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Is $(\mathbb R^2,+,\cdot )$ a field where $(a,b)\cdot (c,d)=(ac-bd, ad+bc)$ ?

I think it is because $\varphi :\mathbb R^2\to \mathbb C$ defined by $\varphi (a,b)=a+ib$ is a ring isomorphism, but my teacher says it's not, and I don't understand why. Any idea ?

Martin Sleziak
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Walace
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  • It is a field, since $-1$ is not a square in $\Bbb R$ (which is needed for the existence of a multiplicative inverse). – Dietrich Burde Aug 04 '20 at 13:55
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    I think you are correct. One can check that $\varphi$ is both an additive and multiplicative group homomorphism. Further, the identity of $\mathbb R^2$ (as defined) is $(1, 0),$ and that maps to the identity of $\mathbb C$ under $\varphi,$ so $\varphi$ is a ring homomorphism. – Dylan C. Beck Aug 04 '20 at 14:02
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    The question is what the teacher really said. – Dietrich Burde Aug 04 '20 at 14:07
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    This is completely answered by https://math.stackexchange.com/a/250770/29335 – rschwieb Aug 04 '20 at 14:24
  • Yes, and this is one possible way to define $\mathbb{C}$ (with defining $i=(0,1)$ and $1=(1,0)$, and writing $a+bi$ instead of $(a,b)$). – Botond Aug 04 '20 at 14:43
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    You are right: That mapping is a ring isomorphism, and a field isomorphism. $\qquad$ – Michael Hardy Aug 04 '20 at 16:04

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To check if it is a field, let's check the requirements for a field: closure with respect to addition, closure with respect to multiplication, additive identity, multiplicative identity, additive commutativity, additive associativity, multiplicative commutativity, multiplicative associativity, additive inverse, and multiplicative inverse.

Our potential field is defined as $(a,b)$, where $a,b\in\mathbb{R}$. Our addition operation is defined as $(a,b)+(c,d)=(a+c,b+d)$. Our multiplication operation is defined as $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$.

The sum and product of real numbers are real numbers, so we have closure with respect to both addition and multiplication.

The additive identity is $(0,0)$.

The multiplicative identity is $(1,0)$.

$$(a,b)+(c,d)=(a+c,b+d)=(c,d)+(a,b)$$ $$((a,b)+(c,d))+(e,f)=(a+c,b+d)+(e,f)=(a+c+e,b+d+f)=(a,b)+(c+e,d+f)=(a,b)+((c,d)+(e,f))$$ $$(a,b)\cdot(c,d)=(ac-bd,ad+bc)=(c,d)\cdot(a,b)$$ $$((a,b)\cdot(c,d))\cdot(e,f)=(ac-bd,ad+bc)\cdot(e,f)=(ace-bde-adf-bcf,acf-bdf+ade+bce)=(a,b)\cdot(ce-df,cf+de)=(a,b)\cdot((c,d)\cdot(e,f))$$

Additive inverse of $(a,b)$ is $(-a,-b)$.

Multiplicative inverse of $(a,b)\neq(0,0)$ is $(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2})$.

This meets all the criteria for a field, so it is a field

Moko19
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