Show that: exists unique ring homomorphism $\varphi : \mathbb{Z} [ i ] \longrightarrow R$ with $\varphi ( i ) = a$, where $a^{2} =-1_{R}$

Question

Problem: Let R be a ring which has an element $a \in R$ such that $a^{2}=-1_{R}$. Prove that: there exists a unique ring homomorphism $\varphi : \mathbb{Z} [ i ] \longrightarrow R$ such that $\varphi ( i ) = a$.

So, $\mathbb{Z} [ i ]$ has the above Universal Property.

Note that: $\mathbb{Z} [ i ]$ denote the ring of Gauss integers, where $\mathbb{Z} [ i ] = \big \{ a + b i : a , b \in \mathbb{Z} \big \}$ with $i^{2}=-1$ the imaginary unit or otherwise is the quotient ring $\mathbb{Z} [ x ] / < x^{2} + 1 > \cong \mathbb{Z} [ i ]$, where $\mathbb{Z} [ x ]$ is the ring of polynomials with integer coefficients and $< x^{2} + 1 > = \big \{ (x^{2} + 1) f ( x ) : f ( x ) \in \mathbb{Z} [ x ] \big \}$ the ideal generated by the polynomial $x^{2} + 1$.

How can I solve this problem? I'm stuck, please give me a hand.

I have done the following: \begin{align*} \varphi \Big( \big( x + y i \big) + \big( z + t i \big) \Big) = \varphi \Big( \big( x + z \big) + \big( y + t \big) i \Big) = \varphi \big( x + z \big) + \varphi \big( y + t \big) a = \\ = \big( \varphi ( x ) + \varphi ( y ) a \big) + \big( \varphi ( z ) + \varphi ( t ) a \big) = \\ = \varphi \big( x + y i \big) + \varphi \big( z + t i \big) \end{align*} and \begin{align*} \varphi \Big( \big( x + y i \big) \big( z + t i \big) \Big) = \varphi \Big( \big( x z - y t \big) + \big( x t + y z \big) i \Big) = \varphi \big( x z - y t \big) + \varphi \big( x t + y z \big) a = \\ = \varphi ( x ) \varphi ( z ) - \varphi ( y ) \varphi( t ) + \Big[ \varphi ( x ) \varphi ( t ) + \varphi ( y ) \varphi ( z ) \Big] a = \\ = \big( \varphi ( x ) + \varphi ( y ) a \big) \big( \varphi ( z ) + \varphi ( t ) a \big) = \\ = \varphi \big( x + y a \big) \varphi \big( z + t a \big). \end{align*}

Am I correct about that $\varphi$ is ring homomorphism?

Answer 1

Preliminary facts.

1. For any ring $R$, there is a unique homomorphism $\mathbb Z\to R$.
2. Fixed a ring $K$, for any choice of a ring $R$, a homomorphism $\phi:K\to R$ and an element $a\in R$, there is a unique homomorphism $\varphi :K[X]\to R$ such that $\varphi(X)=a$ and $\phi(k)=\varphi(k)$ for all $k\in K$.
3. (1) and (2) imply that, for any choice of a ring $R$ and an element $a\in R$, there is a unique homomorphism $\alpha:\mathbb Z[X]\to R$ such that $\alpha(X)=a$.

Existence. Let's prove that, given a ring $R$ and $a\in R$ satisfying $a^2+1=0$, there is a homomorphism $\mathbb Z[X]/(X^2+1)\to R$ sending the residue class of $X$ to $a$. By (3) there is a homomorphism $f:\mathbb Z[X]\to R$ such that $f(X)=a$. Let $\pi:\mathbb Z[X]\to \mathbb Z[X]/(X^2+1)$ be the canonical homomorphism: since $f(X^2+1)=a^2+1=0$, the universal property of the quotient yields a homomorphism $f':\mathbb Z[X]/(X^2+1)\to R$ such that $f'\circ \pi=f$ (so $f'$ sends the residue class of $X$ to $a$).

Uniqueness. If a homomorphism $g:\mathbb Z[X]/(X^2+1)\to R$ sends the residue class of $X$ to $a$, then $g\circ \pi$ sends $X$ to $a$, hence coincides with $f$ by (3). But also $f'\circ \pi=f$, so by the universal property of the quotient $g=f'$.