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