. $\endgroup$ Oct 27, 2020 at 23:26
  • $\begingroup$ Ohh, that makes a lot more sense, thank you! $\endgroup$
    – beginner
    Oct 28, 2020 at 13:37

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