Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

Proof.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

Let $x\in f(\sqrt{I}\,)$ be an arbitrary element. Then there is $a\in \sqrt{I}$ such that $f(a)=x$. As $a\in \sqrt{I}$, there exists a positive integer $n$ such that $a^n\in I$.

It follows that we have
\begin{align*}
x^n=f(a)^n=f(a^n)\in f(I).
\end{align*}

This implies that $x\in \sqrt{f(I)}$.
Hence we have $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

$(\subset)$ Let $x\in \sqrt{f^{-1}(I’)}$. Then there is a positive integer $n$ such that $x^n\in f^{-1}(I’)$ and thus $f(x^n)\in I’$.
As $f$ is a ring homomorphism, it follows that $f(x)^n=f(x^n)\in I’$.

Hence $f(x)\in \sqrt{I’}$, and then $x\in f^{-1}(\sqrt{I’})$.
This proves that $\sqrt{f^{-1}(I’)} \subset f^{-1}(\sqrt{I’})$.

$(\supset)$ Let $x\in f^{-1}(\sqrt{I’})$. Then $f(x)\in \sqrt{I’}$. It follows that there exists a positive integer $n$ such that $f(x^n)=f(x)^n\in I’$.
Hence $x^n\in f^{-1}(I’)$, and we deduce that $x\in \sqrt{f^{-1}(I’)}$.

This proves that $f^{-1}(\sqrt{I’}) \subset \sqrt{f^{-1}(I’)}$.
Combining this with the previous inclusion yields that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$.

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

We now suppose that $f$ is surjective and $\ker(f)\subset I$. We proved $f(\sqrt{I}\,) \subset \sqrt{f(I)}$ in part (a). To show the reverse inclusion, let $x\in \sqrt{f(I)}\subset R’$.
Then there is a positive integer $n$ such that $x^n\in f(I)$.
So there exists $a\in I$ such that $f(a)=x^n$.

Since $f:R\to R’$ is surjective, there exists $y\in R$ such that $f(y)=x$.
Then we have
\begin{align*}
f(a)=x^n=f(y)^n=f(y^n),
\end{align*}
and hence $f(a-y^n)=0$.
Thus $a-y^n\in \ker(f) \subset I$ by assumption.
As $a\in I$, it follows that $y^n\in I$ as well.

We deduce that $y\in \sqrt{I}$ and
\[x=f(y)\in f(\sqrt{I}),\] which completes the proof that $\sqrt{f(I)} \subset f(\sqrt{I})$.

Putting together this inclusion and the inclusion in (a) yields the required equality $f(\sqrt{I}\,) =\sqrt{f(I)}$.

The post Ring Homomorphisms and Radical Ideals first appeared on Problems in Mathematics.

Let $R$ and $S$ be rings. Suppose that $f: R \to S$ is a surjective ring homomorphism.

Prove that every image of an ideal of $R$ under $f$ is an ideal of $S$.
Namely, prove that if $I$ is an ideal of $R$, then $J=f(I)$ is an ideal of $S$.

Proof.

As in the statement of the problem, let $I$ be an ideal of $R$.
Our goal is to show that the image $J=f(I)$ is an ideal of $S$.

For any $a,b\in J$ and $s\in S$, we need to show that

Since $a, b\in J=f(I)$, there exists $a’, b’\in I$ such that
\[f(a’)=a \text{ and } f(b’)=b.\]

Then we have
\begin{align*}
a+b=f(a’)+f(b’)=f(a’+b’)
\end{align*}
since $f$ is a homomorphism.

Since $I$ is an ideal, the sum $a’+b’$ is in $I$.
This yields that $a+b\in f(I)=J$, which proves (1).

Since $f:R\to S$ is surjective, there exists $r\in R$ such that $f(r)=s$.
It follows that
\[sa=f(r)f(a’)=f(ra’)\] since $f$ is a homomorphism.

Since $I$ is an ideal of $R$, the product $ra’$ is in $I$.
Hence $sa\in f(I)=J$, and (2) is proved.

Therefore the image $J=f(I)$ is an ideal of $S$.

The post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal first appeared on Problems in Mathematics.

Let
\[R=\left\{\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\] Then the usual matrix addition and multiplication make $R$ an ring.

Let
\[J=\left\{\, \begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix} \quad \middle | \quad b \in \Q \,\right\}\] be a subset of the ring $R$.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$.

Proof.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

Let
\[\alpha=\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix} \text{ and } \beta=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\] be arbitrary elements in $J$ with $a, b\in \Q$.
Then since we have
\begin{align*}
\alpha-\beta=\begin{bmatrix}
0 & a-b\\
0& 0
\end{bmatrix}\in J,
\end{align*}
the subset $J$ is an additive group.

Now consider any elements
\[\rho=\begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \in R \text{ and } \gamma=\begin{bmatrix}
0 & c\\
0& 0
\end{bmatrix}\in J.\] Then we have
\begin{align*}
\rho \gamma&=\begin{bmatrix}
0 & ac\\
0& 0
\end{bmatrix}\in J \text{ and }\\[6pt] \gamma \rho &=\begin{bmatrix}
0 & ca\\
0& 0
\end{bmatrix}\in J.
\end{align*}
Thus, each element of $J$ multiplied by an element of $R$ is still in $J$.

Hence $J$ is an ideal of the ring $R$.

(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$.

Consider the map $\phi:R\to \Q$ defined by
\[\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)=a,\] for $\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\in R$.

We first show that the map $\phi$ is a ring homomorphism.
First of all, we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} \,\right)=1.
\end{align*}
Thus $\phi$ maps the unity element of $R$ to the unity element of $\Q$.

Take
\[\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix}\in R.\] Then we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix}+\begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix}
a+c & b+d\\
0& a+c
\end{bmatrix} \,\right)=a+c\\[6pt] &=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)+\phi\left(\, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)
\end{align*}
and
\begin{align*}
\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix}
ac & ad+bc\\
0& ac
\end{bmatrix} \,\right)=ac\\[6pt] &=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)\phi\left(\, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right).
\end{align*}
It follows from these computations that $\phi:R\to \Q$ is a ring homomorphism.

Next, we determine the kernel of $\phi$.
We claim that $\ker(\phi)=J$.

If $\rho=\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\in \ker(\phi)$, then we have
\[0=\phi(\rho)=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)=a.\]

So $\rho=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\in J$, and hence $\ker(\phi)\subset J$.

On the other hand, if $\rho=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\in J$, then it follows from the definition of $\phi$ that $\phi(\rho)=0$.
Thus, $J \subset \ker(\phi)$.
Putting these two inclusions together yields $J=\ker(\phi)$.

Observe that the homomorphism $\phi$ is surjective.
In fact, for any $a\in \Q$, we take $\begin{bmatrix}
a & 0\\
0& a
\end{bmatrix}\in R$. Then we have
\[\phi\left(\, \begin{bmatrix}
a & 0\\
0& a
\end{bmatrix} \,\right)=a.\]

In summary, $\phi:R\to \Q$ is a surjective homomorphism with kernel $J$.

It follows from the isomorphism theorem that
\[R/J\cong \Q,\] as required.

Remark.

Recall that the kernel of a ring homomorphism $\phi:R\to S$ is always an ideal of $R$.

Thus, the proof of (b) shows that $J$ is an ideal of $R$. This gives an alternative proof of part (a).

The post The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$. first appeared on Problems in Mathematics.

Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.

Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.

Proof.

Define the map $\psi: R[x,y] \to R[y]$ by sending $f(x,y)\in R[x,y]$ to $f(0,y)$.
Namely, the map $\psi$ is the substitution $x=0$.
It is straightforward to check that $\psi$ is a ring homomorphism.

For any polynomial $g(y)\in R[y]$, let $G(x, y)=g(y)\in R[x,y]$.
Then we have $\psi(G(x,y))=G(0,y)=g(y)$.
This proves that $\psi$ is surjective.

We claim that the kernel of $\psi$ is the ideal $(x)$.

Suppose that $f(x,y) \in \ker(\psi)$.
We write
\[f(x,y)=f_0(y)+f_1(y)x+\cdots +f_n(y)x^n,\] where $f_i\in R[y]$ for $i=1, \dots, n$.

Since $f(x,y)\in \ker(\psi)$, it yields that
\[0=\psi(f(x,y))=f(0,y)=f_0(y).\] Hence
\begin{align*}
f(x,y)&=f_1(y)x+\cdots +f_n(y)x^n\\
&=x\left(f_1(y)+\cdots +f_n(y)x^{n-1}\right) \in (x).
\end{align*}
Thus, $\ker(\psi) \subset (x)$.

On the other hand, suppose $f(x,y)\in (x)$.
Then there exists $g(x,y) \in R[x,y]$ such that
\[f(x,y)=xg(x,y).\] It follows that
\begin{align*}
\psi\left(\, f(x,y) \,\right)&=\psi\left(\, xg(x,y) \,\right)=0g(0,y)=0.
\end{align*}
It implies that $f(x,y) \in \ker(\psi)$, hence $\ker(\psi) \subset (x)$.

Putting two inclusions together gives $(x)=\ker(\psi)$.

In summary, the map $\psi:R[x,y] \to R[y]$ is a surjective ring homomorphism with kernel $(x)$.

Hence by the isomorphism theorem, we obtain the isomorphism
\[R[x,y]/(x)\cong R[y].\] Click here if solved 20

The post Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$ first appeared on Problems in Mathematics.

Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.

Proof.

We give three proofs.
The first two proofs use only the properties of ring homomorphism.

The third proof resort to the units of rings.

If you are familiar with units of $\Z[x]$, then the third proof might be concise and easy to follow.

The First Proof

Assume on the contrary that the rings $\Z[x]$ and $\Q[x]$ are isomorphic.
Let
\[\phi:\Q[x] \to \Z[x]\] be an isomorphism.

The polynomial $x$ in $\Q[x]$ is mapped to the polynomial $\phi(x)\in \Z[x]$.

Note that $\frac{x}{2^n}$ is an element in $\Q[x]$ for any positive integer $n$.
Thus we have
\begin{align*}
\phi(x)&=\phi(2^n\cdot \frac{x}{2^n})\\
&=2^n\phi\left(\frac{x}{2^n}\right)
\end{align*}
since $\phi$ is a homomorphism.

As $\phi$ is injective, the polynomial $\phi(\frac{x}{2^n})\neq 0$.
Since $\phi(\frac{x}{2^n})$ is a nonzero polynomial with integer coefficients, the absolute values of the nonzero coefficients of $2^n\phi(\frac{x}{2^n})$ is at least $2^n$.

However, since this is true for any positive integer, the coefficients of the polynomial $\phi(x)=2^n\phi(\frac{x}{2^n})$ is arbitrarily large, which is impossible.
Thus, there is no isomorphism between $\Q[x]$ and $\Z[x]$.

The Second Proof

Seeking a contradiction, assume that we have an isomorphism
\[\phi:\Q[x] \to \Z[x].\]

Since $\phi$ is a ring homomorphism, we have $\phi(1)=1$.
Then we have
\begin{align*}
1&=\phi(1)=\phi \left(2\cdot \frac{1}{2}\right)\\
&=2\phi\left( \frac{1}{2} \right)
\end{align*}
since $\phi$ is a homomorphism.

Since $\phi\left( \frac{1}{2} \right)\in \Z[x]$, we write
\[\phi\left( \frac{1}{2} \right)=a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0,\] for some integers $a_0, \dots, a_n$.

Since $2\phi\left( \frac{1}{2} \right)=1$, it follows that
\[2a_n=0, \dots, 2a_1=0, 2a_0=1.\] Since $a_0$ is an integer, this is a contradiction.
Thus, such an isomorphism does not exists.
Hence $\Q[x]$ and $\Z[x]$ are not isomorphic.

The Third Proof

Note that in general the units of the polynomial ring $R[x]$ over an integral domain $R$ is the units $R^{\times}$ of $R$.

Since $\Z$ and $\Q$ are both integral domain, the units are
\[\Z[x]^{\times}=\Z^{\times}=\{\pm 1\} \text{ and } \Q[x]^{\times}=\Q^{\times}=\Q\setminus \{0\}.\] Since every ring isomorphism maps units to units, if two rings are isomorphic then the number of units must be the same.

As seen above, $\Z[x]$ contains only two units although $\Q[x]$ contains infinitely many units.
Thus, they cannot be isomorphic.

The post The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic first appeared on Problems in Mathematics.

Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\] be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\] Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?

Solution.

We prove that the ring $\Z[\sqrt{10}]/P$ is isomorphic to the ring $\Zmod{2}$.

We define the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ by sending $a+b\sqrt{10}$ to $\bar{a}=a \pmod 2 \in \Zmod{2}$.
The map $\Psi$ is a ring homomorphism. To see this,
let $a+b\sqrt{10}, c+d\sqrt{10} \in \Z[\sqrt{10}]$ .

We have
\begin{align*}
\Psi\left( (a+b\sqrt{10})(c+d\sqrt{10}) \right) &=\Psi\left(ac+10bd+(ad+bc)\sqrt{10}\right)\\
&=ac+10bd \pmod{2}=ac \pmod{2}\\
&=\Psi(a+b\sqrt{10}) \Psi(c+d\sqrt{10}).
\end{align*}

We also have
\begin{align*}
\Psi\left( (a+b\sqrt{10})+(c+d\sqrt{10}) \right) &=\Psi\left( a+c+(b+d)\sqrt{10}) \right) \\
&=a+c \pmod{2}\\
&=\Psi(a+b\sqrt{10})+\Psi(c+d\sqrt{10}).
\end{align*}

Therefore the map $\Psi$ is a ring homomorphism.

Since $\Psi(0)=\bar{0}$ and $\Psi(1)=\bar{1}$, the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is surjective.

We have $\Psi(a+b\sqrt{10})=\bar{0}$ if and only if $a$ is even.
Thus, the kernel of the homomorphism $\Psi$ is
\[\ker(\Psi)=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}=P.\]

In summary the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is a surjective ring homomorphism with the kernel $P$. Hence by the first isomorphism theorem, we have
\[\Z[\sqrt{10}] /P \cong \Zmod{2}\] as we claimed.

Since $\Zmod{2}$ is a field, the ideal $P$ is a maximal ideal, and in particular $P$ is a prime ideal.

Related Question.

A direct proof that the ideal $P=(2, \sqrt{10})$ is prime in the ring $\Z[\sqrt{10}]$ is given in the post “A prime ideal in the ring $\Z[\sqrt{10}]$“.

The post Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$ first appeared on Problems in Mathematics.

Let $R$ be a ring with $1$. Prove that the following three statements are equivalent.

1. The ring $R$ is a field.
2. The only ideals of $R$ are $(0)$ and $R$.
3. Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.

Proof.

We prove the equivalences $(1) \Leftrightarrow (2)$ and $(2) \Leftrightarrow (3)$.

$(1) \implies (2)$

Suppose that $R$ is a field. Let $I$ be an ideal of $R$.
If $I=(0)$, then there is nothing to prove.
So assume that $I\neq (0)$.

Then there is a nonzero element $x$ in $I$.
Since $R$ is a field, we have $x^{-1}\in R$.

Since $I$ is an ideal, we have
\[1=x^{-1}\cdot x\in I.\] This yields that $I=R$.

$(2) \implies (1)$

Suppose now that the only ideals of $R$ are $(0)$ and $R$.
Let $x$ be a nonzero element of $R$. We show the existence of the inverse of $x$.
Consider the ideal $(x)=xR$ generated by $x$.

Since $x$ is nonzero, the ideal $(x)\neq 0$, and thus we have $(x)=R$ by assumption.
Thus, there exists $y\in R$ such that
\[xy=1.\]

So $y$ is the inverse element of $x$.
Hence $R$ is a field.

$(2)\implies (3)$

Suppose that the only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$ and $f:R\to S$ be any ring homomorphism.
Consider the kernel $\ker(f)$. The kernel $\ker(f)$ is an ideal of $R$, and thus $\ker(f)$ is either $(0)$ or $R$ by assumption.

If $\ker(f)=R$, then the homomorphism $f$ sends $1\in R$ to $0\in S$, which is a contradiction since any ring homomorphism between rings with $1$ sends $1$ to $1$.
Thus, we must have $\ker(f)=0$, and this yields that the homomorphism $f$ is injective.

$(3) \implies (2)$

Suppose that statement 3 is true. That is, any ring homomorphism $f:R\to S$, where $S$ is any ring with $1$, is injective.
Let $I$ be a proper ideal of $R$: an ideal $I\neq R$.
Then the quotient $R/I$ is a ring with $1$ and the natural projection
\[f:R\to R/I\] is a ring homomorphism.

By assumption, the ring homomorphism $f$ is injective, and hence we have
\[(0)=\ker(f)=I.\] This proves that the only ideals of $R$ are $(0)$ and $R$.

The post Three Equivalent Conditions for a Ring to be a Field first appeared on Problems in Mathematics.

Suppose that $f:R\to R’$ is a surjective ring homomorphism.
Prove that if $R$ is a Noetherian ring, then so is $R’$.

Definition.

A ring $S$ is Noetherian if for every ascending chain of ideals of $S$
\[I_1 \subset I_2 \subset \cdots \subset I_k \subset \cdots\] there exists an integer $N$ such that we have
\[I_N=I_{N+1}=I_{N+2}=\dots.\]

Proof.

To prove the ascending chain condition for $R’$, let
\[I_1 \subset I_2 \subset \cdots \subset I_k \subset \cdots\] be an ascending chain of ideals of $R’$.
Note that the preimage $f^{-1}(I_k)$ of the ideal $I_k$ by a ring homomorphism is an ideal of $R$.
(See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a proof.)

Thus we obtain the ascending chain of ideals of $R$
\[f^{-1}(I_1) \subset f^{-1}(I_2) \subset \cdots \subset f^{-1}(I_k) \subset \cdots.\] By assumption $R$ is Noetherian, and hence this ascending chain of ideals terminates. That is, there is an integer $N$ such that
\[f^{-1}(I_N)=f^{-1}(I_{N+1})=f^{-1}(I_{N+2})=\dots.\]

Since $f$ is surjective, we have
\[f\left(\, f^{-1}(I_k) \,\right)=I_k\] for any $k$. Hence it follows that we have
\[I_N=I_{N+1}=I_{N+2}=\dots.\] So each ascending chain of ideals of $R’$ terminates, and thus $R’$ is a Noetherian ring.

The post If $R$ is a Noetherian Ring and $f:R\to R'$ is a Surjective Homomorphism, then $R'$ is Noetherian first appeared on Problems in Mathematics.

Let $f: R\to R’$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R’$.

Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.

Proof.

The preimage of an ideal by a ring homomorphism is an ideal.
(See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a proof.)

Thus, $f^{-1}(P)$ is an ideal of $R$.

We prove that the ideal $f^{-1}(P)$ is prime.
Suppose that we have $ab\in f^{-1}(P)$ for $a, b\in R$. Then we have $f(ab) \in P$.
Since $f$ is a ring homomorphism, we obtain
\begin{align*}
f(a)f(b)=f(ab)\in P.
\end{align*}

Since $P$ is a prime ideal, it follows that either $f(a)\in P$ or $f(b)\in P$.
Hence we have either $a\in f^{-1}(P)$ or $b\in f^{-1}(P)$.
This proves that the ideal $f^{-1}(P)$ is prime.

The post The Preimage of Prime ideals are Prime Ideals first appeared on Problems in Mathematics.

Let $f:R\to R’$ be a ring homomorphism. Let $I’$ be an ideal of $R’$ and let $I=f^{-1}(I)$ be the preimage of $I$ by $f$. Prove that $I$ is an ideal of the ring $R$.

Proof.

To prove $I=f^{-1}(I’)$ is an ideal of $R$, we need to check the following two conditions:

1. For any $a, b\in I$, we have $a-b\in I$.
2. For any $a\in I$ and $r\in R$, we have $ra\in I$.

Let us first prove condition 1. Let $a, b\in I$. Then it follows from the definition of $I$ that $f(a), f(b)\in I’$.
Since $I’$ is an ideal (and hence an additive abelian group) we have $f(a)-f(b)\in I’$.
Since $f$ is a ring homomorphism, it yields that
\[f(a-b)=f(a)-f(b)\in I’.\] Thus we have $a-b \in I$, and condition 1 is met.
This implies that $I$ is an additive abelian group of $R$.

Next, we check condition 2. Let $a\in I$ and $r\in R$. Since $a\in I$, we have $f(a)\in I’$.
Since $I’$ is an ideal of $R’$ and $f(r)\in R’$, we have $f(r)f(a)\in I’$.
Since $f$ is a ring homomorphism, it follows that
\begin{align*}
f(ra)=f(r)f(a)\in I’,
\end{align*}
and hence $ra\in I$. So condition 2 is also met and we conclude that $I$ is an ideal of $R$.

Comment.

Instead of condition 1, we could have used

Condition 1′: For any $a, b\in I$, we have $a+b\in I$.

The reason is that condition 2 guarantee the existence of the additive inverses, and hence condition 1 and 2 are equivalent to condition 1′ and 2.

The post The Inverse Image of an Ideal by a Ring Homomorphism is an Ideal first appeared on Problems in Mathematics.