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I have just started out proofs so I am not fully grasping this concept yet.

Question: Is the set $\mathbb{R}^2$, with addition and multiplication defined below a field? Explain.

  1. $(a, b) + (c, d) = (a + c, b + d)$

  2. $(a, b) (c, d) = (ac, bd)$

Attempted solution:

We know: $f\colon F \times F \rightarrow F$

$$f(x, y) = x + y$$

$g\colon F \times F \rightarrow F$

$$g(x, y) = xy$$

So solving 1:

$$(a, b) + (c, d) = (a + c, b + d)$$

$$ \begin{align*} f(a, b) + (c, d) &= (a + b) + (c + d) &&\text{definition of field} \\ &= (a + c) + (b + d) \\ &= (a + c, b + d) \end{align*} $$ The second part I did almost the same thing using $g(x, y) = xy$.

Sorry for the formatting, I still don't know how to use MathJaX but any help is appreciated. If you can also explain for a beginner that would be great, thanks.

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MathJax just renders basic LaTeX commands. See this question in Meta for pointers on learning some basic LaTeX: meta.math.stackexchange.com/questions/934/… –  Arturo Magidin Jan 30 '11 at 23:54

3 Answers 3

up vote 5 down vote accepted

Your attempted solution apparently amounts to substituting the definition of addition and multiplication, but these are given to you so there's no need to "find them". In fact, there are lots of different ways to define addition and multiplication. Most of them would not result in a field (think of $(a,b)+(c,d) = (0,0)$). On the other hand, there might be more than one which will result in a field. So the question really depends on the definition of addition and multiplication.

In order to check whether this object is a field, you have to verify all the field axioms. For example, in a field $x + y = y + x$. In your case, we need to verify that $$(x_1,x_2) + (y_1,y_2) = (y_1,y_2) + (x_1,x_2).$$ All we need to do is substitute the definition of addition: $$(x_1,x_2) + (y_1,y_2) = (x_1+y_1,x_2+y_2)$$ whereas $$(y_1,y_2) + (x_1,x_2) = (y_1+x_1,y_2+x_2),$$ and both expressions are equal by the commutativity of real addition.

Some of the field axioms require you to identify a zero element and a unit element, i.e. the $0$ and $1$ of the field. What should these be? Do they satisfy all the required axioms? For example, is every non-zero element invertible?

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Exactly what i needed, though i didnt get the last part where you mentioned the 0,1 elements since i was taught that the field must contain 0,1 so i just take that as a fact. –  1337holiday Jan 30 '11 at 22:22
    
@1337holiday: You are right that the field must contain an additive identity (which we call 0) and a multiplicative identity (which we call 1). $\mathbb{R^2}$ only has elements of the form $(a,b)$, so you have to find one to be 0 and another to be 1. The definitions of addition and multiplication should guide you. Then you have to prove the rest of the axioms are satisfied. This includes the existence of inverses and the distributive law. –  Ross Millikan Jan 30 '11 at 22:27

HINT $\rm\ \ \ 0\ =\ x^2 - x\ =\ (x_1^2-x_1,\ x_2^2-x_2)\ $ has more than two roots $\rm\ x = (x_1,x_2)\in \mathbb R^2\:$.

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I liked your answer (though unfortunately I don't have the votes right now to upvote). I have written an answer myself, more like a long comment on this answer and Yuval's. –  Srivatsan Jan 6 '12 at 18:21

I liked Bill Dubuque’s answer, and decided to add my own thoughts on the problem. As Yuval Filmus proceeds, to show that a given set together with some operations does not form a field, one approach is to systematically verify each axiom defining a field and identify one that is violated. Alternatively, we can contradict any consequence of the said axioms, i.e., any theorem that is known to hold for all fields. Bill’s approach is exactly that.

As a preliminary step, note that $(\mathbb R^2, + , \cdot)$ is a ring with additive identity $\mathbf 0 = (0, 0)$ and multiplicative identity $\mathbf 1 = (1, 1)$. We want to show that this is not a field; this can be done in multiple ways:

  1. Existence of a nonzero element without inverse. If $x = (0, 1)$, then for any $y \in \mathbb R^2$, their product $x \cdot y$ has a zero as the first component, and hence $xy$ can never be $\mathbf 1$. In other words, $x$ has no multiplicative inverse, and we are done. $\quad \diamond$

  2. Existence of zero-divisors. Since $$ (0,1) \cdot (1, 0) = (1, 0) \cdot (0, 1) = (0, 0) = \mathbf 0,, $$ it follows that both $(1, 0)$ and $(0,1)$ are nonzero zero-divisors in the ring. Hence the ring is not a field.

    Notice that this is also an alternative way to check that $(0, 1)$ is not invertible. $\quad \diamond$

  3. A polynomial with more roots than degree. Bill shows that there is a polynomial with more roots than its degree, which is impossible in a field. Contradiction!

    Even though we are done, it is worthwhile staring at his example a bit more: the polynomial $x (1 - x)$ has four zeroes: $\{ \mathbf 0, \mathbf 1, (1, 0), (0,1) \}$. The first two roots are the usual ones, corresponding to $x= 0$ and $1-x =0$. On the other hand, at the extra roots, namely $(1, 0)$ and $(0, 1)$, we find that neither $x$ nor $1-x$ vanishes but their product does. Hence it must be the case that $(1, 0)$ and $(0, 1)$ are zero-divisors. $\quad \diamond$

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