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Please help with what I am doing wrong here. It has been awhile since Ive been in school and need some help. The question is:

Let $F$ be a field and let $G=F\times F$. Define operations of addition and multiplication on $G$ by setting $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)*(c,d)=(ac,db)$. Do these operations define the structure of a field on $G$?

In order to be a field, the following conditions must apply:

  1. Associativity of addition and multiplication
  2. commutativity of addition and mulitplication
  3. distributivity of multiplication over addition
  4. existence of identy elements for addition and multiplication
  5. existence of additive inverses
  6. existence of multiplicative inverse 0 cannot equala, a-1*a=1

I started with 1. saying

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

$(a,b)+[(c,d)+(e,f)]=[(a,b)+(c,d)]+(e,f)$

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

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

which is not correct but I'm not sure where I went wrong. Is my logic incorrect?

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If you are trying to prove associativity, you should note that addition is only strictly defined for two elements, so you should be trying to prove the second line. You can do that from your definition so $$(a,b)+[(c,d)+(e,f)]=\\(a,b)+(c+e,d+f)=\\(a+c+e,b+d+f)=\\(a+c,b+d)+(e,f)=\\ [(a,b)+(c,d)]+(e,f)$$ You made a typo going to your fourth line. You can do the same for multiplication. The place you will fail is inverses. What is the inverse of $(1,0)$? Imagine it is $(a,b)$ and derive a contradiction.

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It is true that addition and multiplication are associative and commutative for $F\times F$. So when you say it's "not correct", you must be referring to the larger fact that $F \times F$ isn't a field.

For this, it's because there are some elements without multiplicative inverses. In particular, consider something like $(1,0)$. What is it's inverse?

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  • $\begingroup$ considering that 0 does not have an inverse, it cannot be done so therefore not a field $\endgroup$ – cele Jun 5 '14 at 15:42
  • $\begingroup$ @khernz7 You really need to rewrite $6$ to exclude $0$. $0$ is not required to have an inverse, since otherwise the definition does not result in anything other than the ring $\{0\}$. $\endgroup$ – rschwieb Jun 5 '14 at 15:53
  • $\begingroup$ @rschwieb, yes, i apoligize. the condition does exclude 0. $\endgroup$ – cele Jun 5 '14 at 16:00
  • $\begingroup$ @rschwieb, so if we exclude 0 from 6, that would mean that there is a multiplicative inverse then, correct? $\endgroup$ – cele Jun 5 '14 at 16:33
  • $\begingroup$ @khernz7 In this example? Have you tried to find the inverse for $(1,0)$ as mixedmath suggested? $\endgroup$ – rschwieb Jun 5 '14 at 17:04
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It can help proving some more general result; let $A$ be any non empty set with an operation $\%$ on it (I use a generic symbol). Consider now the operation $?$ on $A\times A$ defined by $$ (a,b)\mathbin{?}(c,d)=(a\mathbin{\%}c,b\mathbin{\%}d) $$

  • If $\%$ is associative, then also $?$ is associative
  • $?$ is commutative if and only if $\%$ is commutative
  • If $\%$ has a neutral element $e$, then $(e,e)$ is a neutral element for $?$
  • An element $(a,b)$ has an inverse (with respect to $?$) if and only if both $a$ and $b$ have an inverse (with respect to $\%$)

Once you have verified these facts, you immediately have 1, 2, 4 and 5. Moreover, the neutral element for addition in $G$ is $(0,0)$. Distributivity of multiplication in $G$ can be verified directly.

However, condition 6 fails, because $(1,0)$ is not the neutral element for addition, but it has no multiplicative inverse.

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  • $\begingroup$ Would 6 be true though if we were to state that a cannot equal 0? @ $\endgroup$ – cele Jun 5 '14 at 16:32
  • $\begingroup$ Im not sure i understand how condition 6 fails if it cannot be 0. Doesnt that void the example you gave above of (1,0)? $\endgroup$ – cele Jun 5 '14 at 16:35
  • $\begingroup$ @khernz7 In a field, every element different from the neutral element of addition must have an inverse with respect to multiplication. In $G$, $(1,0)$ is not the neutral element of addition, which is $(0,0)$, but it has no inverse either. $\endgroup$ – egreg Jun 5 '14 at 16:37
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Hint $\ (1,1) = (a,b)(a',b') = (aa',bb')\iff aa' = 1 = bb'.\,$ Therefore $\,(a,b)\,$ is invertible iff $\,a,b\,$ are. Employing this observation, it should be clear how to construct a noninvertible $\,(a,b)\ne (0,0)$

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If you know a little about Ring Theory, this explanation might help.

Consider the projection maps $\pi_1,\pi_2:R\times R\rightarrow R$. Then from your definitions, we have that

  • $\pi_1(\alpha+\beta)=\pi_1(\alpha)+\pi_1(\beta)$
  • $\pi_2(\alpha+\beta)=\pi_2(\alpha)+\pi_2(\beta)$
  • $\pi_1(\alpha\beta)=\pi_1(\alpha)\pi_1(\beta)$
  • $\pi_2(\alpha\beta)=\pi_2(\alpha)\pi_2(\beta)$
  • $\pi_1((1,1))=1$
  • $\pi_2((1,1))=1$
  • $\pi_1(R\times R)=\pi_2(R\times R)=R$

where $\alpha,\beta\in R\times R$. Thus $\pi_1$ and $\pi_2$ are surjective ring homomorphisms from $R\times R$ to $R$. If $R\times R$ were a field, it would have exactly two ideals, namely itself and $\{0\}$. But we can find at least 4 distinct ideals in $R\times R$, namely itself, $\pi_1^{-1}(R)$, $\pi_2^{-1}(R)$ and $\{0\}$.

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